From c6f69b3f45b81d19b8eeb87184bf16e6de0fad24 Mon Sep 17 00:00:00 2001
From: Juan Marín Noguera <juan.marinn@um.es>
Date: Thu, 20 Feb 2020 16:07:37 +0100
Subject: 2

---
 tem/n.lyx  |  212 ++++++
 tem/n1.lyx | 2320 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 tem/n2.lyx | 1269 +++++++++++++++++++++++++++++++++
 tem/n3.lyx | 1734 +++++++++++++++++++++++++++++++++++++++++++++
 tem/n4.lyx | 2144 +++++++++++++++++++++++++++++++++++++++++++++++++++++++
 tem/n5.lyx | 1516 +++++++++++++++++++++++++++++++++++++++
 6 files changed, 9195 insertions(+)
 create mode 100644 tem/n.lyx
 create mode 100644 tem/n1.lyx
 create mode 100644 tem/n2.lyx
 create mode 100644 tem/n3.lyx
 create mode 100644 tem/n4.lyx
 create mode 100644 tem/n5.lyx

(limited to 'tem')

diff --git a/tem/n.lyx b/tem/n.lyx
new file mode 100644
index 0000000..c17307e
--- /dev/null
+++ b/tem/n.lyx
@@ -0,0 +1,212 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package babel
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
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+\use_non_tex_fonts false
+\font_sc false
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+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize 10
+\spacing single
+\use_hyperref false
+\papersize a5paper
+\use_geometry true
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 0.2cm
+\topmargin 0.7cm
+\rightmargin 0.2cm
+\bottommargin 0.7cm
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
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+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle empty
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Title
+Topología de espacios métricos
+\end_layout
+
+\begin_layout Date
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+def
+\backslash
+cryear{2018}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "../license.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Bibliografía:
+\end_layout
+
+\begin_layout Itemize
+Topología de Espacios Métricos, Grado en Matemáticas, Dr.
+ Luis J.
+ Alías & Dr.
+ Miguel Ángel Javaloyes, Departamento de Matemáticas, Universidad de Murcia
+ (Curso 2017–18).
+\end_layout
+
+\begin_layout Chapter
+Espacios métricos
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n1.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Subconjuntos notables
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n2.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Aplicaciones continuas
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n3.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Espacios compactos
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n4.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Chapter
+Espacios conexos
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset include
+LatexCommand input
+filename "n5.lyx"
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/tem/n1.lyx b/tem/n1.lyx
new file mode 100644
index 0000000..63ebf66
--- /dev/null
+++ b/tem/n1.lyx
@@ -0,0 +1,2320 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Section
+Espacios topológicos
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+espacio topológico
+\series default
+ es un par 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ en el que 
+\begin_inset Formula ${\cal T}\subseteq{\cal P}(X)$
+\end_inset
+
+ y cumple que:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $X,\emptyset\in{\cal T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\{A_{1},\dots,A_{n}\}\subseteq{\cal T}\implies\bigcap_{i=1}^{n}A_{i}\in{\cal T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\{A_{i}\}_{i\in I}\subseteq{\cal T}\implies\bigcup_{i\in I}A_{i}\in{\cal T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Decimos que 
+\begin_inset Formula ${\cal T}$
+\end_inset
+
+ es una 
+\series bold
+topología
+\series default
+ para 
+\begin_inset Formula $X$
+\end_inset
+
+ y sus elementos son 
+\series bold
+conjuntos abiertos
+\series default
+, o simplemente 
+\series bold
+abiertos
+\series default
+, de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+.
+ Llamamos 
+\series bold
+cerrados
+\series default
+ a los complementarios de los abiertos: 
+\begin_inset Formula ${\cal C_{T}}:={\cal C}:=\{X\backslash A\}_{A\in{\cal T}}$
+\end_inset
+
+.
+ Un 
+\series bold
+entorno
+\series default
+ de 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ es un abierto que contiene a 
+\begin_inset Formula $p$
+\end_inset
+
+, y llamamos 
+\begin_inset Formula ${\cal E}(p)$
+\end_inset
+
+ a la familia de todos los entornos de 
+\begin_inset Formula $p$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $A\in{\cal T}$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\forall p\in A,\exists{\cal U}\in{\cal E}(p):{\cal U}\subseteq A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Dado 
+\begin_inset Formula $x\in A$
+\end_inset
+
+, 
+\begin_inset Formula ${\cal U}=A$
+\end_inset
+
+ es un entorno de 
+\begin_inset Formula $x$
+\end_inset
+
+ en 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Para cada 
+\begin_inset Formula $x\in A$
+\end_inset
+
+, sea 
+\begin_inset Formula ${\cal U}_{x}\in{\cal E}(x)$
+\end_inset
+
+ tal que 
+\begin_inset Formula ${\cal U}_{x}\subseteq A$
+\end_inset
+
+, se afirma que 
+\begin_inset Formula $\bigcup_{x\in A}{\cal U}_{x}=A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\subseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula ${\cal U}_{x}\subseteq A\forall x\in A\implies\bigcup_{x\in A}{\cal U}_{x}\subseteq A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\supseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $\forall x\in A,x\in{\cal U}_{x}\subseteq\bigcup_{x\in A}{\cal U}_{x}\implies A\subseteq\bigcup_{x\in A}{\cal U}_{x}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Propiedades de los cerrados:
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $X,\emptyset\in{\cal C_{T}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\{C_{1},\dots,C_{n}\}\subseteq{\cal C_{T}}\implies\bigcup_{i=1}^{n}C_{i}\in{\cal C_{T}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\{C_{i}\}_{i\in I}\subseteq{\cal C_{T}}\implies\bigcap_{i\in I}C_{i}\in{\cal C_{T}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $A$
+\end_inset
+
+ es un abierto y 
+\begin_inset Formula $C$
+\end_inset
+
+ un cerrado, entonces 
+\begin_inset Formula $A\backslash C$
+\end_inset
+
+ es abierto y 
+\begin_inset Formula $C\backslash A$
+\end_inset
+
+ es cerrado.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $X\backslash C$
+\end_inset
+
+ es abierto, por lo que 
+\begin_inset Formula $A\backslash C=A\cap(X\backslash C)$
+\end_inset
+
+ también.
+ Por otro lado, 
+\begin_inset Formula $X\backslash(C\backslash A)=(X\backslash C)\cup A$
+\end_inset
+
+, que es abierto, por lo que 
+\begin_inset Formula $C\backslash A$
+\end_inset
+
+ es cerrado.
+\end_layout
+
+\begin_layout Standard
+Algunas topologías:
+\end_layout
+
+\begin_layout Itemize
+La 
+\series bold
+topología discreta
+\series default
+: 
+\begin_inset Formula ${\cal T}_{D}:={\cal P}(X)$
+\end_inset
+
+, la topología más grande que se puede definir sobre 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+La 
+\series bold
+topología trivial
+\series default
+ o 
+\series bold
+indiscreta
+\series default
+: 
+\begin_inset Formula ${\cal T}_{T}=\{\emptyset,X\}$
+\end_inset
+
+, la topología más pequeña que se puede definir sobre 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+La 
+\series bold
+topología cofinita
+\series default
+: 
+\begin_inset Formula ${\cal T}_{CF}=\{\emptyset\}\cup\{A\subseteq X:X\backslash A\text{ es finito}\}$
+\end_inset
+
+.
+ Esta se define sobre conjuntos infinitos, pues de lo contrario es 
+\begin_inset Formula ${\cal T}_{CF}={\cal T}_{D}$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Sean 
+\begin_inset Formula $A,B\in{\cal T}$
+\end_inset
+
+ no vacíos, 
+\begin_inset Formula $X\backslash A$
+\end_inset
+
+ y 
+\begin_inset Formula $X\backslash B$
+\end_inset
+
+ son finitos, por lo que 
+\begin_inset Formula $(X\backslash A)\cup(X\backslash B)=X\backslash(A\cap B)$
+\end_inset
+
+ también lo es y 
+\begin_inset Formula $A\cap B\in{\cal T}$
+\end_inset
+
+.
+ Si, por ejemplo, 
+\begin_inset Formula $B=\emptyset$
+\end_inset
+
+, entonces 
+\begin_inset Formula $A\cap B=\emptyset\in{\cal T}$
+\end_inset
+
+.
+ Por otro lado, si 
+\begin_inset Formula $\{A_{i}\}_{i\in I}\subseteq{\cal T}$
+\end_inset
+
+ es tal que 
+\begin_inset Formula $\bigcup_{i\in I}A_{i}\neq\emptyset$
+\end_inset
+
+, entonces 
+\begin_inset Formula $X\backslash\bigcup_{i\in I}A_{i}=\bigcap_{i\in I}(X\backslash A_{i})$
+\end_inset
+
+ es finito.
+\end_layout
+
+\begin_layout Standard
+Dado el espacio topológico 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+, definimos la 
+\series bold
+topología inducida
+\series default
+ por 
+\begin_inset Formula ${\cal T}$
+\end_inset
+
+ en 
+\begin_inset Formula $H\subseteq X$
+\end_inset
+
+, 
+\series bold
+topología relativa
+\series default
+ o 
+\series bold
+topología de subespacio
+\series default
+ como 
+\begin_inset Formula ${\cal T}|_{H}:={\cal T}_{H}:=\{A\cap H\}_{A\in{\cal T}}$
+\end_inset
+
+.
+ Los abiertos de 
+\begin_inset Formula ${\cal T}_{H}$
+\end_inset
+
+ se llaman 
+\series bold
+abiertos relativos
+\series default
+, y 
+\begin_inset Formula $(H,{\cal T}_{H})$
+\end_inset
+
+ es un 
+\series bold
+subespacio topológico
+\series default
+ de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+.
+ Todo subespacio topológico es un espacio topológico.
+ 
+\series bold
+Demostración:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\emptyset=\emptyset\cap H$
+\end_inset
+
+ y 
+\begin_inset Formula $H=X\cap H$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $A',B'\in{\cal T}_{H}$
+\end_inset
+
+, existen 
+\begin_inset Formula $A,B\in{\cal T}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $A'=A\cap H$
+\end_inset
+
+ y 
+\begin_inset Formula $B'=B\cap H$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $A'\cap B'=A\cap B\cap H\in{\cal T}_{H}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $\{A'_{i}\}_{i\in I}\subseteq{\cal T}_{H}$
+\end_inset
+
+, para cada 
+\begin_inset Formula $i\in I$
+\end_inset
+
+ existe un 
+\begin_inset Formula $A_{i}\in{\cal T}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $A'_{i}=A_{i}\cap H$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $\bigcup_{i\in I}A'_{i}=\bigcup_{i\in I}(A_{i}\cap H)=\left(\bigcup_{i\in I}A_{i}\right)\cap H\in{\cal T}_{H}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $H$
+\end_inset
+
+ es abierto en 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ entonces todo abierto relativo 
+\begin_inset Formula $A'\in{\cal T}_{H}$
+\end_inset
+
+ también es abierto en el total.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $A\in{\cal T}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $A'=A\cap H$
+\end_inset
+
+, como 
+\begin_inset Formula $A,H\in{\cal T}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $A'\in{\cal T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dado 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+, un subconjunto 
+\begin_inset Formula $C'\subseteq H\subseteq X$
+\end_inset
+
+ es cerrado relativo (
+\begin_inset Formula $C'\in{\cal C}_{H})$
+\end_inset
+
+ si y sólo si existe 
+\begin_inset Formula $C\in{\cal C}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $C'=C\cap H$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $C'\in{\cal C}_{H}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $H\backslash C'\in{\cal T}_{H}$
+\end_inset
+
+, por lo que existe 
+\begin_inset Formula $A\in{\cal T}$
+\end_inset
+
+ con 
+\begin_inset Formula $H\backslash C'=A\cap H$
+\end_inset
+
+.
+ Pero si 
+\begin_inset Formula $C:=X\backslash A$
+\end_inset
+
+, entonces 
+\begin_inset Formula $C'=H\backslash(H\backslash C')=H\backslash(A\cap H)=H\backslash A=H\cap(X\backslash A)=H\cap C$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $C'=C\cap H$
+\end_inset
+
+ con 
+\begin_inset Formula $C\in{\cal C}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $H\backslash C'=H\backslash(C\cap H)=H\backslash C=H\cap(X\backslash C)$
+\end_inset
+
+, y como 
+\begin_inset Formula $X\backslash C\in{\cal T}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $H\backslash C'\in{\cal T}_{H}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $C'\in{\cal C}_{H}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Primer axioma de numerabilidad y condición de Hausdorff
+\end_layout
+
+\begin_layout Standard
+Una 
+\series bold
+base de entornos
+\series default
+ de 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ es una subfamilia 
+\begin_inset Formula ${\cal B}(p)\subseteq{\cal E}(p)$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall V\in{\cal E}(p),\exists U\in{\cal B}(p):U\subseteq V$
+\end_inset
+
+.
+ A partir de aquí, un espacio topológico 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ satisface el 
+\series bold
+primer axioma de numerabilidad
+\series default
+, o es 
+\series bold
+1AN
+\series default
+, si todo punto posee una base de entornos numerable, es decir, si 
+\begin_inset Formula $\forall p\in X,\exists{\cal B}(p)\text{ base de }p:|{\cal B}(p)|\leq|\mathbb{N}|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así, 
+\begin_inset Formula $(X,{\cal T}_{T})$
+\end_inset
+
+ es 1AN, pues cada punto posee la base 
+\begin_inset Formula ${\cal B}(p)=\{X\}$
+\end_inset
+
+.
+ Sin embargo, 
+\begin_inset Formula $(\mathbb{R},{\cal T}_{CF})$
+\end_inset
+
+ no es 1AN.
+ 
+\series bold
+Demostración:
+\series default
+ Si lo fuera, tendríamos 
+\begin_inset Formula ${\cal B}(0)=\{U_{n}\}_{n\in\mathbb{N}}$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $U_{n}=\mathbb{R}\backslash F_{n}$
+\end_inset
+
+, con 
+\begin_inset Formula $F_{n}$
+\end_inset
+
+ finito, para cada 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+.
+ Ahora bien, como la unión numerable de conjuntos finitos es numerable y
+ 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ no lo es, podemos elegir un punto 
+\begin_inset Formula $x\in\mathbb{R}\backslash\left(\bigcup_{n\in\mathbb{N}}F_{n}\right)=\bigcap_{n\in\mathbb{N}}(\mathbb{R}\backslash F_{n})=\bigcap_{n\in\mathbb{N}}U_{n}$
+\end_inset
+
+ con 
+\begin_inset Formula $x\neq0$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $A=\mathbb{R}\backslash\{x\}\in{\cal E}(0)$
+\end_inset
+
+, existirá un 
+\begin_inset Formula $U_{i}\subseteq A$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $x\in U_{i}\subseteq A=\mathbb{R}\backslash\{x\}\#$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+La propiedad 1AN es hereditaria, es decir, si 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es 1AN, también lo es cualquier 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+subes
+\backslash
+-pa
+\backslash
+-cio
+\end_layout
+
+\end_inset
+
+ topológico de este.
+ 
+\series bold
+Demostración:
+\series default
+ Debemos probar que si 
+\begin_inset Formula $Y\subseteq X$
+\end_inset
+
+, dado 
+\begin_inset Formula $y\in Y$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}(y)$
+\end_inset
+
+ una base de entornos de 
+\begin_inset Formula $y$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+, debemos probar que 
+\begin_inset Formula ${\cal B}_{Y}(y)=\{B\cap Y\}_{B\in{\cal B}(y)}$
+\end_inset
+
+ es base de entornos de 
+\begin_inset Formula $y$
+\end_inset
+
+ en 
+\begin_inset Formula $Y$
+\end_inset
+
+, pues entonces 
+\begin_inset Formula $|{\cal B}_{Y}(y)|\leq|{\cal B}(y)|\leq|\mathbb{N}|$
+\end_inset
+
+.
+ Para ello, vemos que todo 
+\begin_inset Formula $A\in{\cal B}_{Y}(y)$
+\end_inset
+
+ es entorno de 
+\begin_inset Formula $y$
+\end_inset
+
+ en 
+\begin_inset Formula $Y$
+\end_inset
+
+, pues 
+\begin_inset Formula $A=B\cap Y\in{\cal T}_{Y}$
+\end_inset
+
+ con 
+\begin_inset Formula $B$
+\end_inset
+
+ un entorno de 
+\begin_inset Formula $y$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+.
+ Ahora, si 
+\begin_inset Formula $V$
+\end_inset
+
+ es un entorno de 
+\begin_inset Formula $y$
+\end_inset
+
+ en 
+\begin_inset Formula $Y$
+\end_inset
+
+, entonces 
+\begin_inset Formula $V$
+\end_inset
+
+ es abierto en 
+\begin_inset Formula $Y$
+\end_inset
+
+, por lo que existe un 
+\begin_inset Formula $A\in{\cal T}$
+\end_inset
+
+ abierto en 
+\begin_inset Formula $X$
+\end_inset
+
+ tal que 
+\begin_inset Formula $V=A\cap Y$
+\end_inset
+
+, y como 
+\begin_inset Formula $A$
+\end_inset
+
+ es entorno de 
+\begin_inset Formula $y$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+, existe un 
+\begin_inset Formula $B\in{\cal B}(y)$
+\end_inset
+
+ con 
+\begin_inset Formula $B\subseteq A$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $y\in B\cap Y\subseteq A\cap Y=V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Un espacio topológico 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es 
+\series bold
+de Hausdorff
+\series default
+ o 
+\begin_inset Formula $T_{2}$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall p,q\in X,p\neq q;\exists U\in{\cal E}(p),V\in{\cal E}(q):U\cap V=\emptyset$
+\end_inset
+
+.
+ Así, por ejemplo, 
+\begin_inset Formula $(X,{\cal T}_{T})$
+\end_inset
+
+ no es de Hausdorff para 
+\begin_inset Formula $|X|\geq2$
+\end_inset
+
+, pues dados 
+\begin_inset Formula $x,y\in X$
+\end_inset
+
+ con 
+\begin_inset Formula $x\neq y$
+\end_inset
+
+, el único entorno de 
+\begin_inset Formula $x$
+\end_inset
+
+ es 
+\begin_inset Formula $X$
+\end_inset
+
+ y contiene a 
+\begin_inset Formula $y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Espacios métricos
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+espacio métrico
+\series default
+ es un par 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ formado por un conjunto 
+\begin_inset Formula $X\neq\emptyset$
+\end_inset
+
+ y una aplicación 
+\begin_inset Formula $d:X\times X\rightarrow\mathbb{R}$
+\end_inset
+
+ que cumple que 
+\begin_inset Formula $\forall x,y,z\in X:$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $d(x,y)\geq0\land(d(x,y)=0\iff x=y)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Simetría:
+\series default
+ 
+\begin_inset Formula $d(y,x)=d(x,y)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Desigualdad triangular:
+\series default
+ 
+\begin_inset Formula $d(x,z)\leq d(x,y)+d(y,z)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Decimos que 
+\begin_inset Formula $d$
+\end_inset
+
+ es una 
+\series bold
+métrica
+\series default
+ o 
+\series bold
+distancia
+\series default
+ sobre 
+\begin_inset Formula $X$
+\end_inset
+
+.
+ Ejemplos de métricas:
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Métrica usual
+\series default
+ sobre 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+: 
+\begin_inset Formula $d_{u}(x,y)=d_{|\,|}(x,y)=|x-y|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Métrica del ascensor
+\series default
+ sobre 
+\begin_inset Formula $\mathbb{R}^{2}$
+\end_inset
+
+
+\series bold
+:
+\series default
+ 
+\begin_inset Formula 
+\[
+d((x_{1},x_{2}),(y_{1},y_{2}))=\begin{cases}
+|x_{2}-y_{2}| & \text{si }x_{1}=y_{1}\\
+|x_{1}-y_{1}|+|x_{2}|+|y_{2}| & \text{si }x_{1}\neq y_{1}
+\end{cases}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Métrica discreta
+\series default
+: 
+\begin_inset Formula $d_{D}(x,y)=\begin{cases}
+0 & \text{si }x=y\\
+1 & \text{si }x\neq y
+\end{cases}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Espacios métricos producto
+\series default
+: Dados los espacios métricos 
+\begin_inset Formula $(X_{1},d_{1}),\dots,(X_{n},d_{n})$
+\end_inset
+
+, sean 
+\begin_inset Formula $x=(x_{1},\dots,x_{n}),y=(y_{1},\dots,y_{n})\in\prod_{i=1}^{n}X_{i}$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+
+\series bold
+Métrica del taxi:
+\series default
+ 
+\begin_inset Formula $d_{T}(x,y)=\sum_{i=1}^{n}d_{i}(x_{i},y_{i})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Métrica euclídea:
+\series default
+ 
+\begin_inset Formula $d_{E}(x,y)=\sqrt{\sum_{i=1}^{n}d_{i}(x_{i},y_{i})^{2}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Métrica del ajedrez:
+\series default
+ 
+\begin_inset Formula $d_{\infty}(x,y)=\max\{d_{i}(x_{i},y_{i})\}_{1\leq i\leq n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $d_{k}(x,y)=(\sum_{i=1}^{n}d_{i}(x_{i}y_{i})^{k})^{\frac{1}{k}}$
+\end_inset
+
+.
+ Entonces se tiene que 
+\begin_inset Formula $d_{T}=d_{1}$
+\end_inset
+
+, 
+\begin_inset Formula $d_{E}=d_{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $d_{\infty}$
+\end_inset
+
+ tiene un nombre apropiado.
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+
+\series bold
+Métrica estándar acotada
+\series default
+: 
+\begin_inset Formula $\overline{d}(x,y)=\min\{1,d(x,y)\}$
+\end_inset
+
+.
+ En general, obtenemos las mismas propiedades cambiando el 1 por cualquier
+ otro número real positivo.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Métrica estándar acotada (bis)
+\series default
+: 
+\begin_inset Formula $d'(x,y)=\frac{d(x,y)}{1+d(x,y)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+
+\series bold
+Métrica inducida
+\series default
+ por 
+\begin_inset Formula $d$
+\end_inset
+
+ en 
+\begin_inset Formula $H\subseteq X$
+\end_inset
+
+
+\series bold
+:
+\series default
+ 
+\begin_inset Formula $d_{H}:H\times H\rightarrow\mathbb{R}$
+\end_inset
+
+ con 
+\begin_inset Formula $d_{H}(x,y)=d(x,y)$
+\end_inset
+
+ para cualesquiera 
+\begin_inset Formula $x,y\in H$
+\end_inset
+
+.
+ Decimos que 
+\begin_inset Formula $(H,d_{H})$
+\end_inset
+
+ es un 
+\series bold
+subespacio métrico
+\series default
+ de 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{sloppypar}
+\end_layout
+
+\end_inset
+
+Se define la distancia de un punto 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ a un subconjunto 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+ como 
+\begin_inset Formula $d(p,S)=\inf\{d(p,x)\}_{x\in S}$
+\end_inset
+
+.
+ Así, si 
+\begin_inset Formula $p\in S$
+\end_inset
+
+ entonces 
+\begin_inset Formula $d(p,S)=0$
+\end_inset
+
+, si bien el recíproco no es cierto.
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{sloppypar}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Círculos y bolas
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+círculo
+\series default
+ en 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ centrado en 
+\begin_inset Formula $p$
+\end_inset
+
+ con radio 
+\begin_inset Formula $r$
+\end_inset
+
+ es el conjunto 
+\begin_inset Formula $C_{d}(p;r):=C(p;r):=\{x\in X:d(p,x)=r\}$
+\end_inset
+
+.
+ Del mismo modo, la 
+\series bold
+bola abierta
+\series default
+ en 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ centrada en 
+\begin_inset Formula $p$
+\end_inset
+
+ con radio 
+\begin_inset Formula $r$
+\end_inset
+
+ es el conjunto 
+\begin_inset Formula $B_{d}(p;r):=B(p;r):=\{x\in X:d(p,x)<r\}$
+\end_inset
+
+, y la 
+\series bold
+bola cerrada
+\series default
+ en 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ centrada en 
+\begin_inset Formula $p$
+\end_inset
+
+ con radio 
+\begin_inset Formula $r$
+\end_inset
+
+ es el conjunto 
+\begin_inset Formula $\overline{B}_{d}(p;r):=\overline{B}(p;r):=B[p;r]:=\{x\in X:d(p,x)\leq r\}$
+\end_inset
+
+.
+ Se tiene que 
+\begin_inset Formula $B_{d}(p;r)=\bigcup_{0<s<r}C_{d}(p;s)$
+\end_inset
+
+, y 
+\begin_inset Formula $\overline{B}_{d}(p;r)=\bigcup_{0<s\leq r}C_{d}(p;s)$
+\end_inset
+
+.
+ Dado el espacio métrico 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ y 
+\begin_inset Formula $H\subseteq X$
+\end_inset
+
+, 
+\begin_inset Formula $B_{d_{H}}(p;r)=B_{d}(p;r)\cap H$
+\end_inset
+
+ para cualesquiera 
+\begin_inset Formula $p\in H$
+\end_inset
+
+ y 
+\begin_inset Formula $r>0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ es 
+\series bold
+acotado
+\series default
+ si 
+\begin_inset Formula $\exists k>0:\forall x,y\in X,d(x,y)\leq k$
+\end_inset
+
+, y decimos entonces que 
+\begin_inset Formula $d$
+\end_inset
+
+ es una 
+\series bold
+métrica acotada
+\series default
+.
+ Esto sucede si y sólo si 
+\begin_inset Formula $\exists k>0,x_{0}\in X:B(x_{0};k)=X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $x_{0}\in X$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\forall x\in X,d(x_{0},x)\leq k<k+1\implies x\in B_{d}(x_{0},k+1)\implies X\subseteq B_{d}(x_{0},k+1)\subseteq X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Por la desigualdad triangular, 
+\begin_inset Formula $\forall p,q\in X,d(p,q)\leq d(p,x_{0})+d(x_{0},q)<k+k=2k$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ es acotado por 
+\begin_inset Formula $2k$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+También se dice que 
+\begin_inset Formula $H\subseteq X$
+\end_inset
+
+ es acotado si 
+\begin_inset Formula $(H,d_{H})$
+\end_inset
+
+ es acotado, o equivalentemente, si 
+\begin_inset Formula $\exists k>0,x_{0}\in X:H\subseteq B_{d}(x_{0};k)$
+\end_inset
+
+.
+ Por tanto las bolas son subconjuntos acotados, pues 
+\begin_inset Formula $B(p;r)$
+\end_inset
+
+ está acotado por 
+\begin_inset Formula $r$
+\end_inset
+
+ y 
+\begin_inset Formula $\overline{B}_{d}(x;r)$
+\end_inset
+
+ por (al menos) 
+\begin_inset Formula $2r$
+\end_inset
+
+.
+ Definimos el 
+\series bold
+diámetro
+\series default
+ de un espacio métrico acotado 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ como 
+\begin_inset Formula $\text{diám}(X)=\sup\{d(x,y)\}_{x,y\in X}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Subconjuntos abiertos y cerrados
+\end_layout
+
+\begin_layout Standard
+En un espacio métrico 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+, 
+\begin_inset Formula $A\subseteq X$
+\end_inset
+
+ es un 
+\series bold
+subconjunto abierto
+\series default
+, o simplemente un 
+\series bold
+abierto
+\series default
+, si 
+\begin_inset Formula $\forall x\in A,\exists r_{x}>0:B(x;r_{x})\subseteq A$
+\end_inset
+
+.
+ Toda bola abierta es un abierto.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $B(x;r)$
+\end_inset
+
+ una bola abierta en 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ e 
+\begin_inset Formula $y\in B(x;r)$
+\end_inset
+
+, si tomamos 
+\begin_inset Formula $\delta$
+\end_inset
+
+ tal que 
+\begin_inset Formula $0<\delta\leq r-d(x,y)$
+\end_inset
+
+ y 
+\begin_inset Formula $z\in B(y;\delta)$
+\end_inset
+
+, por la desigualdad triangular, 
+\begin_inset Formula $d(x,z)\leq d(x,y)+d(y,z)<d(x,y)+\delta\leq r$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $B(y;\delta)\subseteq B(x;r)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+La condición de ser abierto depende de la métrica y del conjunto sobre el
+ que esta se define, si bien el conjunto total 
+\begin_inset Formula $X$
+\end_inset
+
+ y el vacío 
+\begin_inset Formula $\emptyset$
+\end_inset
+
+ son abiertos en cualquier espacio métrico.
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $A_{1},\dots,A_{n}$
+\end_inset
+
+ abiertos en 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+, la intersección finita 
+\begin_inset Formula $\bigcap_{i=1}^{n}A_{i}$
+\end_inset
+
+ también lo es.
+ 
+\series bold
+Demostración:
+\series default
+ Si tomamos un 
+\begin_inset Formula $p\in\bigcap_{i=1}^{n}A_{i}$
+\end_inset
+
+ arbitrario, para cada 
+\begin_inset Formula $i$
+\end_inset
+
+ con 
+\begin_inset Formula $1\leq i\leq n$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $p\in A_{i}$
+\end_inset
+
+ y existe un 
+\begin_inset Formula $r_{i}>0$
+\end_inset
+
+ tal que 
+\begin_inset Formula $B(p;r_{i})\subseteq A_{i}$
+\end_inset
+
+.
+ Ahora bien, si tomamos 
+\begin_inset Formula $r:=\min\{r_{1},\dots,r_{n}\}$
+\end_inset
+
+, vemos que 
+\begin_inset Formula $B(p;r)\subseteq B(p;r_{i})\subseteq A_{i}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $B(p;r)\subseteq\bigcap_{i=1}^{n}A_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dada la familia 
+\begin_inset Formula $\{A_{i}\}_{i\in I}$
+\end_inset
+
+ de abiertos en 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\bigcup_{i\in I}A_{i}$
+\end_inset
+
+ también es un abierto.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $p\in\bigcup_{i\in I}A_{i}$
+\end_inset
+
+ arbitrario.
+ Entonces existe un 
+\begin_inset Formula $i_{0}\in I$
+\end_inset
+
+ tal que 
+\begin_inset Formula $p\in A_{i_{0}}$
+\end_inset
+
+, y como 
+\begin_inset Formula $A_{i_{0}}$
+\end_inset
+
+ es abierto, existe un 
+\begin_inset Formula $r>0$
+\end_inset
+
+ tal que 
+\begin_inset Formula $B(p;r)\subseteq A_{i_{0}}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $B(p;r)\subseteq A_{i_{0}}\subseteq\bigcup_{i\in I}A_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así pues, todo espacio métrico 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ lleva asociado un espacio topológico 
+\begin_inset Formula $(X,{\cal T}_{d})$
+\end_inset
+
+, donde 
+\begin_inset Formula ${\cal T}_{d}$
+\end_inset
+
+ es el conjunto de abiertos de 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Espacios metrizables
+\end_layout
+
+\begin_layout Standard
+Un espacio topológico 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es 
+\series bold
+metrizable
+\series default
+ si existe una métrica 
+\begin_inset Formula $d$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+ tal que 
+\begin_inset Formula ${\cal T}={\cal T}_{d}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+La métrica discreta lleva asociada la topología discreta (
+\begin_inset Formula ${\cal T}_{D}={\cal T}_{d_{D}}$
+\end_inset
+
+).
+\begin_inset Newline newline
+\end_inset
+
+Todo subconjunto de 
+\begin_inset Formula $X$
+\end_inset
+
+ es abierto en 
+\begin_inset Formula $(X,d_{D})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+La topología indiscreta solo es metrizable si 
+\begin_inset Formula $X$
+\end_inset
+
+ es 
+\series bold
+unipuntual
+\series default
+ (
+\begin_inset Formula $|X|=1$
+\end_inset
+
+).
+\begin_inset Newline newline
+\end_inset
+
+De lo contrario tendríamos 
+\begin_inset Formula $p,q\in X$
+\end_inset
+
+ con 
+\begin_inset Formula $p\neq q$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $d(p,q)=r>0$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $q\notin B(p;\frac{r}{2})$
+\end_inset
+
+, pero esta bola sería un abierto distinto del vacío y del total, lo que
+ no existe en 
+\begin_inset Formula ${\cal T}_{T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dado el espacio métrico 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ y 
+\begin_inset Formula $H\subseteq X$
+\end_inset
+
+, entonces 
+\begin_inset Formula ${\cal T}_{d}|_{H}={\cal T}_{d_{H}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\subseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $A'\in{\cal T}_{d}|_{H}$
+\end_inset
+
+, existe 
+\begin_inset Formula $A\in{\cal T}_{d}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $A'=A\cap H$
+\end_inset
+
+.
+ Entonces para todo 
+\begin_inset Formula $p\in A'\subseteq A$
+\end_inset
+
+ existe un 
+\begin_inset Formula $r>0$
+\end_inset
+
+ tal que 
+\begin_inset Formula $B_{d}(p;r)\subseteq A$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $B_{d}(p;r)\cap H\subseteq A'$
+\end_inset
+
+, pero como 
+\begin_inset Formula $B_{d}(p;r)\cap H=B_{d_{H}}(p;r)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $A'\in{\cal T}_{d_{H}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\supseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $A'\in{\cal T}_{d_{H}}$
+\end_inset
+
+, entonces para todo 
+\begin_inset Formula $p\in A'$
+\end_inset
+
+ existe un 
+\begin_inset Formula $r>0$
+\end_inset
+
+ tal que 
+\begin_inset Formula $B_{d_{H}}(p;r)=B_{d}(p;r)\cap H\subseteq A'$
+\end_inset
+
+, y si llamamos 
+\begin_inset Formula $A=\bigcup_{p\in A'}B_{d}(p;r)$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $A'\subseteq A\cap H=\left(\bigcup_{p\in A'}B_{d}(p;r)\right)\cap H=\bigcup_{p\in A'}(B_{d}(p;r)\cap H)=\bigcup_{p\in A'}B_{d_{H}}(p;r)\subseteq A'$
+\end_inset
+
+ y 
+\begin_inset Formula $A'=A\cap H$
+\end_inset
+
+ con 
+\begin_inset Formula $A\in{\cal T}_{d}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $A'\in{\cal T}_{d}|_{H}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Todo espacio metrizable es 1AN, pues cada punto 
+\begin_inset Formula $x\in X$
+\end_inset
+
+ posee la base de entornos 
+\begin_inset Formula ${\cal B}(x)=\{B(x;\frac{1}{n})\}_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ También es 
+\begin_inset Formula $T_{2}$
+\end_inset
+
+, pues dados 
+\begin_inset Formula $p,q\in X$
+\end_inset
+
+ con 
+\begin_inset Formula $p\neq q$
+\end_inset
+
+, si 
+\begin_inset Formula $r=d(p,q)>0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $B(p;\frac{r}{2})\cap B(q;\frac{r}{2})=\emptyset$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Métricas equivalentes
+\end_layout
+
+\begin_layout Standard
+Dos métricas 
+\begin_inset Formula $d$
+\end_inset
+
+ y 
+\begin_inset Formula $d'$
+\end_inset
+
+ sobre 
+\begin_inset Formula $X$
+\end_inset
+
+ son 
+\series bold
+equivalentes
+\series default
+ si 
+\begin_inset Formula ${\cal T}_{d}={\cal T}_{d'}$
+\end_inset
+
+.
+ Equivalentemente, lo son si 
+\begin_inset Formula $\forall p\in X,r>0;(\exists\delta>0:B_{d}(p;\delta)\subseteq B_{d'}(p;r)\land\exists\delta'>0:B_{d'}(p;\delta')\subseteq B_{d}(p;r))$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sean 
+\begin_inset Formula $d$
+\end_inset
+
+ y 
+\begin_inset Formula $d'$
+\end_inset
+
+ equivalentes, dados 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ y 
+\begin_inset Formula $r>0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $B_{d'}(p;r)$
+\end_inset
+
+ es un abierto en 
+\begin_inset Formula ${\cal T}_{d'}$
+\end_inset
+
+ y por tanto en 
+\begin_inset Formula ${\cal T}_{d}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $\exists\delta>0:B_{d}(p;\delta)\subseteq B_{d'}(p;r)$
+\end_inset
+
+.
+ La otra condición se prueba de forma análoga.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $A$
+\end_inset
+
+ un abierto de 
+\begin_inset Formula ${\cal T}_{d}$
+\end_inset
+
+ y 
+\begin_inset Formula $p\in A$
+\end_inset
+
+, existe pues un 
+\begin_inset Formula $r>0$
+\end_inset
+
+ tal que 
+\begin_inset Formula $B_{d}(p;r)\subseteq A$
+\end_inset
+
+ y por tanto un 
+\begin_inset Formula $\delta'>0$
+\end_inset
+
+ tal que 
+\begin_inset Formula $B_{d'}(p;\delta')\subseteq B_{d}(p;r)$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $A$
+\end_inset
+
+ es abierto en 
+\begin_inset Formula ${\cal T}_{d'}$
+\end_inset
+
+.
+ El otro contenido se prueba de forma análoga.
+\end_layout
+
+\begin_layout Standard
+Dadas dos métricas 
+\begin_inset Formula $d$
+\end_inset
+
+ y 
+\begin_inset Formula $d'$
+\end_inset
+
+ sobre 
+\begin_inset Formula $X$
+\end_inset
+
+, si existen 
+\begin_inset Formula $m,M>0$
+\end_inset
+
+ tales que 
+\begin_inset Formula $\forall x,y\in X,md(x,y)\leq d'(x,y)\leq Md(x,y)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $d$
+\end_inset
+
+ y 
+\begin_inset Formula $d'$
+\end_inset
+
+ son equivalentes.
+ 
+\series bold
+Demostración:
+\series default
+ Dados 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ y 
+\begin_inset Formula $r>0$
+\end_inset
+
+, tomando 
+\begin_inset Formula $\delta=\frac{r}{M}$
+\end_inset
+
+, se tiene que si 
+\begin_inset Formula $d(p,q)\leq\delta$
+\end_inset
+
+ entonces 
+\begin_inset Formula $d'(p,q)\leq Md(p,q)\leq M\delta=r$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $B_{d}(p;\delta)\subseteq B_{d'}(p;r)$
+\end_inset
+
+.
+ Análogamente, tomando 
+\begin_inset Formula $\delta'=mr$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $B_{d'}(p;\delta')\subseteq B_{d}(p;r)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así, las métricas 
+\begin_inset Formula $d_{E}$
+\end_inset
+
+, 
+\begin_inset Formula $d_{T}$
+\end_inset
+
+ y 
+\begin_inset Formula $d_{\infty}$
+\end_inset
+
+ sobre un mismo conjunto 
+\begin_inset Formula $X=X_{1}\times\dots\times X_{n}$
+\end_inset
+
+ y métricas 
+\begin_inset Formula $d_{1},\dots,d_{n}$
+\end_inset
+
+ son equivalentes, y si un subconjunto es acotado para alguna de las tres
+ métricas también lo es para las otras dos.
+ 
+\series bold
+Demostración:
+\series default
+ Se deduce de que 
+\begin_inset Formula $\frac{1}{n}d_{T}(x,y)\leq d_{\infty}(x,y)\leq d_{T}(x,y)$
+\end_inset
+
+ y 
+\begin_inset Formula $\frac{1}{\sqrt{n}}d_{E}(x,y)\leq d_{\infty}(x,y)\leq d_{E}(x,y)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+No obstante, las métricas euclídea y discreta no tienen por qué ser equivalentes
+, pues en 
+\begin_inset Formula $\mathbb{R}^{2}$
+\end_inset
+
+, 
+\begin_inset Formula $\{(0,0)\}$
+\end_inset
+
+ es abierto en la discreta pero no en la euclídea.
+ Llamamos 
+\begin_inset Formula $(\mathbb{R}^{n},d_{u})=(\mathbb{R}^{n},d_{E})$
+\end_inset
+
+ con 
+\begin_inset Formula $d_{E}$
+\end_inset
+
+ definido sobre 
+\begin_inset Formula $d_{|\,|}$
+\end_inset
+
+ en 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+, y 
+\begin_inset Formula ${\cal T}_{u}$
+\end_inset
+
+ a la topología asociada a 
+\begin_inset Formula $d_{u}$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/tem/n2.lyx b/tem/n2.lyx
new file mode 100644
index 0000000..02c4d59
--- /dev/null
+++ b/tem/n2.lyx
@@ -0,0 +1,1269 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Section
+Clausura
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ un espacio topológico y 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+, la 
+\series bold
+clausura
+\series default
+ o 
+\series bold
+adherencia
+\series default
+ de 
+\begin_inset Formula $S$
+\end_inset
+
+ es el menor cerrado que contiene a 
+\begin_inset Formula $S$
+\end_inset
+
+, es decir, la intersección de todos los cerrados que lo contienen, y se
+ denota 
+\begin_inset Formula 
+\[
+\overline{S}:=\text{cl}(S):=\text{ad}(S):=\bigcap\{C\in{\cal C}_{{\cal T}}:S\subseteq C\}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Dado 
+\begin_inset Formula $p\in X$
+\end_inset
+
+, 
+\begin_inset Formula $p\in\overline{S}\iff\forall V\in{\cal E}(p),V\cap S\neq\emptyset$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $p\in\overline{S}$
+\end_inset
+
+ y supongamos que existe 
+\begin_inset Formula $V\in{\cal E}(p)$
+\end_inset
+
+ con 
+\begin_inset Formula $V\cap S=\emptyset$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $S\subseteq X\backslash V\in{\cal C_{T}}$
+\end_inset
+
+, luego 
+\begin_inset Formula $p\in\overline{S}\subseteq X\backslash V$
+\end_inset
+
+.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ tal que 
+\begin_inset Formula $V\cap S\neq\emptyset\forall V\in{\cal E}(x)$
+\end_inset
+
+ y supongamos 
+\begin_inset Formula $p\notin\overline{S}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $p\in X\backslash\overline{S}\in{\cal E}(p)$
+\end_inset
+
+, pero 
+\begin_inset Formula $(X\backslash\overline{S})\cap S=\emptyset$
+\end_inset
+
+.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ es un espacio métrico y 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+, dado 
+\begin_inset Formula $p\in X$
+\end_inset
+
+, 
+\begin_inset Formula $p\in\overline{S}\iff d(p,S)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $p\in\overline{S}$
+\end_inset
+
+, si suponemos 
+\begin_inset Formula $d(p,S)=r>0$
+\end_inset
+
+, entonces 
+\begin_inset Formula $B(p;r)\cap S=\emptyset$
+\end_inset
+
+, lo que contradice 
+\begin_inset Formula $p\in\overline{S}$
+\end_inset
+
+.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $d(p,S)=0$
+\end_inset
+
+, 
+\begin_inset Formula $\forall n\in\mathbb{N},\exists q\in S:d(p,q)<\frac{1}{n}$
+\end_inset
+
+, luego 
+\begin_inset Formula $\forall n\in\mathbb{N},B(p;\frac{1}{n})\cap S\neq\emptyset$
+\end_inset
+
+ y 
+\begin_inset Formula $p\in\overline{S}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Propiedades:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $S\subseteq T\implies\overline{S}\subseteq\overline{T}$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula $S\subseteq T\subseteq\overline{T}\in{\cal C_{T}}$
+\end_inset
+
+, por lo que 
+\begin_inset Formula $\overline{T}$
+\end_inset
+
+ es un cerrado que contiene a 
+\begin_inset Formula $S$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $\overline{S}\subseteq\overline{T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\bigcup_{i\in I}\overline{S_{i}}\subseteq\overline{\bigcup_{i\in I}S_{i}}$
+\end_inset
+
+; 
+\begin_inset Formula $\bigcup_{i=1}^{n}\overline{S_{i}}=\overline{\bigcup_{i=1}^{n}S_{i}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\subseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $\forall j\in I,S_{j}\subseteq\bigcup_{i\in I}S_{i}\implies\overline{S_{j}}\subseteq\overline{\bigcup_{i\in I}S_{i}}\implies\bigcup_{i\in I}\overline{S_{i}}\subseteq\overline{\bigcup_{i\in I}S_{i}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\supseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $\overline{\bigcup_{i\in I}S_{i}}\subseteq\overline{\bigcup_{i\in I}\overline{S_{i}}}\overset{\text{\textbf{SI \ensuremath{I} es finito}}}{=}\bigcup_{i\in I}\overline{S_{i}}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\overline{\bigcap_{i\in I}S_{i}}\subseteq\bigcap_{i\in I}\overline{S_{i}}$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall i\in I,S_{i}\subseteq\overline{S_{i}}\implies\bigcap_{i\in I}S_{i}\subseteq\bigcap_{i\in I}\overline{S_{i}}\implies\overline{\bigcap_{i\in I}S_{i}}\subseteq\overline{\bigcap_{i\in I}\overline{S_{i}}}=\bigcap_{i\in I}\overline{S_{i}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $S\in{\cal C_{T}}\iff\overline{S}=S$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $S\in{\cal C_{T}}\implies\overline{S}\subseteq S\overset{S\subseteq\overline{S}}{\implies}S=\overline{S}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $S=\overline{S}\in{\cal C_{T}}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\overline{\overline{S}}=\overline{S}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $D\subseteq X$
+\end_inset
+
+ es 
+\series bold
+denso
+\series default
+ en 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ si 
+\begin_inset Formula $\overline{D}=X$
+\end_inset
+
+, si y sólo si cualquier abierto no vacío corta a 
+\begin_inset Formula $D$
+\end_inset
+
+.
+ 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es 
+\series bold
+separable
+\series default
+ si admite un subconjunto denso y numerable.
+\end_layout
+
+\begin_layout Standard
+Todo espacio numerable es separable pero el recíproco no se cumple, pues
+ por ejemplo, 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+ es denso en 
+\begin_inset Formula $(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ y numerable y por tanto 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ es separable, pero no es numerable.
+ Igualmente 
+\begin_inset Formula $(X,{\cal T}_{D})$
+\end_inset
+
+ es separable si y sólo si es numerable, mientras que 
+\begin_inset Formula $(X,{\cal T}_{CF})$
+\end_inset
+
+ es siempre separable (basta tomar un subconjunto numerable no finito).
+\end_layout
+
+\begin_layout Section
+Puntos de acumulación y aislados
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+, 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ es un 
+\series bold
+punto de acumulación
+\series default
+ de 
+\begin_inset Formula $S$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall U\in{\cal E}(p),(U\backslash\{p\})\cap S\neq\emptyset$
+\end_inset
+
+.
+ Llamamos 
+\series bold
+acumulación
+\series default
+ o 
+\series bold
+conjunto derivado
+\series default
+ de 
+\begin_inset Formula $S$
+\end_inset
+
+ (
+\begin_inset Formula $\text{ac}(S)$
+\end_inset
+
+ o 
+\begin_inset Formula $S'$
+\end_inset
+
+) al conjunto de todos los puntos de acumulación de 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ Por otro lado, 
+\begin_inset Formula $p\in S$
+\end_inset
+
+ es un 
+\series bold
+punto aislado
+\series default
+ de 
+\begin_inset Formula $S$
+\end_inset
+
+ si 
+\begin_inset Formula $\exists U\in{\cal E}(p):U\cap S=\{p\}$
+\end_inset
+
+, y el conjunto de todos los puntos aislados de 
+\begin_inset Formula $S$
+\end_inset
+
+ es 
+\begin_inset Formula $\text{ais}(S)=S\backslash S'$
+\end_inset
+
+, y se tiene que 
+\begin_inset Formula $\overline{S}=S\cup S'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Frontera
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+, 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ es un 
+\series bold
+punto frontera
+\series default
+ de 
+\begin_inset Formula $S$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall U\in{\cal E}(p),(U\cap S\neq\emptyset\land U\cap(X\backslash S)\neq\emptyset)$
+\end_inset
+
+.
+ Llamamos 
+\series bold
+frontera
+\series default
+ de 
+\begin_inset Formula $S$
+\end_inset
+
+ (
+\begin_inset Formula $\partial S$
+\end_inset
+
+ o 
+\begin_inset Formula $\text{fr}(S)$
+\end_inset
+
+) al conjunto de todos los puntos frontera de 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ Propiedades:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\partial S=\overline{S}\cap\overline{X\backslash S}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\partial S\in{\cal C_{T}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Además, en un espacio métrico,
+\begin_inset Formula 
+\begin{eqnarray*}
+p\in\partial S & \iff & \forall r>0,(B(p;r)\cap S\neq\emptyset\land B(p;r)\cap(X\backslash S)\neq\emptyset)\\
+ & \iff & \forall n\in\mathbb{N},(B(p;\frac{1}{n})\cap S\neq\emptyset\land B(p;\frac{1}{n})\cap(X\backslash S)\neq\emptyset)\\
+ & \iff & d(p,S)=d(p,X\backslash S)=0
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Interior
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ un espacio topológico y 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+, el 
+\series bold
+interior
+\series default
+ de 
+\begin_inset Formula $S$
+\end_inset
+
+ es el mayor abierto contenido en 
+\begin_inset Formula $S$
+\end_inset
+
+, es decir, la unión de todos los abiertos contenidos en 
+\begin_inset Formula $S$
+\end_inset
+
+, y se denota
+\begin_inset Formula 
+\[
+\mathring{S}:=\text{int}S:=\bigcup\{A\in{\cal T}:A\subseteq S\}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Newpage newpage
+\end_inset
+
+Propiedades:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\mathring{S}=X\backslash\overline{X\backslash S}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\subseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $p\in\mathring{S}\implies\exists A\in{\cal T}:p\in A\subseteq\mathring{S}\subseteq S\implies A\cap(X\backslash S)=\emptyset\implies p\notin\overline{X\backslash S}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\supseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $\begin{array}{c}
+X\backslash S\subseteq\overline{X\backslash S}\implies X\backslash\overline{X\backslash S}\subseteq S\\
+X\backslash\overline{X\backslash S}\in{\cal T}
+\end{array}\implies X\backslash\overline{X\backslash S}\subseteq\mathring{S}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $S\in{\cal T}\iff S=\mathring{S}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\partial S=\overline{S}\backslash\mathring{S}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\partial S=\overline{S}\cap\overline{X\backslash S}=\overline{S}\cap(X\backslash\mathring{S})=\overline{S}\backslash\mathring{S}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $S\in{\cal T}\iff S\cap\partial S=\emptyset$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $S\in{\cal T}\implies\partial S=\overline{S}\backslash\mathring{S}=\overline{S}\backslash S\implies\partial S\cap S=\emptyset$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $\emptyset=\partial S\cap S=(\overline{S}\backslash\mathring{S})\cap S=S\backslash\mathring{S}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $p\in\mathring{S}\iff\exists U\in{\cal E}(p):U\subseteq S$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $S\subseteq T\implies\mathring{S}\subseteq\mathring{T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\bigcap_{i=1}^{n}\mathring{S_{i}}=\mathring{\overbrace{\bigcap_{i=1}^{n}S_{i}}}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\begin{array}{c}
+\mathring{S}\cap\mathring{T}=(X\backslash\overline{X\backslash S})\cap(X\backslash\overline{X\backslash T})=X\backslash(\overline{X\backslash S}\cup\overline{X\backslash T})=\\
+=X\backslash\overline{(X\backslash S)\cup(X\backslash T)}=X\backslash\overline{X\backslash(S\cap T)}=\mathring{\overbrace{S\cap T}}
+\end{array}
+\]
+
+\end_inset
+
+Esto NO se cumple para la unión.
+\end_layout
+
+\begin_layout Standard
+Además, en un espacio métrico,
+\begin_inset Formula 
+\begin{eqnarray*}
+p\in\mathring{S} & \iff & \exists r>0:B(p;r)\subseteq S\\
+ & \iff & \exists n\in\mathbb{N}:B(p;\frac{1}{n})\subseteq S\\
+ & \iff & d(p,X\backslash S)>0
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Clausura, frontera e interior relativos
+\end_layout
+
+\begin_layout Standard
+Escribimos 
+\begin_inset Formula $\text{cl}_{X}(S)$
+\end_inset
+
+, 
+\begin_inset Formula $\text{int}_{X}(S)$
+\end_inset
+
+ y 
+\begin_inset Formula $\partial_{X}(S)$
+\end_inset
+
+ en 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ y 
+\begin_inset Formula $\text{cl}_{H}(S)$
+\end_inset
+
+, 
+\begin_inset Formula $\text{int}_{H}(S)$
+\end_inset
+
+ y 
+\begin_inset Formula $\partial_{H}(S)$
+\end_inset
+
+ en 
+\begin_inset Formula $(H,{\cal T}|_{H})$
+\end_inset
+
+.
+ Así, sea 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ un espacio topológico y 
+\begin_inset Formula $S\subseteq H\subseteq X$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{cl}_{H}(S)=\text{cl}_{X}(S)\cap H$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\subseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sabemos que 
+\begin_inset Formula $S\subseteq\text{cl}_{X}(S)\cap H\in{\cal C}_{H}$
+\end_inset
+
+, y como 
+\begin_inset Formula $\text{cl}_{H}(S)$
+\end_inset
+
+ es el menor cerrado en 
+\begin_inset Formula $H$
+\end_inset
+
+ que contiene a 
+\begin_inset Formula $S$
+\end_inset
+
+, 
+\begin_inset Formula $\text{cl}_{H}(S)\subseteq\text{cl}_{X}(S)\cap H$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\supseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $p\in\text{cl}_{X}(S)\cap H$
+\end_inset
+
+ y 
+\begin_inset Formula $U'\in{\cal E}_{H}(p)$
+\end_inset
+
+, entonces existe 
+\begin_inset Formula $U\in{\cal E}_{X}(p)$
+\end_inset
+
+ tal que 
+\begin_inset Formula $U'=U\cap H$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $p\in\text{cl}_{X}(S)$
+\end_inset
+
+, 
+\begin_inset Formula $U\cap S\neq\emptyset$
+\end_inset
+
+, ahora bien, 
+\begin_inset Formula $U'\cap S=U\cap H\cap S=U\cap S\neq\emptyset$
+\end_inset
+
+, luego 
+\begin_inset Formula $p\in\text{cl}_{H}(S)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\text{int}_{X}(S)\cap H\subseteq\text{int}_{H}(S)$
+\end_inset
+
+, y esta inclusión suele ser estricta.
+\series bold
+
+\begin_inset Newline newline
+\end_inset
+
+
+\series default
+
+\begin_inset Formula $\text{int}_{X}(S)\cap H$
+\end_inset
+
+ es un abierto de 
+\begin_inset Formula $H$
+\end_inset
+
+ contenido en 
+\begin_inset Formula $S$
+\end_inset
+
+, y por tanto 
+\begin_inset Formula $\text{int}_{X}(S)\cap H\subseteq\text{int}_{H}(S)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\partial_{H}(S)\subseteq\partial_{X}(S)\cap H$
+\end_inset
+
+.
+\begin_inset Formula 
+\begin{multline*}
+\begin{array}{c}
+\partial_{H}(S)=\text{cl}_{H}(S)\backslash\text{int}_{H}(S)\subseteq(\text{cl}_{X}(S)\cap H)\backslash(\text{int}_{X}(S)\cap H)=\\
+=(\text{cl}_{X}(S)\backslash\text{int}_{X}(S))\cap H=\partial_{X}(S)\cap H
+\end{array}
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Convergencia
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $\{x_{n}\}_{n=1}^{\infty}$
+\end_inset
+
+ una sucesión de puntos de 
+\begin_inset Formula $X$
+\end_inset
+
+, 
+\begin_inset Formula $\{x_{n}\}_{n=1}^{\infty}$
+\end_inset
+
+ 
+\series bold
+converge
+\series default
+ o 
+\series bold
+tiende
+\series default
+ a 
+\begin_inset Formula $x$
+\end_inset
+
+ (
+\begin_inset Formula $x_{n}\rightarrow x$
+\end_inset
+
+ o 
+\begin_inset Formula $\lim x_{n}=x$
+\end_inset
+
+) si 
+\begin_inset Formula $\forall U\in{\cal E}(x),\exists n_{U}\in\mathbb{N}:\forall n\geq n_{U},x_{n}\in U$
+\end_inset
+
+.
+ En particular, en un espacio métrico 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+, 
+\begin_inset Formula $x_{n}\rightarrow x\iff\forall\varepsilon>0,\exists n_{\varepsilon}\in\mathbb{N}:\forall n\geq n_{\varepsilon},x_{n}\in B(x;r)$
+\end_inset
+
+, o lo que es lo mismo, si la sucesión 
+\begin_inset Formula $\{d(x_{n},x)\}_{n=1}^{\infty}$
+\end_inset
+
+ converge a 0 en 
+\begin_inset Formula $(\mathbb{R},d_{u})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ un espacio métrico, 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+ y 
+\begin_inset Formula $x\in X$
+\end_inset
+
+, entonces 
+\begin_inset Formula $x\in\overline{S}\iff\exists\{x_{n}\}_{n=1}^{\infty}\subseteq S:x_{n}\rightarrow x$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $x\in\overline{S}$
+\end_inset
+
+, para cada 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+, 
+\begin_inset Formula $B(x;\frac{1}{n})\cap S\neq\emptyset$
+\end_inset
+
+, luego podemos tomar 
+\begin_inset Formula $x_{n}\in B(x;\frac{1}{n})\cap S$
+\end_inset
+
+ y construir así la sucesión.
+ Entonces 
+\begin_inset Formula $d(x_{n},x)<\frac{1}{n}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $x_{n}\rightarrow x$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Cualquier 
+\begin_inset Formula $U\in{\cal E}(x)$
+\end_inset
+
+ contiene puntos de la sucesión, de forma que 
+\begin_inset Formula $U\cap S\neq\emptyset$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $x\in\overline{S}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así pues, en un espacio métrico 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+, 
+\begin_inset Formula $S$
+\end_inset
+
+ es denso en 
+\begin_inset Formula $X$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\forall x\in X,\exists\{x_{n}\}_{n=1}^{\infty}\subseteq S:x_{n}\rightarrow x$
+\end_inset
+
+, y 
+\begin_inset Formula $x\in\partial S$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\exists\{x_{n}\}_{n=1}^{\infty}\subseteq S,\{y_{n}\}_{n=1}^{\infty}\subseteq X\backslash S:x_{n},y_{n}\rightarrow x$
+\end_inset
+
+.
+ Estas caracterizaciones sólo son ciertas en espacios métricos, pero no
+ es espacios topológicos arbitrarios.
+\end_layout
+
+\end_body
+\end_document
diff --git a/tem/n3.lyx b/tem/n3.lyx
new file mode 100644
index 0000000..245e95c
--- /dev/null
+++ b/tem/n3.lyx
@@ -0,0 +1,1734 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Una aplicación 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es 
+\series bold
+continua
+\series default
+ en 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall V\in{\cal E}(f(p)),\exists U\in{\cal E}(p):f(U)\subseteq V$
+\end_inset
+
+.
+ Equivalentemente, si 
+\begin_inset Formula ${\cal B}(p)$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}(f(p))$
+\end_inset
+
+ son bases de entornos de 
+\begin_inset Formula $p$
+\end_inset
+
+ y 
+\begin_inset Formula $f(p)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\forall V\in{\cal B}(f(p)),\exists U\in{\cal B}(p):f(U)\subseteq V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+, dado 
+\begin_inset Formula $V\in{\cal B}(f(p))$
+\end_inset
+
+, existe 
+\begin_inset Formula $U\in{\cal E}(p)$
+\end_inset
+
+ con 
+\begin_inset Formula $f(U)\subseteq V$
+\end_inset
+
+, pero entonces existe 
+\begin_inset Formula $U'\in{\cal B}(p)$
+\end_inset
+
+ con 
+\begin_inset Formula $U'\subseteq U$
+\end_inset
+
+, luego 
+\begin_inset Formula $f(U')\subseteq f(U)\subseteq V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Dado 
+\begin_inset Formula $V\in{\cal E}(f(p))$
+\end_inset
+
+, existe 
+\begin_inset Formula $V'\in{\cal B}(f(p))$
+\end_inset
+
+ con 
+\begin_inset Formula $V'\subseteq V$
+\end_inset
+
+, pero existe 
+\begin_inset Formula $U\in{\cal B}(p)\subseteq{\cal E}(p)$
+\end_inset
+
+ con 
+\begin_inset Formula $f(U)\subseteq V'\subseteq V$
+\end_inset
+
+, luego 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+De aquí que 
+\begin_inset Formula $f:(X,d)\rightarrow(Y,d')$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+ respecto a las topologías métricas 
+\begin_inset Formula ${\cal T}_{d}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal T}_{d'}$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\forall\varepsilon>0,\exists\delta>0:\forall x\in X,(d(x,p)<\delta\implies d'(f(x),f(p))<\varepsilon)$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Tomando 
+\begin_inset Formula ${\cal B}(p)=\{B(p;\delta):\delta>0\}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal B}(f(p))=\{B(f(p);r)\}_{r>0}$
+\end_inset
+
+, la equivalencia es consecuencia de lo anterior y de que 
+\begin_inset Formula $x\in B(p;\delta)\iff d(x,p)<\delta$
+\end_inset
+
+ y 
+\begin_inset Formula $f(p)\in B(f(p);\varepsilon)\iff d(f(x),f(p))<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es 1AN, 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\forall\{x_{n}\}_{n=1}^{\infty}\subseteq X,(x_{n}\rightarrow p\implies f(x_{n})\rightarrow f(p))$
+\end_inset
+
+.
+ Además, la implicación a la derecha se cumple para espacios topológicos
+ arbitrarios.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+, dada una sucesión 
+\begin_inset Formula $\{x_{n}\}_{n=1}^{\infty}\subseteq X$
+\end_inset
+
+ que converge a 
+\begin_inset Formula $p$
+\end_inset
+
+ y 
+\begin_inset Formula $V\in{\cal E}(f(p))$
+\end_inset
+
+, existe 
+\begin_inset Formula $U\in{\cal E}(p)$
+\end_inset
+
+ con 
+\begin_inset Formula $f(U)\subseteq V$
+\end_inset
+
+, y por la convergencia de 
+\begin_inset Formula $\{x_{n}\}_{n=1}^{\infty}$
+\end_inset
+
+, existe un 
+\begin_inset Formula $n_{U}$
+\end_inset
+
+ tal que si 
+\begin_inset Formula $n>n_{U}$
+\end_inset
+
+ entonces 
+\begin_inset Formula $x_{n}\in U$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $f(x_{n})\in f(U)\subseteq V$
+\end_inset
+
+, luego 
+\begin_inset Formula $f(x_{n})\rightarrow f(p)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula ${\cal B}(p)$
+\end_inset
+
+ una base de entornos de 
+\begin_inset Formula $p$
+\end_inset
+
+ numerable, si suponemos que 
+\begin_inset Formula $f$
+\end_inset
+
+ no es continua, entonces 
+\begin_inset Formula $\exists V\in{\cal B}(f(p)):\forall U\in{\cal B}(p),f(U)\nsubseteq V$
+\end_inset
+
+.
+ Sea ahora 
+\begin_inset Formula $U_{1}\in{\cal B}(p)$
+\end_inset
+
+ y 
+\begin_inset Formula $V_{1}$
+\end_inset
+
+ un entorno de 
+\begin_inset Formula $p$
+\end_inset
+
+ que no contiene a 
+\begin_inset Formula $U_{1}$
+\end_inset
+
+.
+ Podemos tomar 
+\begin_inset Formula $V'_{1}:=V_{1}\cap U_{1}\in{\cal E}(p)$
+\end_inset
+
+ y existirá 
+\begin_inset Formula $U_{2}\in{\cal B}(p)$
+\end_inset
+
+ con 
+\begin_inset Formula $U_{2}\subseteq V'_{1}$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula ${\cal B}(p)$
+\end_inset
+
+ es numerable, podemos hacer esto sucesivamente ordenando así sus elementos
+ en una sucesión 
+\begin_inset Formula $\{U_{n}\}_{n=1}^{\infty}$
+\end_inset
+
+ de entornos con 
+\begin_inset Formula $U_{1}\supseteq U_{2}\supseteq\dots$
+\end_inset
+
+.
+ Con esto formamos una sucesión 
+\begin_inset Formula $\{x_{n}\}_{n=1}^{\infty}$
+\end_inset
+
+ con 
+\begin_inset Formula $x_{i}\in U_{i}$
+\end_inset
+
+ y 
+\begin_inset Formula $f(x_{i})\notin V$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $x_{n}\rightarrow p$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+ mientras que 
+\begin_inset Formula $f(x_{n})\not\rightarrow f(p)$
+\end_inset
+
+ en 
+\begin_inset Formula $Y$
+\end_inset
+
+, lo que contradice la hipótesis.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $(X,{\cal T})\overset{f}{\rightarrow}(Y,{\cal T}')\overset{g}{\rightarrow}(Z,{\cal T}'')$
+\end_inset
+
+ aplicaciones continuas en 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ y 
+\begin_inset Formula $f(p)$
+\end_inset
+
+, respectivamente, entonces 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Dado 
+\begin_inset Formula $W\in{\cal E}(g(f(p)))$
+\end_inset
+
+, como 
+\begin_inset Formula $g$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $f(p)$
+\end_inset
+
+, existe 
+\begin_inset Formula $V\in{\cal E}(f(p))$
+\end_inset
+
+ con 
+\begin_inset Formula $g(V)\subseteq W$
+\end_inset
+
+, y como 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+, existe 
+\begin_inset Formula $U\in{\cal E}(p)$
+\end_inset
+
+ con 
+\begin_inset Formula $f(U)\subseteq V$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $g(f(U))\subseteq g(V)\subseteq W$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dado 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+, si 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p\in\overline{S}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f(p)\in\overline{f(S)}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $V\in{\cal E}(f(p))$
+\end_inset
+
+, como 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+, existe 
+\begin_inset Formula $U\in{\cal E}(p)$
+\end_inset
+
+ con 
+\begin_inset Formula $f(U)\subseteq V$
+\end_inset
+
+, pero como 
+\begin_inset Formula $p\in\overline{S}$
+\end_inset
+
+ entonces 
+\begin_inset Formula $U\cap S\neq\emptyset$
+\end_inset
+
+, luego 
+\begin_inset Formula $\emptyset\neq f(U\cap S)\subseteq f(U)\cap f(S)\subseteq V\cap f(S)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Continuidad global
+\end_layout
+
+\begin_layout Standard
+Una aplicación 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua si lo es en cualquier punto de 
+\begin_inset Formula $X$
+\end_inset
+
+.
+ Equivalentemente, 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua si y sólo si 
+\begin_inset Formula $\forall A\in{\cal T}',f^{-1}(A)\in{\cal T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $f$
+\end_inset
+
+ continua, 
+\begin_inset Formula $A\in{\cal T}'$
+\end_inset
+
+.
+ Dado 
+\begin_inset Formula $p\in f^{-1}(A)$
+\end_inset
+
+ arbitrario, entonces 
+\begin_inset Formula $f(p)\in A\in{\cal E}(f(p))$
+\end_inset
+
+, y como 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua, existe 
+\begin_inset Formula $V_{p}\in{\cal E}(p)$
+\end_inset
+
+ con 
+\begin_inset Formula $f(V_{p})\subseteq A$
+\end_inset
+
+, luego 
+\begin_inset Formula $V_{p}\subseteq f^{-1}(A)$
+\end_inset
+
+.
+ Pero entonces 
+\begin_inset Formula $\bigcup_{p\in f^{-1}(A)}V_{p}=f^{-1}(A)\in{\cal T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sean 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ y 
+\begin_inset Formula $A\in{\cal E}(f(p))$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $p\in f^{-1}(A)$
+\end_inset
+
+, y como por hipótesis 
+\begin_inset Formula $f^{-1}(A)\in{\cal T}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f^{-1}(A)\in{\cal E}(p)$
+\end_inset
+
+ es pues el entorno de 
+\begin_inset Formula $p$
+\end_inset
+
+ buscado para que 
+\begin_inset Formula $f$
+\end_inset
+
+ sea continua en 
+\begin_inset Formula $p$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua si y sólo si 
+\begin_inset Formula $\forall p\in X,V\in{\cal E}(f(p));f^{-1}(V)\in{\cal E}(p)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Trivial.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Cada 
+\begin_inset Formula $A\in{\cal T}'$
+\end_inset
+
+ se puede escribir como 
+\begin_inset Formula $A=\bigcup_{q\in A}V_{q}$
+\end_inset
+
+ con 
+\begin_inset Formula $V_{q}\in{\cal E}(q)$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $f^{-1}(A)=f^{-1}(\bigcup_{q\in A}V_{q})=\bigcup_{q\in A}f^{-1}(V_{q})$
+\end_inset
+
+.
+ Por tanto, si los 
+\begin_inset Formula $f^{-1}(V_{q})$
+\end_inset
+
+ son abiertos, 
+\begin_inset Formula $f^{-1}(A)$
+\end_inset
+
+ también lo es por ser unión arbitraria de abiertos.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua si y sólo si 
+\begin_inset Formula $\forall C\in{\cal C}_{{\cal T}'},f^{-1}(C)\in{\cal C_{T}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua y 
+\begin_inset Formula $C$
+\end_inset
+
+ es cerrado en 
+\begin_inset Formula $(Y,{\cal T}')$
+\end_inset
+
+, entonces 
+\begin_inset Formula $X\backslash f^{-1}(C)=f^{-1}(Y\backslash C)\in{\cal T}$
+\end_inset
+
+, luego 
+\begin_inset Formula $f^{-1}(C)\in{\cal C_{T}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Análoga.
+\end_layout
+
+\begin_layout Standard
+Algunas aplicaciones continuas:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $id:(X,{\cal T})\rightarrow(X,{\cal T}')$
+\end_inset
+
+ es continua si y sólo si 
+\begin_inset Formula ${\cal T}'\subseteq{\cal T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Una aplicación constante siempre es continua.
+\end_layout
+
+\begin_layout Enumerate
+Toda 
+\begin_inset Formula $f:(X,{\cal T}_{D})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua.
+\end_layout
+
+\begin_layout Enumerate
+Toda 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}_{T})$
+\end_inset
+
+ es continua.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $f,g:(X,{\cal T})\rightarrow(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ son continuas entonces 
+\begin_inset Formula $f+g,fg:(X,{\cal T})\rightarrow(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ también lo son.
+ Si además 
+\begin_inset Formula $g(x)\neq0\forall x\in X$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\frac{f}{g}:(X,{\cal T})\rightarrow(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ es continua.
+\end_layout
+
+\begin_layout Enumerate
+Las proyecciones 
+\begin_inset Formula $\pi_{i}:(\mathbb{R}^{n},d_{u})\rightarrow(\mathbb{R},d_{u})$
+\end_inset
+
+ con 
+\begin_inset Formula $\pi_{i}(x_{1},\dots,x_{n})=x_{i}$
+\end_inset
+
+ son continuas.
+\end_layout
+
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(\mathbb{R}^{n},{\cal T}_{u})$
+\end_inset
+
+ dada por 
+\begin_inset Formula $f(x)=(f_{1}(x),\dots,f_{n}(x))$
+\end_inset
+
+, siendo 
+\begin_inset Formula $f_{1},\dots,f_{n}:(X,{\cal T})\rightarrow(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ las llamadas 
+\series bold
+funciones coordenadas
+\series default
+ de 
+\begin_inset Formula $f$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua si y sólo si 
+\begin_inset Formula $f_{1},\dots,f_{n}$
+\end_inset
+
+ lo son.
+\end_layout
+
+\begin_layout Enumerate
+Las funciones polinómicas 
+\begin_inset Formula $f:(\mathbb{R}^{n},{\cal T}_{u})\rightarrow(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ sobre una o varias variables son siempre continuas.
+\end_layout
+
+\begin_layout Standard
+Para toda aplicación continua 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ y todo 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+ se tiene que 
+\begin_inset Formula $f(\overline{S})\subseteq\overline{f(S)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Homeomorfismos
+\end_layout
+
+\begin_layout Standard
+Un 
+\series bold
+homeomorfismo
+\series default
+ es una aplicación 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ biyectiva, continua y con aplicación inversa continua.
+ Dos espacios topológicos son 
+\series bold
+homeomorfos
+\series default
+ si existe un homeomorfismo entre ellos, y una 
+\series bold
+propiedad topológica
+\series default
+ es una propiedad de los espacios topológicos invariante por homomorfismos.
+ Ejemplos:
+\end_layout
+
+\begin_layout Itemize
+Dos espacios topológicos triviales, o dos discretos, son homeomorfos si
+ y sólo si existe una aplicación biyectiva entre ellos.
+\end_layout
+
+\begin_layout Itemize
+En 
+\begin_inset Formula $(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+, son homeomorfos todos los intervalos de la forma 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ y 
+\begin_inset Formula $[c,d]$
+\end_inset
+
+; 
+\begin_inset Formula $(a,b)$
+\end_inset
+
+ y 
+\begin_inset Formula $(c,d)$
+\end_inset
+
+; 
+\begin_inset Formula $(a,+\infty)$
+\end_inset
+
+ y 
+\begin_inset Formula $(b,+\infty)$
+\end_inset
+
+; 
+\begin_inset Formula $(-\infty,a)$
+\end_inset
+
+ y 
+\begin_inset Formula $(-\infty,b)$
+\end_inset
+
+, y 
+\begin_inset Formula $(a,+\infty)$
+\end_inset
+
+ y 
+\begin_inset Formula $(-\infty,b)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ es homeomorfo a cualquier intervalo abierto y acotado, por ejemplo, por
+ 
+\begin_inset Formula $\tan:(-\frac{\pi}{2},\frac{\pi}{2})\rightarrow\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dada una aplicación 
+\emph on
+biyectiva
+\emph default
+ 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+, son equivalentes:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f$
+\end_inset
+
+ es un homeomorfismo.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A\in{\cal T}\iff f(A)\in{\cal T}'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $C\in{\cal C_{T}}\iff f(C)\in{\cal C}_{{\cal T}'}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $1\implies2]$
+\end_inset
+
+ Sea 
+\begin_inset Formula $g:=f^{-1}:Y\rightarrow X$
+\end_inset
+
+ continua y 
+\begin_inset Formula $A\in{\cal T}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f(A)=(f^{-1})^{-1}(A)=g^{-1}(A)\in{\cal T}'$
+\end_inset
+
+.
+ Recíprocamente, si 
+\begin_inset Formula $f(A)\in{\cal T}'$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f^{-1}(f(A))=A\in{\cal T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $2\implies1]$
+\end_inset
+
+ Para ver que 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua, dado 
+\begin_inset Formula $A\subseteq X$
+\end_inset
+
+, si 
+\begin_inset Formula $f(A)\in{\cal T}'$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f^{-1}(f(A))=A\in{\cal T}$
+\end_inset
+
+.
+ Para ver que 
+\begin_inset Formula $g:=f^{-1}$
+\end_inset
+
+ es continua, dado 
+\begin_inset Formula $A\subseteq X$
+\end_inset
+
+, si 
+\begin_inset Formula $A\in{\cal T}$
+\end_inset
+
+ entonces 
+\begin_inset Formula $g^{-1}(A)=f(A)\in{\cal T}'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $1\iff3]$
+\end_inset
+
+ Análogo usando la caracterización de continuidad por cerrados.
+\end_layout
+
+\begin_layout Standard
+Una aplicación 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es 
+\series bold
+abierta
+\series default
+ si 
+\begin_inset Formula $\forall A\in{\cal T},f(A)\in{\cal T}'$
+\end_inset
+
+, y es 
+\series bold
+cerrada
+\series default
+ si 
+\begin_inset Formula $\forall C\in{\cal C_{T}},f(C)\in{\cal C}_{{\cal T}'}$
+\end_inset
+
+.
+ Así, una aplicación biyectiva es un homeomorfismo si y sólo si es continua
+ y abierta (o continua y cerrada).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es abierta si y sólo si 
+\begin_inset Formula $\forall S\subseteq X,f(\mathring{S})\subseteq\mathring{\overbrace{f(S)}}$
+\end_inset
+
+, es un homeomorfismo si y sólo si es biyectiva y 
+\begin_inset Formula $\forall S\subseteq X,f(\mathring{S})=\mathring{\overbrace{f(S)}}$
+\end_inset
+
+, y es cerrada si y sólo si 
+\begin_inset Formula $\forall S\subseteq X,\overline{f(S)}\subseteq f(\overline{S})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Continuidad en subespacios
+\end_layout
+
+\begin_layout Standard
+La aplicación inclusión 
+\begin_inset Formula $i:(H,{\cal T}_{H})\looparrowright(X,{\cal T})$
+\end_inset
+
+ es continua.
+ 
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $A\in{\cal T}$
+\end_inset
+
+, 
+\begin_inset Formula $i^{-1}(A)=A\cap H\in{\cal T}_{H}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una aplicación 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ con 
+\begin_inset Formula $f(X)\subseteq H\subseteq Y$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $\hat{f}:(X,{\cal T})\rightarrow(H,{\cal T}_{H})$
+\end_inset
+
+ con 
+\begin_inset Formula $\hat{f}(x)=f(x)$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+.
+ En particular, 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua si y sólo si 
+\begin_inset Formula $\hat{f}$
+\end_inset
+
+ es continua.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+, dado 
+\begin_inset Formula $V'\in{\cal E}_{{\cal T}'_{H}}(f(p))$
+\end_inset
+
+, existe 
+\begin_inset Formula $V\in{\cal E}_{{\cal T}'}(f(p))$
+\end_inset
+
+ con 
+\begin_inset Formula $V'=V\cap H$
+\end_inset
+
+, luego existe 
+\begin_inset Formula $U\in{\cal E}_{{\cal T}}(p)$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f(U)\subseteq V$
+\end_inset
+
+, y entonces 
+\begin_inset Formula $f'(U)=f(U)=f(U)\cap H\subseteq V\cap H=V'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(H,{\cal T}'_{H})$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+, como la inclusión es continua en 
+\begin_inset Formula $f(p)$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f=i\circ\hat{f}$
+\end_inset
+
+ es también continua en 
+\begin_inset Formula $p$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p\in H\subseteq X$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f|_{H}:(H,{\cal T}_{H})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ también es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+.
+ En particular, si 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua también lo es 
+\begin_inset Formula $f|_{H}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Como la inclusión es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+, 
+\begin_inset Formula $f|_{H}=f\circ i$
+\end_inset
+
+ también lo es.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ si y sólo si existe 
+\begin_inset Formula $U\in{\cal E}(p)$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f|_{U}$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Basta tomar 
+\begin_inset Formula $U=X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $f|_{U}:(U,{\cal T}_{U})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+, sea 
+\begin_inset Formula $V\in{\cal E}(f(p))$
+\end_inset
+
+, por la continuidad de 
+\begin_inset Formula $f|_{U}$
+\end_inset
+
+ existe 
+\begin_inset Formula $U'\in{\cal E}(p)$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f|_{U}(U')\subseteq V$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $f(U')=f|_{U}(U')\subseteq V$
+\end_inset
+
+, lo que prueba la continuidad de 
+\begin_inset Formula $f$
+\end_inset
+
+ en 
+\begin_inset Formula $p$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ y 
+\begin_inset Formula $\{A_{i}\}_{i\in I}$
+\end_inset
+
+ una familia de abiertos de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ con 
+\begin_inset Formula $X=\bigcup_{i\in I}A_{i}$
+\end_inset
+
+, si 
+\begin_inset Formula $f|_{A_{i}}$
+\end_inset
+
+ es continua para todo 
+\begin_inset Formula $i\in I$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua.
+ 
+\series bold
+Demostración:
+\series default
+ Dado 
+\begin_inset Formula $p\in X$
+\end_inset
+
+, existe un 
+\begin_inset Formula $i_{0}\in I$
+\end_inset
+
+ tal que 
+\begin_inset Formula $p\in A_{i_{0}}\in{\cal E}(p)$
+\end_inset
+
+ y por la propiedad anterior, 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua en 
+\begin_inset Formula $p$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sea 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ y 
+\begin_inset Formula $\{C_{1},\dots,C_{n}\}$
+\end_inset
+
+ una familia finita de cerrados de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ con 
+\begin_inset Formula $X=\bigcup_{i=1}^{n}C_{i}$
+\end_inset
+
+, si 
+\begin_inset Formula $f|_{C_{i}}$
+\end_inset
+
+ es continua para todo 
+\begin_inset Formula $i\in1,\dots,n$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua.
+ 
+\series bold
+Demostración:
+\series default
+ Dado 
+\begin_inset Formula $C'\in(Y,{\cal T}')$
+\end_inset
+
+, 
+\begin_inset Formula $f^{-1}(C')=f^{-1}(C')\cap X=f^{-1}(C')\cap\left(\bigcup_{i=1}^{n}C_{i}\right)=\bigcup_{i=1}^{n}(C_{i}\cap f^{-1}(C'))=\bigcup_{i=1}^{n}f|_{C_{i}}^{-1}(C')$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $f|_{C_{i}}$
+\end_inset
+
+ es continua para cualquier 
+\begin_inset Formula $i\in\{1,\dots,n\}$
+\end_inset
+
+, 
+\begin_inset Formula $f|_{C_{i}}^{-1}(C')$
+\end_inset
+
+ es cerrado en 
+\begin_inset Formula $(C_{i},{\cal T}_{C_{i}})$
+\end_inset
+
+, y como 
+\begin_inset Formula $C_{i}$
+\end_inset
+
+ es cerrado en 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f|_{C_{i}}^{-1}(C')$
+\end_inset
+
+ es cerrado en 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $f^{-1}(C')$
+\end_inset
+
+ es cerrado en 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+, luego 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua.
+\end_layout
+
+\begin_layout Section
+Continuidad uniforme e isometrías
+\end_layout
+
+\begin_layout Standard
+Definimos la 
+\series bold
+oscilación
+\series default
+ de una función 
+\begin_inset Formula $f:D\subseteq\mathbb{R}\rightarrow\mathbb{R}$
+\end_inset
+
+ en un intervalo 
+\begin_inset Formula $I\subseteq D$
+\end_inset
+
+ como
+\begin_inset Formula 
+\[
+\theta(f,J)=\begin{cases}
+\sup\{f(I)\}-\inf\{f(I)\} & \text{si }f(I)\text{ está acotado}\\
++\infty & \text{si }f(I)\text{ no está acotado}
+\end{cases}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Una aplicación 
+\begin_inset Formula $f:(X,d)\rightarrow(Y,d')$
+\end_inset
+
+ es 
+\series bold
+uniformemente continua
+\series default
+ si 
+\begin_inset Formula $\forall\varepsilon>0,\exists\delta>0:\forall x_{1},x_{2}\in X,(d(x_{1},x_{2})<\delta\implies d'(f(x_{1}),f(x_{2}))<\varepsilon)$
+\end_inset
+
+.
+ Toda aplicación uniformemente continua es continua.
+\end_layout
+
+\begin_layout Standard
+Llamamos 
+\series bold
+isometría
+\series default
+ a una aplicación 
+\begin_inset Formula $f:(X,d)\rightarrow(Y,d')$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall x_{1},x_{2}\in X,(d(x_{1},x_{2})=d'(f(x_{1}),f(x_{2})))$
+\end_inset
+
+.
+ Toda isometría es inyectiva y uniformemente continua.
+ Finalmente, una aplicación 
+\begin_inset Formula $f:(X,d)\rightarrow(X,d')$
+\end_inset
+
+ es 
+\series bold
+lipschitziana
+\series default
+ si 
+\begin_inset Formula $\exists M>0:\forall x,y\in X,d'(f(x),f(y))\leq Md(x,y)$
+\end_inset
+
+, y es además 
+\series bold
+contráctil
+\series default
+ si podemos encontrar un 
+\begin_inset Formula $M<1$
+\end_inset
+
+ para el que se cumpla la propiedad.
+\end_layout
+
+\end_body
+\end_document
diff --git a/tem/n4.lyx b/tem/n4.lyx
new file mode 100644
index 0000000..574a4a5
--- /dev/null
+++ b/tem/n4.lyx
@@ -0,0 +1,2144 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Un 
+\series bold
+recubrimiento
+\series default
+ de 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+ es una familia 
+\begin_inset Formula ${\cal A}=\{A_{i}\}_{i\in I}$
+\end_inset
+
+ de subconjuntos de 
+\begin_inset Formula $X$
+\end_inset
+
+ con 
+\begin_inset Formula $S\subseteq\bigcup_{i\in I}A_{i}$
+\end_inset
+
+, y un 
+\series bold
+subrecubrimiento
+\series default
+ es una familia 
+\begin_inset Formula ${\cal B}\subseteq{\cal A}$
+\end_inset
+
+ que es también recubrimiento de 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ Un recubrimiento 
+\begin_inset Formula $\{A_{i}\}_{i\in I}$
+\end_inset
+
+ de 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+ es 
+\series bold
+finito
+\series default
+ si está formado por una cantidad finita de conjuntos, y es 
+\series bold
+abierto
+\series default
+ en 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ si cada 
+\begin_inset Formula $A_{i}$
+\end_inset
+
+ lo es.
+ Con esto, un espacio topológico 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es 
+\series bold
+compacto
+\series default
+ si todo recubrimiento abierto de 
+\begin_inset Formula $X$
+\end_inset
+
+ admite un subrecubrimiento finito.
+\end_layout
+
+\begin_layout Section
+Subespacios compactos
+\end_layout
+
+\begin_layout Standard
+El subespacio 
+\begin_inset Formula $(K,{\cal T}_{K})$
+\end_inset
+
+ de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es compacto si y sólo si todo recubrimiento de 
+\begin_inset Formula $K$
+\end_inset
+
+ por abiertos de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ admite un subrecubrimiento finito.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $\{A_{i}\}_{i\in I}$
+\end_inset
+
+ un recubrimiento de 
+\begin_inset Formula $K$
+\end_inset
+
+ por abiertos de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\{A_{i}\cap K\}_{i\in I}$
+\end_inset
+
+ es un recubrimiento de 
+\begin_inset Formula $K$
+\end_inset
+
+ por abiertos de 
+\begin_inset Formula $(K,{\cal T}_{K})$
+\end_inset
+
+, por lo que existe una familia finita 
+\begin_inset Formula $A_{i_{1}},\dots,A_{i_{r}}$
+\end_inset
+
+ con 
+\begin_inset Formula $K=(A_{i_{1}}\cap K)\cup\dots\cup(A_{i_{r}}\cap K)$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $K\subseteq A_{i_{1}}\cup\dots\cup A_{i_{r}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $\{A'_{i}\}_{i\in I}$
+\end_inset
+
+ un recubrimiento de 
+\begin_inset Formula $K$
+\end_inset
+
+ por abiertos de 
+\begin_inset Formula $(K,{\cal T}_{K})$
+\end_inset
+
+, y sea por tanto 
+\begin_inset Formula $\{A_{i}\}_{i\in I}$
+\end_inset
+
+ una familia de abiertos de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ con 
+\begin_inset Formula $A'_{i}=A_{i}\cap K$
+\end_inset
+
+, entonces 
+\begin_inset Formula $K\subseteq\bigcup_{i\in I}A_{i}$
+\end_inset
+
+ y por hipótesis existen 
+\begin_inset Formula $A_{i_{1}},\dots,A_{i_{r}}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $K\subseteq A_{i_{1}}\cup\dots\cup A_{i_{r}}$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $K=(A_{i_{1}}\cap K)\cup\dots\cup(A_{i_{r}}\cap K)=A'_{i_{1}}\cup\dots\cup A'_{i_{r}}$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $K$
+\end_inset
+
+ es compacto.
+\end_layout
+
+\begin_layout Standard
+Por tanto el concepto de compacidad es intrínseco del espacio topológico,
+ pues no depende del espacio total donde se considere.
+\end_layout
+
+\begin_layout Standard
+Todo cerrado 
+\begin_inset Formula $C$
+\end_inset
+
+ de un compacto 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es compacto.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula ${\cal A}=\{A_{i}\}_{i\in I}$
+\end_inset
+
+ un recubrimiento de 
+\begin_inset Formula $C$
+\end_inset
+
+ por abiertos de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+, entonces 
+\begin_inset Formula ${\cal A}\cup\{X\backslash C\}$
+\end_inset
+
+ es un recubrimiento abierto de 
+\begin_inset Formula $X$
+\end_inset
+
+, del que extraemos un subrecubrimiento finito 
+\begin_inset Formula $\{A_{i_{1}},\dots,A_{i_{n}}\}$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $C\subseteq X=\{A_{i_{1}},\dots,A_{i_{n}}\}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+teorema de Heine-Borel
+\series default
+ afirma que todo intervalo cerrado y acotado 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ en 
+\begin_inset Formula $(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ es compacto.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula ${\cal A}=\{A_{i}\}_{i\in I}$
+\end_inset
+
+ un recubrimiento de 
+\begin_inset Formula $[a,b]$
+\end_inset
+
+ por abiertos de 
+\begin_inset Formula $(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ y definimos 
+\begin_inset Formula $G=\{x\in[a,b]|\exists\{A_{i_{1}},\dots,A_{i_{n}}\}\in{\cal P}_{0}({\cal A}):[a,x]\subseteq A_{i_{1}}\cup\dots\cup A_{i_{n}}\}$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $a\in[a,b]$
+\end_inset
+
+, existe 
+\begin_inset Formula $i_{0}\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $a\in A_{i_{0}}\in{\cal T}_{u}$
+\end_inset
+
+, luego 
+\begin_inset Formula $\exists\varepsilon>0:[a,a+\varepsilon)\subseteq A_{i_{0}}$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $[a,a+\varepsilon)\subseteq G$
+\end_inset
+
+ y 
+\begin_inset Formula $G\neq\emptyset$
+\end_inset
+
+.
+ Ahora veamos que 
+\begin_inset Formula $G$
+\end_inset
+
+ es cerrado.
+ Sea 
+\begin_inset Formula $y\in[a,b]\backslash G$
+\end_inset
+
+, y como 
+\begin_inset Formula $y\in[a,b]$
+\end_inset
+
+, existe 
+\begin_inset Formula $j_{0}\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $y\in A_{j_{0}}\in{\cal T}_{u}$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $\exists\delta>0:(y-\delta,y+\delta)\subseteq A_{j_{0}}$
+\end_inset
+
+, e 
+\begin_inset Formula $(y-\delta,y+\delta)\subseteq[a,b]\backslash G$
+\end_inset
+
+.
+ En efecto, si existiera un 
+\begin_inset Formula $z\in(y-\delta,y+\delta)\cap G$
+\end_inset
+
+, como 
+\begin_inset Formula $z\in G$
+\end_inset
+
+, entonces 
+\begin_inset Formula $[a,z]\subseteq\bigcup_{j=1}^{n}A_{i_{j}}$
+\end_inset
+
+, y como 
+\begin_inset Formula $\{A_{i_{0}},\dots,A_{i_{n}},A_{j_{0}}\}\in{\cal P}_{0}({\cal A})$
+\end_inset
+
+, entonces para 
+\begin_inset Formula $t\in(y-\delta,y+\delta)$
+\end_inset
+
+ se tendría 
+\begin_inset Formula $[a,t]\subseteq\bigcup_{j=1}^{n}A_{i_{j}}\cup A_{j_{0}}$
+\end_inset
+
+, llegando así a la contradicción de que 
+\begin_inset Formula $y\in G$
+\end_inset
+
+.
+ En consecuencia, 
+\begin_inset Formula $(y-\delta,y+\delta)\subseteq[a,b]\backslash G$
+\end_inset
+
+, y como 
+\begin_inset Formula $y$
+\end_inset
+
+ es un elemento arbitrario de 
+\begin_inset Formula $[a,b]\backslash G$
+\end_inset
+
+, se tiene que 
+\begin_inset Formula $[a,b]\backslash G$
+\end_inset
+
+ es abierto y por tanto 
+\begin_inset Formula $G$
+\end_inset
+
+ es cerrado.
+ Finalmente, vemos que 
+\begin_inset Formula $G=[a,b]$
+\end_inset
+
+.
+ En efecto, sea 
+\begin_inset Formula $s=\sup(G)$
+\end_inset
+
+, y como 
+\begin_inset Formula $G$
+\end_inset
+
+ es cerrado entonces 
+\begin_inset Formula $s\in G$
+\end_inset
+
+.
+ Supongamos que 
+\begin_inset Formula $s<b$
+\end_inset
+
+, entonces existe 
+\begin_inset Formula $k_{0}\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $s\in A_{k_{0}}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $[s,s+\varepsilon)\subseteq A_{k_{0}}$
+\end_inset
+
+ contradiciendo que sea 
+\begin_inset Formula $s$
+\end_inset
+
+ el supremo de 
+\begin_inset Formula $G$
+\end_inset
+
+.
+ Y como 
+\begin_inset Formula $s\in G\implies[a,s]\subseteq G$
+\end_inset
+
+, entonces 
+\begin_inset Formula $G=[a,b]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+En un espacio métrico 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ donde las bolas cerradas son siempre compactas (como sabemos que ocurre
+ en 
+\begin_inset Formula $(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ por el teorema anterior), todo subespacio cerrado y acotado es compacto.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $C\subseteq X$
+\end_inset
+
+ cerrado y acotado, entonces existen 
+\begin_inset Formula $x_{0}\in X$
+\end_inset
+
+ y 
+\begin_inset Formula $r>0$
+\end_inset
+
+ con 
+\begin_inset Formula $C\subseteq B_{d}(x_{0};r)\subseteq B_{d}[x_{0};r]$
+\end_inset
+
+, y como 
+\begin_inset Formula $C$
+\end_inset
+
+ es un cerrado contenido en el compacto 
+\begin_inset Formula $B_{d}[x_{0};r]$
+\end_inset
+
+, es también compacto.
+\end_layout
+
+\begin_layout Standard
+Todo subespacio compacto 
+\begin_inset Formula $K$
+\end_inset
+
+ de un espacio topológico Hausdorff 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es cerrado.
+ 
+\series bold
+Demostración:
+\series default
+ Probamos que 
+\begin_inset Formula $X\backslash K$
+\end_inset
+
+ es abierto, para lo cual vemos que todos sus puntos son interiores, es
+ decir, 
+\begin_inset Formula $\forall p\in X\backslash K,\exists A\in{\cal E}(p):A\subseteq X\backslash K$
+\end_inset
+
+.
+ Dado 
+\begin_inset Formula $p\in X\backslash K$
+\end_inset
+
+, para cada 
+\begin_inset Formula $x\in K$
+\end_inset
+
+, como 
+\begin_inset Formula $p\neq x$
+\end_inset
+
+, la condición de Hausdorff nos asegura que existen 
+\begin_inset Formula $A_{x}\in{\cal E}(p)$
+\end_inset
+
+ y 
+\begin_inset Formula $B_{x}\in{\cal E}(x)$
+\end_inset
+
+ disjuntos.
+ Ahora bien, 
+\begin_inset Formula $\{B_{x}\}_{x\in K}$
+\end_inset
+
+ es un recubrimiento de 
+\begin_inset Formula $K$
+\end_inset
+
+ por abiertos de 
+\begin_inset Formula $X$
+\end_inset
+
+ del que podemos extraer un subrecubrimiento finito 
+\begin_inset Formula $B_{x_{1}},\dots,B_{x_{r}}$
+\end_inset
+
+ para ciertos 
+\begin_inset Formula $x_{1},\dots,x_{r}\in K$
+\end_inset
+
+.
+ Sea entonces 
+\begin_inset Formula $A:=\bigcap_{i=1}^{r}A_{x_{i}}\in{\cal E}(p)$
+\end_inset
+
+, dado 
+\begin_inset Formula $a\in A$
+\end_inset
+
+, para cada 
+\begin_inset Formula $i\in\{1,\dots,r\}$
+\end_inset
+
+ se tiene que 
+\begin_inset Formula $a\in A_{x_{i}}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $a\notin B_{x_{i}}$
+\end_inset
+
+, luego 
+\begin_inset Formula $a\notin K$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $A\subseteq X\backslash K$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Todo subespacio compacto 
+\begin_inset Formula $K$
+\end_inset
+
+ de un espacio métrico 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ es acotado.
+ 
+\series bold
+Demostración:
+\series default
+ Dado 
+\begin_inset Formula $a\in X$
+\end_inset
+
+, para todo 
+\begin_inset Formula $x\in K$
+\end_inset
+
+ existe un 
+\begin_inset Formula $n_{x}\in\mathbb{N}$
+\end_inset
+
+ con 
+\begin_inset Formula $d(x,a)<n_{x}$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $\{B(a;n)\}_{n=1}^{\infty}$
+\end_inset
+
+ es un recubrimiento abierto de 
+\begin_inset Formula $K$
+\end_inset
+
+ del que podemos extraer un subrecubrimiento finito 
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula $\{B(a;n_{1}),\dots,B(a;n_{r})\}$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $K\subseteq B(a;n_{1})\cup\dots\cup B(a;n_{r})=B(a;\max\{n_{1},\dots,n_{r}\})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+De las tres últimas proposiciones se tiene que si 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ es un espacio métrico donde las bolas cerradas son siempre compactas, entonces
+ un subespacio de 
+\begin_inset Formula $(X,{\cal T}_{d})$
+\end_inset
+
+ es compacto si y sólo si es cerrado y acotado en 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Productos finitos
+\end_layout
+
+\begin_layout Standard
+Dados dos espacios topológicos 
+\begin_inset Formula $(X_{1},{\cal T}_{1})$
+\end_inset
+
+ y 
+\begin_inset Formula $(X_{2},{\cal T}_{2})$
+\end_inset
+
+, llamamos 
+\series bold
+espacio topológico producto
+\series default
+ 
+\begin_inset Formula $(X_{1},{\cal T}_{1})\times(X_{2},{\cal T}_{2})=(X_{1}\times X_{2},{\cal T}_{1}\times{\cal T}_{2})$
+\end_inset
+
+ a aquel en el que 
+\begin_inset Formula $G\in{\cal T}_{1}\times{\cal T}_{2}\iff\forall(x_{1},x_{2})\in G,\exists A_{1}\in{\cal T}_{1},A_{2}\in{\cal T}_{2}:(x_{1},x_{2})\in A_{1}\times A_{2}\subseteq G$
+\end_inset
+
+.
+ Veamos que en efecto este es un espacio topológico.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\emptyset,X_{1}\times X_{2}\in{\cal T}_{1}\times{\cal T}_{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Sean 
+\begin_inset Formula $G,G'\in{\cal T}_{1}\times{\cal T}_{2}$
+\end_inset
+
+, dado 
+\begin_inset Formula $(x_{1},x_{2})\in G\cap G'$
+\end_inset
+
+, existen 
+\begin_inset Formula $A_{1}\in{\cal T}_{1},A_{2}\in{\cal T}_{2}$
+\end_inset
+
+ con 
+\begin_inset Formula $(x_{1},x_{2})\in A_{1}\times A_{2}\subseteq G$
+\end_inset
+
+, y análogamente, existen 
+\begin_inset Formula $A'_{1}\in{\cal T}_{1},A'_{2}\in{\cal T}_{2}$
+\end_inset
+
+ con 
+\begin_inset Formula $A'_{1}\times A'_{2}\subseteq G'$
+\end_inset
+
+.
+ Por tanto 
+\begin_inset Formula $(x_{1},x_{2})\in(A_{1}\cap A'_{1})\times(A_{2}\cap A'_{2})\subseteq G\cap G'$
+\end_inset
+
+.
+ En efecto, si 
+\begin_inset Formula $(p_{1},p_{2})\in(A_{1}\cap A'_{1})\times(A_{2}\cap A'_{2})$
+\end_inset
+
+ entonces 
+\begin_inset Formula $p_{1}\in A_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $p_{2}\in A_{2}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $(p_{1},p_{2})\in A_{1}\times A_{2}\subseteq G$
+\end_inset
+
+, y análogamente 
+\begin_inset Formula $(p_{1},p_{2})\in G'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Sea 
+\begin_inset Formula $\{G_{i}\}_{i\in I}$
+\end_inset
+
+ una familia de abiertos de 
+\begin_inset Formula ${\cal T}_{1}\times{\cal T}_{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $(x_{1},x_{2})\in\bigcup_{i\in I}G_{i}$
+\end_inset
+
+, entonces existe 
+\begin_inset Formula $j\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $(x_{1},x_{2})\in G_{j}\in{\cal T}_{1}\times{\cal T}_{2}$
+\end_inset
+
+, de modo que existen 
+\begin_inset Formula $A_{j1}\in{\cal T}_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $A_{j2}\in{\cal T}_{2}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $(x_{1},x_{2})\in A_{j1}\times A_{j2}\subseteq G_{j}\subseteq\bigcup_{i\in I}G_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $G\subseteq X_{1}\times X_{2}$
+\end_inset
+
+ es abierto en 
+\begin_inset Formula $(X_{1}\times X_{2},{\cal T}_{1}\times{\cal T}_{2})$
+\end_inset
+
+ si y sólo si existen un conjunto de índices 
+\begin_inset Formula $I$
+\end_inset
+
+ y abiertos 
+\begin_inset Formula $A_{i1}\in{\cal T}_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $A_{i2}\in{\cal T}_{2}$
+\end_inset
+
+ para cada 
+\begin_inset Formula $i\in I$
+\end_inset
+
+ tales que 
+\begin_inset Formula $G=\bigcup_{i\in I}(A_{i1}\times A_{i2})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $G\in{\cal T}_{1}\times{\cal T}_{2}$
+\end_inset
+
+, para cada 
+\begin_inset Formula $x=(x_{1},x_{2})\in G$
+\end_inset
+
+ existen 
+\begin_inset Formula $A_{x_{1}}\in{\cal T}_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $A_{x_{2}}\in{\cal T}_{2}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $x=A_{x_{1}}\times A_{x_{2}}\subseteq G$
+\end_inset
+
+, luego 
+\begin_inset Formula $G=\bigcup_{x\in G}\{x\}\subseteq\bigcup_{x\in G}(A_{x_{1}}\times A_{x_{2}})\subseteq G$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $G=\bigcup_{i\in I}(A_{i1}\times A_{i2})$
+\end_inset
+
+, entonces todo punto de 
+\begin_inset Formula $G$
+\end_inset
+
+ se encuentra en algún 
+\begin_inset Formula $A_{j1}\times A_{j2}$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $G$
+\end_inset
+
+ cumple la definición de abierto de la topología producto.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+teorema de Tíjonov
+\series default
+ o 
+\series bold
+Tychonoff
+\series default
+ afirma que 
+\begin_inset Formula $(X\times Y,{\cal T}\times{\cal T}')$
+\end_inset
+
+ es compacto si y sólo si 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ e 
+\begin_inset Formula $(Y,{\cal T}')$
+\end_inset
+
+ son compactos.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $(X\times Y,{\cal T}\times{\cal T}')$
+\end_inset
+
+ es compacto, sea 
+\begin_inset Formula ${\cal A}=\{A_{i}\}_{i\in I}$
+\end_inset
+
+ un recubrimiento abierto de 
+\begin_inset Formula $X$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\{A_{i}\times Y\}_{i\in I}$
+\end_inset
+
+ es un recubrimiento abierto de 
+\begin_inset Formula $X\times Y$
+\end_inset
+
+ del que podemos extraer un subrecubrimiento finito 
+\begin_inset Formula $\{A_{1}\times Y,\dots,A_{r}\times Y\}$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $\{A_{1},\dots,A_{r}\}$
+\end_inset
+
+ es un subrecubrimiento finito de 
+\begin_inset Formula ${\cal A}$
+\end_inset
+
+ para 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sean 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ e 
+\begin_inset Formula $(Y,{\cal T}')$
+\end_inset
+
+ compactos, 
+\begin_inset Formula ${\cal W}$
+\end_inset
+
+ un recubrimiento abierto de 
+\begin_inset Formula $X\times Y$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal G}$
+\end_inset
+
+ la familia de subconjuntos 
+\begin_inset Formula $S\subseteq X$
+\end_inset
+
+ tales que 
+\begin_inset Formula $S\times Y$
+\end_inset
+
+ puede ser recubierto por una cantidad finita de abiertos de 
+\begin_inset Formula ${\cal W}$
+\end_inset
+
+, hemos de demostrar que 
+\begin_inset Formula $X\in{\cal G}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Sean 
+\begin_inset Formula $S,S'\in{\cal G}$
+\end_inset
+
+, entonces existen 
+\begin_inset Formula ${\cal X}$
+\end_inset
+
+ y 
+\begin_inset Formula ${\cal X}'$
+\end_inset
+
+ subrecubrimientos finitos de 
+\begin_inset Formula $S\times Y$
+\end_inset
+
+ y 
+\begin_inset Formula $S'\times Y$
+\end_inset
+
+, respectivamente, por lo que 
+\begin_inset Formula ${\cal X}\cup{\cal X}'$
+\end_inset
+
+ es un subrecubrimiento finito de 
+\begin_inset Formula $(S\cup S')\times Y$
+\end_inset
+
+ y 
+\begin_inset Formula $S\cup S'\in{\cal G}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Dado 
+\begin_inset Formula $x\in X$
+\end_inset
+
+, para cada 
+\begin_inset Formula $y\in Y$
+\end_inset
+
+, como 
+\begin_inset Formula ${\cal W}$
+\end_inset
+
+ es un recubrimiento de 
+\begin_inset Formula $X\times Y$
+\end_inset
+
+, existe 
+\begin_inset Formula $W_{y}\in{\cal W}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $(x,y)\in W_{y}$
+\end_inset
+
+, de modo que podemos encontrar 
+\begin_inset Formula $A_{y}\in{\cal T}$
+\end_inset
+
+ y 
+\begin_inset Formula $B_{y}\in{\cal T}'$
+\end_inset
+
+ tales que 
+\begin_inset Formula $(x,y)\in A_{y}\times B_{y}\subseteq W_{y}$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $\{B_{y}\}_{y\in Y}$
+\end_inset
+
+ es un recubrimiento abierto de 
+\begin_inset Formula $Y$
+\end_inset
+
+ del que podemos obtener un subrecubrimiento finito 
+\begin_inset Formula $\{B_{y_{1}},\dots,B_{y_{r}}\}$
+\end_inset
+
+.
+ Sea entonces 
+\begin_inset Formula $A_{x}=A_{y_{1}}\cap\dots\cap A_{y_{r}}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $A_{x}\times Y=A_{x}\times(\bigcup_{i=1}^{r}B_{y_{i}})=\bigcup_{i=1}^{r}(A_{x}\times B_{y_{i}})\subseteq\bigcup_{i=1}^{r}(A_{y_{i}}\times B_{y_{i}})\subseteq\bigcup_{i=1}^{r}W_{y_{i}}$
+\end_inset
+
+, de modo que 
+\begin_inset Formula $A_{x}\in{\cal G}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Por lo segundo, tenemos un recubrimiento abierto de 
+\begin_inset Formula $X$
+\end_inset
+
+ de la forma 
+\begin_inset Formula $\{A_{x}\}_{x\in X}$
+\end_inset
+
+ donde cada 
+\begin_inset Formula $A_{x}\in{\cal G}$
+\end_inset
+
+, por lo que podemos encontrar un subrecubrimiento finito 
+\begin_inset Formula $X=A_{x_{1}}\cup\dots\cup A_{x_{k}}$
+\end_inset
+
+, y por lo primero esto implica 
+\begin_inset Formula $X\in{\cal G}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+De esto, junto con el apartado anterior, se obtiene la versión general del
+ teorema de Heine-Borel, que afirma que un subespacio de 
+\begin_inset Formula $(\mathbb{R}^{n},{\cal T}_{u})$
+\end_inset
+
+ es compacto si y sólo si es cerrado y acotado para alguna de las métricas
+ 
+\begin_inset Formula $d_{T}$
+\end_inset
+
+, 
+\begin_inset Formula $d_{E}$
+\end_inset
+
+ y 
+\begin_inset Formula $d_{\infty}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Compacidad y continuidad
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua y 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es compacto entonces 
+\begin_inset Formula $f(X)$
+\end_inset
+
+ es compacto.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula ${\cal A}=\{A_{i}\}_{i\in I}$
+\end_inset
+
+ un recubrimiento de 
+\begin_inset Formula $f(X)$
+\end_inset
+
+ por abiertos de 
+\begin_inset Formula $(Y,{\cal T}')$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\{f^{-1}(A_{i})\}_{i\in I}$
+\end_inset
+
+ es un recubrimiento abierto de 
+\begin_inset Formula $X$
+\end_inset
+
+, que admite pues un subrecubrimiento finito 
+\begin_inset Formula $\{f^{-1}(A_{1}),\dots,f^{-1}(A_{r})\}$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $\{A_{1},\dots,A_{r}\}$
+\end_inset
+
+ es un subrecubrimiento finito de 
+\begin_inset Formula ${\cal A}$
+\end_inset
+
+ para 
+\begin_inset Formula $f(X)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Esto significa que la compacidad es una propiedad topológica, es decir,
+ si 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ e 
+\begin_inset Formula $(Y,{\cal T}')$
+\end_inset
+
+ son homeomorfos y 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es compacto, 
+\begin_inset Formula $(Y,{\cal T}')$
+\end_inset
+
+ también lo es.
+ También significa que, si 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es compacto, toda función continua 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,d')$
+\end_inset
+
+ es cerrada y acotada.
+ En particular toda 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ continua alcanza su máximo y su mínimo en 
+\begin_inset Formula $X$
+\end_inset
+
+, y si 
+\begin_inset Formula $(X,{\cal T})=([a,b],{\cal T}_{u})$
+\end_inset
+
+ entonces 
+\begin_inset Formula $f([a,b])$
+\end_inset
+
+ es un intervalo cerrado y acotado.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+teorema de la continuidad de la función inversa
+\series default
+ afirma que toda 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ biyectiva y continua, siendo 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ compacto e 
+\begin_inset Formula $(Y,{\cal T}')$
+\end_inset
+
+ Hausdorff, es un homeomorfismo.
+ 
+\series bold
+Demostración:
+\series default
+ Basta probar que 
+\begin_inset Formula $g:=f^{-1}$
+\end_inset
+
+ es continua.
+ Así, 
+\begin_inset Formula $f$
+\end_inset
+
+ lleva compactos de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ a compactos de 
+\begin_inset Formula $(Y,{\cal T}')$
+\end_inset
+
+, pero dado 
+\begin_inset Formula $C\in{\cal C_{T}}$
+\end_inset
+
+, 
+\begin_inset Formula $C$
+\end_inset
+
+ es compacto, 
+\begin_inset Formula $f(C)$
+\end_inset
+
+ también y por ser 
+\begin_inset Formula $(Y,{\cal T}')$
+\end_inset
+
+ Hausdorff, 
+\begin_inset Formula $f(C)$
+\end_inset
+
+ es cerrado.
+ Hemos probado que dado 
+\begin_inset Formula $C\subseteq X$
+\end_inset
+
+ cerrado, 
+\begin_inset Formula $g^{-1}(C)=f(C)$
+\end_inset
+
+ es cerrado, luego 
+\begin_inset Formula $g$
+\end_inset
+
+ es continua.
+\end_layout
+
+\begin_layout Standard
+Por tanto toda aplicación 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ inyectiva y continua, siendo 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ compacto e 
+\begin_inset Formula $(Y,{\cal T}')$
+\end_inset
+
+ Hausdorff, es un homeomorfismo entre 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ y 
+\begin_inset Formula $(f(X),{\cal T}'_{f(X)})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Toda 
+\begin_inset Formula $f:(X,d)\rightarrow(Y,d')$
+\end_inset
+
+ continua, siendo 
+\begin_inset Formula $(X,{\cal T}_{d})$
+\end_inset
+
+ compacto, es uniformemente con
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+ti
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+-
+\end_layout
+
+\end_inset
+
+nua.
+ 
+\series bold
+Demostración:
+\series default
+ Dado 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+, para 
+\begin_inset Formula $p\in X$
+\end_inset
+
+, existe 
+\begin_inset Formula $\delta_{p}>0$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\forall y\in X,(d(p,y)<\delta_{p}\implies d'(f(p),f(y))<\frac{\varepsilon}{2})$
+\end_inset
+
+.
+ Sea ahora 
+\begin_inset Formula $\delta'_{p}:=\frac{\delta_{p}}{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $\{B(p;\delta'_{p})\}_{p\in X}$
+\end_inset
+
+ un recubrimiento abierto de 
+\begin_inset Formula $X$
+\end_inset
+
+, podemos extraer un subrecubrimiento finito 
+\begin_inset Formula $\{B(p_{1};\delta'_{p_{1}}),\dots,B(p_{r};\delta'_{p_{r}})\}$
+\end_inset
+
+, y llamamos 
+\begin_inset Formula $\delta:=\min\{\delta'_{p_{1}},\dots,\delta'_{p_{r}}\}$
+\end_inset
+
+.
+ Sean 
+\begin_inset Formula $x,y\in X$
+\end_inset
+
+ con 
+\begin_inset Formula $d(x,y)<\delta$
+\end_inset
+
+, entonces existe 
+\begin_inset Formula $i\in\{1,\dots r\}$
+\end_inset
+
+ con 
+\begin_inset Formula $d(x,p_{i})<\delta_{p_{i}}$
+\end_inset
+
+, luego 
+\begin_inset Formula $d(y,p_{i})\leq d(y,x)+d(x,p_{i})<\delta+\delta'_{p_{i}}\leq2\delta'_{p_{i}}=\delta_{p_{i}}$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $d'(f(x),f(p_{i}))<\frac{\varepsilon}{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $d'(f(y),f(p_{i}))<\frac{\varepsilon}{2}$
+\end_inset
+
+, y por tanto 
+\begin_inset Formula $d'(f(y),f(x))\leq d'(f(y),f(p_{i}))+d'(f(p_{i}),f(x))<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Compacidad por sucesiones
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es 
+\series bold
+compacto por sucesiones 
+\series default
+si toda sucesión admite una subsucesión convergente.
+ Ahora probaremos que todo espacio métrico compacto es compacto por sucesiones,
+ y viceversa.
+\end_layout
+
+\begin_layout Standard
+Primero probamos que si 
+\begin_inset Formula $\{x_{n}\}_{n=1}^{\infty}$
+\end_inset
+
+ es una sucesión en 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ y 
+\begin_inset Formula $p$
+\end_inset
+
+ es un punto de acumulación de ella, 
+\begin_inset Formula $\{x_{n}\}_{n=1}^{\infty}$
+\end_inset
+
+ posee una subsucesión convergente a 
+\begin_inset Formula $p$
+\end_inset
+
+.
+ En efecto, sea 
+\begin_inset Formula $S=\{x_{n}\}_{n\in\mathbb{N}}$
+\end_inset
+
+ el conjunto de puntos, para todo 
+\begin_inset Formula $r>0$
+\end_inset
+
+ debe ser 
+\begin_inset Formula $(B(p;r)\backslash\{p\})\cap S$
+\end_inset
+
+ infinito, pues si fuera finito 
+\begin_inset Formula $\{x_{n_{1}},\dots,x_{n_{r}}\}$
+\end_inset
+
+ podríamos escoger 
+\begin_inset Formula $r'>0$
+\end_inset
+
+ con 
+\begin_inset Formula $r'<d(p,x_{n_{i}})\forall i$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $(B(p;r')\backslash\{p\})\cap S=\emptyset$
+\end_inset
+
+, lo que contradice que 
+\begin_inset Formula $p$
+\end_inset
+
+ sea punto de acumulación.
+ Ahora bien, si para 
+\begin_inset Formula $k=1$
+\end_inset
+
+ tomamos 
+\begin_inset Formula $r=1$
+\end_inset
+
+ existirá 
+\begin_inset Formula $x_{n_{1}}\in B(p;1)$
+\end_inset
+
+, y si tenemos 
+\begin_inset Formula $x_{n_{k}}\in B(p;\frac{1}{k})$
+\end_inset
+
+ con 
+\begin_inset Formula $n_{k}>n_{k-1}$
+\end_inset
+
+ entonces como 
+\begin_inset Formula $B(p;\frac{1}{k+1})\cap S$
+\end_inset
+
+ es infinito, podemos tomar 
+\begin_inset Formula $x_{n_{k+1}}\in B(p;\frac{1}{k+1})$
+\end_inset
+
+ con 
+\begin_inset Formula $n_{k+1}>n_{k}$
+\end_inset
+
+, formando una subsucesión 
+\begin_inset Formula $\{x_{n_{k}}\}_{k}$
+\end_inset
+
+ que converge a 
+\begin_inset Formula $p$
+\end_inset
+
+.
+ Esto también vale para cualquier espacio topológico 1AN y Hausdorff.
+\end_layout
+
+\begin_layout Standard
+Ahora vemos que todo subconjunto infinito 
+\begin_inset Formula $S$
+\end_inset
+
+ de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ compacto tiene al menos un punto de acumulación.
+ Supongamos que no los tiene, es decir, 
+\begin_inset Formula $\forall p\in X,\exists U_{p}\in{\cal E}(p):(U_{p}\backslash\{p\})\cap S=\emptyset$
+\end_inset
+
+.
+ Entonces podríamos considerar el recubrimiento abierto 
+\begin_inset Formula $\{U_{p}\}_{p\in X}$
+\end_inset
+
+ de 
+\begin_inset Formula $X$
+\end_inset
+
+, del que podemos extraer un subrecubrimiento finito 
+\begin_inset Formula $\{U_{p_{1}},\dots,U_{p_{r}}\}$
+\end_inset
+
+, pero 
+\begin_inset Formula $S=S\cap X=S\cap(U_{p_{1}}\cup\dots\cup U_{p_{r}})=(S\cap U_{p_{1}})\cup\dots\cup(S\cap U_{p_{r}})\subseteq\{p_{1},\dots,p_{r}\}$
+\end_inset
+
+.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Con esto podemos probar que todo espacio métrico compacto es compacto por
+ sucesiones.
+ Supongamos que 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ es compacto y sea 
+\begin_inset Formula $\{x_{n}\}_{n=1}^{\infty}$
+\end_inset
+
+ una sucesión en 
+\begin_inset Formula $X$
+\end_inset
+
+.
+ Ahora sea 
+\begin_inset Formula $S=\{x_{n}\}_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $S$
+\end_inset
+
+ es finito, debe existir 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ que se repite infinitas veces en la sucesión, y estos términos forman una
+ subsucesión constante y por tanto convergente.
+ Si es infinito, posee un punto de acumulación y por tanto tiene una subsucesión
+ convergente.
+\end_layout
+
+\begin_layout Standard
+Observamos que toda sucesión acotada en 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+ con 
+\begin_inset Formula $d_{T}$
+\end_inset
+
+, 
+\begin_inset Formula $d_{E}$
+\end_inset
+
+ o 
+\begin_inset Formula $d_{\infty}$
+\end_inset
+
+ posee una subsucesión convergente.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ es 
+\series bold
+precompacto
+\series default
+ o 
+\series bold
+totalmente acotado
+\series default
+ si para cada 
+\begin_inset Formula $r>0$
+\end_inset
+
+ existe una cantidad finita de puntos 
+\begin_inset Formula $\{x_{1},\dots,x_{m}\}$
+\end_inset
+
+ de 
+\begin_inset Formula $X$
+\end_inset
+
+ tales que 
+\begin_inset Formula $X=B(x_{1};r)\cup\dots\cup B(x_{m};r)$
+\end_inset
+
+.
+ Esta definición es casi igual a la de compacto, pero no se considera un
+ recubrimiento abierto cualquiera sino solo los de la forma 
+\begin_inset Formula $\{B(p;r)\}_{p\in X}$
+\end_inset
+
+.
+ Así, todo espacio métrico compacto es precompacto, y todo espacio precompacto
+ es acotado.
+\end_layout
+
+\begin_layout Standard
+Todo espacio métrico compacto por sucesiones es precompacto.
+ Sea 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ un espacio métrico compacto por sucesiones tal que 
+\begin_inset Formula $\exists r>0:\forall S\subseteq X,X\neq\bigcup_{x\in S}B(x;r)$
+\end_inset
+
+, y construiremos una sucesión 
+\begin_inset Formula $\{x_{n}\}_{n=1}^{\infty}$
+\end_inset
+
+ en 
+\begin_inset Formula $X$
+\end_inset
+
+ de la siguiente forma.
+ Sea 
+\begin_inset Formula $x_{1}\in X$
+\end_inset
+
+ cualquiera y supongamos que hemos construido 
+\begin_inset Formula $x_{1},\dots,x_{m}$
+\end_inset
+
+ de modo que 
+\begin_inset Formula $d(x_{i},x_{j})>r\forall i,j\leq m,i\neq j$
+\end_inset
+
+, y como por la hipótesis 
+\begin_inset Formula $X\neq\bigcup_{i=1}^{m}B(x_{i};r)$
+\end_inset
+
+, existe 
+\begin_inset Formula $x_{m+1}\in X\backslash\bigcup_{i=1}^{m}B(x_{i};r)$
+\end_inset
+
+ y tenemos por inducción una sucesión tal que 
+\begin_inset Formula $d(x_{i},x_{j})>r\forall i\neq j$
+\end_inset
+
+.
+ Ahora bien, por la compacidad por sucesiones ha de existir una subsucesión
+ 
+\begin_inset Formula $\{x_{n_{k}}\}_{k=1}^{\infty}$
+\end_inset
+
+ convergente a un 
+\begin_inset Formula $p\in X$
+\end_inset
+
+, pero entonces existe 
+\begin_inset Formula $k_{0}\in\mathbb{N}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $d(p,x_{n_{k}})<\frac{r}{2}$
+\end_inset
+
+ para 
+\begin_inset Formula $k\geq k_{0}$
+\end_inset
+
+ y entonces 
+\begin_inset Formula $d(x_{n_{k}},x_{n_{k+1}})\leq r$
+\end_inset
+
+, lo cual es absurdo.
+\end_layout
+
+\begin_layout Standard
+Todo espacio métrico precompacto es separable.
+ Si 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ es precompacto, para 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ existen 
+\begin_inset Formula $\{x_{1n},\dots,x_{r_{n}n}\}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $X=\bigcup_{i=1}^{r_{n}}B(x_{in};\frac{1}{n})$
+\end_inset
+
+.
+ El conjunto 
+\begin_inset Formula $D=\{x_{in}\}_{n\in\mathbb{N},1\leq i\leq r_{n}}$
+\end_inset
+
+ es numerable por ser unión numerable de conjuntos finitos.
+ Probaremos que es denso viendo que, dado 
+\begin_inset Formula $p\in X$
+\end_inset
+
+, se tiene 
+\begin_inset Formula $p\in\overline{D}$
+\end_inset
+
+.
+ Para todo 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ existe 
+\begin_inset Formula $x_{in}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $p\in B(x_{in};\frac{1}{n})$
+\end_inset
+
+, pero entonces 
+\begin_inset Formula $x_{in}\in B(p;\frac{1}{n})$
+\end_inset
+
+ y 
+\begin_inset Formula $B(p;\frac{1}{n})$
+\end_inset
+
+ corta a 
+\begin_inset Formula $D$
+\end_inset
+
+, luego 
+\begin_inset Formula $D$
+\end_inset
+
+ corta a todos los entornos de la base 
+\begin_inset Formula $\{B(p;\frac{1}{n})\}_{n\in\mathbb{N}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dado un recubrimiento abierto 
+\begin_inset Formula ${\cal A}$
+\end_inset
+
+ de 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+, 
+\begin_inset Formula $r>0$
+\end_inset
+
+ es un 
+\series bold
+número de Lebesgue
+\series default
+ de 
+\begin_inset Formula ${\cal A}$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall p\in X,\exists A_{p}\in{\cal A}:B(p;r)\subseteq A_{p}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+lema de Lebesgue
+\series default
+ afirma que si 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ es compacto por sucesiones entonces todo recubrimiento abierto admite un
+ número de Lebesgue.
+ Sea 
+\begin_inset Formula ${\cal A}=\{A_{i}\}_{i\in I}$
+\end_inset
+
+ un recubrimiento abierto de 
+\begin_inset Formula $X$
+\end_inset
+
+ que no admite un número de Lebesgue.
+ Entonces 
+\begin_inset Formula $\forall r>0,\exists p\in X:\forall i\in I,B(p;r)\nsubseteq A_{i}$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $\{x_{n}\}_{n\in\mathbb{N}}\subseteq X$
+\end_inset
+
+ tal que 
+\begin_inset Formula $B(x_{n};\frac{1}{n})\nsubseteq A_{i}\forall i\in I$
+\end_inset
+
+, como 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ es compacto por sucesiones, existirá 
+\begin_inset Formula $\{x_{n_{k}}\}_{k\in\mathbb{N}}$
+\end_inset
+
+ convergente a un 
+\begin_inset Formula $p\in X$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $i_{0}\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $p\in A_{i_{0}}\in{\cal T}_{d}$
+\end_inset
+
+, existe 
+\begin_inset Formula $r_{0}>0$
+\end_inset
+
+ con 
+\begin_inset Formula $B(p;r_{0})\subseteq A_{i_{0}}$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $N\in\mathbb{N}$
+\end_inset
+
+ con 
+\begin_inset Formula $d(p,x_{N})<\frac{r_{0}}{2}$
+\end_inset
+
+ y 
+\begin_inset Formula $\frac{1}{N}<\frac{r_{0}}{2}$
+\end_inset
+
+.
+ Ahora, tomando 
+\begin_inset Formula $t\in B(x_{N};\frac{1}{N})$
+\end_inset
+
+ vemos que 
+\begin_inset Formula $d(p,y)\leq d(p,x_{N})+d(x_{N},y)<r_{0}$
+\end_inset
+
+, luego 
+\begin_inset Formula $y\in B(p;r_{0})\subseteq A_{i_{0}}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $B(x_{N};\frac{1}{N})\subseteq B(p;r_{0})$
+\end_inset
+
+, lo cual es absurdo.
+\end_layout
+
+\begin_layout Standard
+De aquí que todo espacio métrico compacto por sucesiones es compacto.
+ Sean 
+\begin_inset Formula $(X,d)$
+\end_inset
+
+ un espacio métrico compacto por sucesiones y 
+\begin_inset Formula ${\cal A}=\{A_{i}\}_{i\in I}$
+\end_inset
+
+ un recubrimiento abierto de 
+\begin_inset Formula $X$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ un número de Lebesgue para 
+\begin_inset Formula ${\cal A}$
+\end_inset
+
+.
+ Entonces existe un recubrimiento finito de 
+\begin_inset Formula $X$
+\end_inset
+
+ mediante bolas 
+\begin_inset Formula $\{B(x_{1};\varepsilon),\dots,B(x_{r};\varepsilon)\}$
+\end_inset
+
+ de radio 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+.
+ Pero como cada bola 
+\begin_inset Formula $B(x_{i};\varepsilon)$
+\end_inset
+
+ ha de estar contenida en un abierto 
+\begin_inset Formula $A_{i}$
+\end_inset
+
+ de 
+\begin_inset Formula ${\cal A}$
+\end_inset
+
+, tendremos que 
+\begin_inset Formula $\{A_{1},\dots,A_{r}\}$
+\end_inset
+
+ es un subrecubrimiento finito de 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/tem/n5.lyx b/tem/n5.lyx
new file mode 100644
index 0000000..93b40a3
--- /dev/null
+++ b/tem/n5.lyx
@@ -0,0 +1,1516 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+Una 
+\series bold
+separación por abiertos
+\series default
+ o 
+\series bold
+partición por abiertos
+\series default
+ de un espacio topológico 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es un par 
+\begin_inset Formula $\{A,B\}$
+\end_inset
+
+ de subconjuntos abiertos no vacíos con 
+\begin_inset Formula $A\dot{\cup}B=X$
+\end_inset
+
+.
+ 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es 
+\series bold
+conexo
+\series default
+ si no admite ninguna separación por abiertos, y de lo contrario es 
+\series bold
+disconexo
+\series default
+\SpecialChar endofsentence
+ Equivalentemente, 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es conexo si y sólo si no existe ningún par de cerrados 
+\begin_inset Formula $\{C,D\}$
+\end_inset
+
+ no vacíos con 
+\begin_inset Formula $C\dot{\cup}D=X$
+\end_inset
+
+, si y sólo si los únicos subconjuntos de 
+\begin_inset Formula $X$
+\end_inset
+
+ abiertos y cerrados al mismo tiempo son el total y el vacío.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es conexo si y sólo si toda aplicación continua 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(\{0,1\},{\cal T}_{D})$
+\end_inset
+
+ es constante.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ conexo y supongamos que existe 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(\{0,1\},{\cal T}_{D})$
+\end_inset
+
+ continua no constante.
+ Entonces existen 
+\begin_inset Formula $p,q\in X$
+\end_inset
+
+ con 
+\begin_inset Formula $f(p)=0$
+\end_inset
+
+ y 
+\begin_inset Formula $f(q)=1$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $\{0\}$
+\end_inset
+
+ y 
+\begin_inset Formula $\{1\}$
+\end_inset
+
+ son abiertos, 
+\begin_inset Formula $A=f^{-1}(\{0\})$
+\end_inset
+
+ y 
+\begin_inset Formula $B=f^{-1}(\{1\})$
+\end_inset
+
+ forman una separación por abiertos de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ disconexo y 
+\begin_inset Formula $A$
+\end_inset
+
+ y 
+\begin_inset Formula $B$
+\end_inset
+
+ abiertos no vacíos de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ con 
+\begin_inset Formula $A\dot{\cup}B=X$
+\end_inset
+
+.
+ Si definimos 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(\{0,1\},{\cal T}_{D})$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f(p)=0$
+\end_inset
+
+ si 
+\begin_inset Formula $p\in A$
+\end_inset
+
+ y 
+\begin_inset Formula $f(p)=1$
+\end_inset
+
+ si 
+\begin_inset Formula $p\in B$
+\end_inset
+
+, entonces 
+\begin_inset Formula $f$
+\end_inset
+
+ es continua porque la imagen inversa de todo abierto es abierto, pero no
+ es constante.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es conexo si y sólo si toda aplicación continua cumple que 
+\begin_inset Formula $\forall x,y\in X,c\in(f(x),f(y));\exists z\in X:f(z)=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Supongamos que existe 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ tal que existen 
+\begin_inset Formula $x,y\in X$
+\end_inset
+
+ y 
+\begin_inset Formula $c\in\mathbb{R}$
+\end_inset
+
+ con 
+\begin_inset Formula $f(x)<c<f(y)$
+\end_inset
+
+ pero 
+\begin_inset Formula $c\notin f(X)$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $f^{-1}(-\infty,c)$
+\end_inset
+
+ y 
+\begin_inset Formula $f^{-1}(c,+\infty)$
+\end_inset
+
+ forman una separación por abiertos de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ (ningún 
+\begin_inset Formula $x\in X$
+\end_inset
+
+ va a parar a 
+\begin_inset Formula $(-\infty,c)$
+\end_inset
+
+ y a 
+\begin_inset Formula $(c,+\infty)$
+\end_inset
+
+ a la vez).
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ disconexo, entonces existe 
+\begin_inset Formula $g:(X,{\cal T})\rightarrow(\{0,1\},{\cal T}_{D})$
+\end_inset
+
+ no constante.
+ Si componemos esto con la inclusión 
+\begin_inset Formula $\{0,1\}\looparrowright\mathbb{R}$
+\end_inset
+
+, obtenemos 
+\begin_inset Formula $x,y\in X$
+\end_inset
+
+ con 
+\begin_inset Formula $f(x)=0$
+\end_inset
+
+ y 
+\begin_inset Formula $f(y)=1$
+\end_inset
+
+, y si tomamos 
+\begin_inset Formula $c=\frac{1}{2}$
+\end_inset
+
+ entre 
+\begin_inset Formula $f(x)$
+\end_inset
+
+ y 
+\begin_inset Formula $f(y)$
+\end_inset
+
+ entonces 
+\begin_inset Formula $c\notin f(X)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es conexo y 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua entonces 
+\begin_inset Formula $(f(X),{\cal T}'|_{f(X)})$
+\end_inset
+
+ es conexo.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $B$
+\end_inset
+
+ abierto y cerrado en 
+\begin_inset Formula $(f(X),{\cal T}'|_{f(X)})$
+\end_inset
+
+, como 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua, 
+\begin_inset Formula $f^{-1}(B)$
+\end_inset
+
+ es abierto y cerrado en 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+, por lo que es el total o el vacío y por tanto 
+\begin_inset Formula $B=f(X)$
+\end_inset
+
+ o 
+\begin_inset Formula $B=\emptyset$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+De aquí que si 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ e 
+\begin_inset Formula $(Y,{\cal T}')$
+\end_inset
+
+ son homeomorfos y uno es conexo, el otro también, y si 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(Y,{\cal T}')$
+\end_inset
+
+ es continua y 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es conexo, la gráfica 
+\begin_inset Formula $\{(x,f(x))\}_{x\in X}$
+\end_inset
+
+ es conexa, pues es homeomorfa a 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Subespacios conexos de 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+El 
+\series bold
+teorema de Bolzano
+\series default
+ afirma que si 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ es continua y 
+\begin_inset Formula $f(a)$
+\end_inset
+
+ y 
+\begin_inset Formula $f(b)$
+\end_inset
+
+ son de signos opuestos, existe 
+\begin_inset Formula $x\in[a,b]$
+\end_inset
+
+ con 
+\begin_inset Formula $f(x)=0$
+\end_inset
+
+.
+ El 
+\series bold
+teorema de los valores intermedios
+\series default
+ o 
+\series bold
+primer teorema de Weierstrass
+\series default
+ afirma que si 
+\begin_inset Formula $f:[a,b]\rightarrow\mathbb{R}$
+\end_inset
+
+ es continua y 
+\begin_inset Formula $c\in\mathbb{R}$
+\end_inset
+
+ cumple 
+\begin_inset Formula $f(a)\leq c\leq f(b)$
+\end_inset
+
+ entonces existe 
+\begin_inset Formula $x\in[a,b]$
+\end_inset
+
+ con 
+\begin_inset Formula $f(x)=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $S\subseteq\mathbb{R}$
+\end_inset
+
+ no vacío es un 
+\series bold
+intervalo
+\series default
+ si y sólo si 
+\begin_inset Formula $\forall x,y\in S,z\in\mathbb{R};(x<z<y\implies z\in S)$
+\end_inset
+
+.
+ Un subespacio de 
+\begin_inset Formula $(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ es conexo si y sólo si es un intervalo.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $S$
+\end_inset
+
+ no es un intervalo, existen 
+\begin_inset Formula $x,y\in S$
+\end_inset
+
+ y 
+\begin_inset Formula $z\in\mathbb{R}\backslash S$
+\end_inset
+
+ con 
+\begin_inset Formula $x<z<y$
+\end_inset
+
+, luego 
+\begin_inset Formula $S\cap(-\infty,z)$
+\end_inset
+
+ y 
+\begin_inset Formula $S\cap(z,+\infty)$
+\end_inset
+
+ es una separación por abiertos no vacíos de 
+\begin_inset Formula $(S,{\cal T}_{u}|_{S})$
+\end_inset
+
+ y por tanto es disconexo.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sea 
+\begin_inset Formula $S$
+\end_inset
+
+ un intervalo y supongamos que no es conexo.
+ Entonces existe 
+\begin_inset Formula $f:(S,{\cal T}_{u}|_{S})\rightarrow(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ tal que existen 
+\begin_inset Formula $x,y\in S$
+\end_inset
+
+ con 
+\begin_inset Formula $x<y$
+\end_inset
+
+ y 
+\begin_inset Formula $f(x)\neq f(y)$
+\end_inset
+
+ y existe 
+\begin_inset Formula $c\in(f(x),f(y))$
+\end_inset
+
+ con 
+\begin_inset Formula $c\notin f(S)$
+\end_inset
+
+.
+ Pero entonces existe 
+\begin_inset Formula $f:([x,y],{\cal T}_{u}|_{[x,y]})\rightarrow(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ que no cumple el teorema de los valores intermedios.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+De aquí que si 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es conexo y 
+\begin_inset Formula $f:(X,{\cal T})\rightarrow(\mathbb{R},{\cal T}_{u})$
+\end_inset
+
+ es continua entonces 
+\begin_inset Formula $f(X)$
+\end_inset
+
+ es un intervalo.
+\end_layout
+
+\begin_layout Section
+Propiedades
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $H$
+\end_inset
+
+ es un subespacio conexo de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ entonces todo 
+\begin_inset Formula $S$
+\end_inset
+
+ con 
+\begin_inset Formula $H\subseteq S\subseteq\overline{H}$
+\end_inset
+
+ es conexo.
+ 
+\series bold
+Demostración:
+\series default
+ Supongamos que existen abiertos no vacíos 
+\begin_inset Formula $A'$
+\end_inset
+
+ y 
+\begin_inset Formula $B'$
+\end_inset
+
+ en 
+\begin_inset Formula $(S,{\cal T}_{S})$
+\end_inset
+
+ con 
+\begin_inset Formula $A'\dot{\cup}B'=S$
+\end_inset
+
+.
+ Entonces existen 
+\begin_inset Formula $A,B\in{\cal T}$
+\end_inset
+
+ no vacíos con 
+\begin_inset Formula $A'=A\cap S$
+\end_inset
+
+ y 
+\begin_inset Formula $B'=B\cap S$
+\end_inset
+
+, luego 
+\begin_inset Formula $(A\cap H)\cap(B\cap H)=(A'\cap H)\cap(B'\cap H)=A'\cap B'\cap H=\emptyset$
+\end_inset
+
+ y 
+\begin_inset Formula $(A\cap H)\cup(B\cap H)=(A'\cap H)\cup(B'\cap H)=(A'\cup B')\cap H=S\cap H=H$
+\end_inset
+
+, y como 
+\begin_inset Formula $(H,{\cal T}|_{H})$
+\end_inset
+
+ es conexo, debe ser 
+\begin_inset Formula $A\cap H=\emptyset$
+\end_inset
+
+ o 
+\begin_inset Formula $B\cap H=\emptyset$
+\end_inset
+
+.
+ Si por ejemplo 
+\begin_inset Formula $A\cap H=\emptyset$
+\end_inset
+
+, como 
+\begin_inset Formula $A'\neq\emptyset$
+\end_inset
+
+, existe 
+\begin_inset Formula $p\in A'=A\cap S\subseteq A\cap\overline{H}$
+\end_inset
+
+, pero 
+\begin_inset Formula $A$
+\end_inset
+
+ es un entorno de 
+\begin_inset Formula $p$
+\end_inset
+
+ que no corta a 
+\begin_inset Formula $H$
+\end_inset
+
+ por lo que 
+\begin_inset Formula $p$
+\end_inset
+
+ no puede estar en 
+\begin_inset Formula $\overline{H}$
+\end_inset
+
+.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+De aquí que la clausura de un subespacio conexo es conexa.
+ Veamos ahora el 
+\series bold
+criterio del peine
+\series default
+, que afirma que dado un espacio topológico 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ y 
+\begin_inset Formula $\{H_{i}\}_{i\in I}$
+\end_inset
+
+ una familia de subespacios conexos para la que existe 
+\begin_inset Formula $i_{0}\in I$
+\end_inset
+
+ con 
+\begin_inset Formula $H_{i}\cap H_{i_{0}}\neq\emptyset$
+\end_inset
+
+ para todo 
+\begin_inset Formula $i\in I$
+\end_inset
+
+, entonces 
+\begin_inset Formula $H:=\bigcup_{i\in I}H_{i}$
+\end_inset
+
+ es conexo.
+ 
+\series bold
+Demostración:
+\series default
+ Primero vemos que dado 
+\begin_inset Formula $C\subseteq X$
+\end_inset
+
+ conexo y 
+\begin_inset Formula $\{A,B\}$
+\end_inset
+
+ una separación de 
+\begin_inset Formula $X$
+\end_inset
+
+ por abiertos entonces 
+\begin_inset Formula $C\subseteq A$
+\end_inset
+
+ o 
+\begin_inset Formula $C\subseteq B$
+\end_inset
+
+.
+ En efecto, si no fuera así se tendría 
+\begin_inset Formula $C\cap A\neq\emptyset$
+\end_inset
+
+ y 
+\begin_inset Formula $C\cap B\neq\emptyset$
+\end_inset
+
+, y 
+\begin_inset Formula $C=(C\cap A)\cup(C\cap B)$
+\end_inset
+
+ siendo 
+\begin_inset Formula $C\cap A$
+\end_inset
+
+ y 
+\begin_inset Formula $C\cap B$
+\end_inset
+
+ abiertos en 
+\begin_inset Formula $(C,{\cal T}_{C})$
+\end_inset
+
+ no vacíos con 
+\begin_inset Formula $(C\cap A)\cap(C\cap B)=C\cap(A\cap B)=\emptyset$
+\end_inset
+
+, contradiciendo que 
+\begin_inset Formula $C$
+\end_inset
+
+ sea conexo.
+ Ahora bien, si 
+\begin_inset Formula $\{A,B\}$
+\end_inset
+
+ es una separación por abiertos de 
+\begin_inset Formula $(H,{\cal T}_{H})$
+\end_inset
+
+, dado 
+\begin_inset Formula $i\in I$
+\end_inset
+
+, 
+\begin_inset Formula $H_{i}\subseteq B$
+\end_inset
+
+ o 
+\begin_inset Formula $H_{i}\subseteq A$
+\end_inset
+
+.
+ Supongamos 
+\begin_inset Formula $H_{i_{0}}\subseteq A$
+\end_inset
+
+.
+ Como para cualquier 
+\begin_inset Formula $i$
+\end_inset
+
+ se tiene 
+\begin_inset Formula $H_{i}\cap H_{i_{0}}\neq\emptyset$
+\end_inset
+
+ y 
+\begin_inset Formula $A\cap B=\emptyset$
+\end_inset
+
+ no puede ser 
+\begin_inset Formula $H_{i}\subseteq B$
+\end_inset
+
+, luego cada 
+\begin_inset Formula $H_{i}\subseteq A$
+\end_inset
+
+ y 
+\begin_inset Formula $H=\bigcup_{i\in I}H_{i}\subseteq A$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $B=\emptyset$
+\end_inset
+
+ y 
+\begin_inset Formula $H$
+\end_inset
+
+ es conexo.
+\end_layout
+
+\begin_layout Standard
+En particular, si 
+\begin_inset Formula $\{H_{i}\}_{i\in I}$
+\end_inset
+
+ es una familia de subespacios conexos de 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ con 
+\begin_inset Formula $\bigcap_{i\in I}H_{i}\neq\emptyset$
+\end_inset
+
+ entonces 
+\begin_inset Formula $H:=\bigcup_{i\in I}H_{i}$
+\end_inset
+
+ es conexo, y si 
+\begin_inset Formula $H_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $H_{2}$
+\end_inset
+
+ son conexos con 
+\begin_inset Formula $H_{1}\cap H_{2}\neq\emptyset$
+\end_inset
+
+ entonces 
+\begin_inset Formula $H_{1}\cup H_{2}$
+\end_inset
+
+ es conexo.
+\end_layout
+
+\begin_layout Standard
+El espacio producto 
+\begin_inset Formula $(X_{1}\times X_{2},{\cal T}_{1}\times{\cal T}_{2})$
+\end_inset
+
+ es conexo si y sólo si 
+\begin_inset Formula $(X_{1},{\cal T}_{1})$
+\end_inset
+
+ y 
+\begin_inset Formula $(X_{2},{\cal T}_{2})$
+\end_inset
+
+ son conexos.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Se deriva de que las proyecciones son continuas.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sean 
+\begin_inset Formula $X_{1},X_{2}\neq\emptyset$
+\end_inset
+
+ (de lo contrario 
+\begin_inset Formula $X_{1}\times X_{2}=\emptyset$
+\end_inset
+
+ y la propiedad es cierta), dado 
+\begin_inset Formula $p_{2}\in X_{2}$
+\end_inset
+
+, 
+\begin_inset Formula $X_{1}\times\{p_{2}\}$
+\end_inset
+
+ es homeomorfo a 
+\begin_inset Formula $(X_{1},{\cal T}_{1})$
+\end_inset
+
+ y por tanto conexo, y lo mismo ocurre con 
+\begin_inset Formula $\{p_{1}\}\times X_{2}$
+\end_inset
+
+ dado 
+\begin_inset Formula $p_{1}\in X_{1}$
+\end_inset
+
+.
+ La unión de espacios conexos 
+\begin_inset Formula $\bigcup_{p_{1}\in X_{1}}\{p_{1}\}\times X_{2}\cup\bigcup_{p_{2}\in X_{2}}X_{1}\times\{p_{2}\}$
+\end_inset
+
+ da 
+\begin_inset Formula $(X_{1}\times X_{2},{\cal T}_{1}\times{\cal T}_{2})$
+\end_inset
+
+, y basta aplicar el criterio del peine.
+\end_layout
+
+\begin_layout Section
+Conexión por arcos
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $p,q\in X$
+\end_inset
+
+, un 
+\series bold
+arco
+\series default
+ de 
+\begin_inset Formula $p$
+\end_inset
+
+ a 
+\begin_inset Formula $q$
+\end_inset
+
+ en 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es una aplicación continua 
+\begin_inset Formula $\sigma:([0,1],{\cal T}_{u})\rightarrow(X,{\cal T})$
+\end_inset
+
+ con 
+\begin_inset Formula $\sigma(0)=p$
+\end_inset
+
+ y 
+\begin_inset Formula $\sigma(1)=q$
+\end_inset
+
+.
+ Un espacio topológico 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es 
+\series bold
+conexo por arcos
+\series default
+ o 
+\series bold
+por caminos
+\series default
+ si cualquier par de puntos pueden ser conectados por un arco.
+\end_layout
+
+\begin_layout Standard
+Un subconjunto 
+\begin_inset Formula $S\subseteq\mathbb{R}^{n}$
+\end_inset
+
+ es 
+\series bold
+convexo
+\series default
+ si para cualesquiera 
+\begin_inset Formula $x,y\in S$
+\end_inset
+
+, el 
+\series bold
+segmento
+\series default
+ 
+\begin_inset Formula $L_{xy}:=\{(1-t)x+ty\}_{t\in[0,1]}$
+\end_inset
+
+ es un subconjunto de 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ Todo subconjunto convexo de 
+\begin_inset Formula $(\mathbb{R}^{n},{\cal T}_{u})$
+\end_inset
+
+ es conexo por arcos.
+ Por tanto las bolas en 
+\begin_inset Formula $d_{T}$
+\end_inset
+
+, 
+\begin_inset Formula $d_{E}$
+\end_inset
+
+ y 
+\begin_inset Formula $d_{\infty}$
+\end_inset
+
+, tanto abiertas como cerradas, y los rectángulos, son conexos por arcos.
+\end_layout
+
+\begin_layout Standard
+Todo espacio topológico conexo por arcos es conexo.
+ 
+\series bold
+Demostración:
+\series default
+ Supongamos que existe una separación 
+\begin_inset Formula $\{A,B\}$
+\end_inset
+
+ por abiertos no vacíos del espacio 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ conexo por arcos.
+ Entonces podemos tomar 
+\begin_inset Formula $a\in A$
+\end_inset
+
+ y 
+\begin_inset Formula $b\in B$
+\end_inset
+
+ y 
+\begin_inset Formula $\sigma$
+\end_inset
+
+ un arco de 
+\begin_inset Formula $a$
+\end_inset
+
+ hasta 
+\begin_inset Formula $b$
+\end_inset
+
+.
+ Pero como 
+\begin_inset Formula $\sigma([0,1])$
+\end_inset
+
+ es conexo, debe estar contenido en 
+\begin_inset Formula $A$
+\end_inset
+
+ o en 
+\begin_inset Formula $B$
+\end_inset
+
+.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+El recíproco no se cumple, pues 
+\begin_inset Formula $([0,1]\times\{0\})\cup(\{\frac{1}{n}\}_{n\in\mathbb{N}}\times[0,1])\cup\{(0,1)\}$
+\end_inset
+
+ en 
+\begin_inset Formula $(\mathbb{R}^{2},{\cal T}_{u})$
+\end_inset
+
+ es conexo pero no conexo por arcos.
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $\sigma_{1},\sigma_{2}:[0,1]\rightarrow X$
+\end_inset
+
+ dos arcos en 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ que unen, respectivamente, 
+\begin_inset Formula $x$
+\end_inset
+
+ con 
+\begin_inset Formula $y$
+\end_inset
+
+ e 
+\begin_inset Formula $y$
+\end_inset
+
+ con 
+\begin_inset Formula $z$
+\end_inset
+
+, llamamos 
+\series bold
+unión
+\series default
+, 
+\series bold
+producto
+\series default
+ o 
+\series bold
+composición de arcos
+\series default
+, escrito 
+\begin_inset Formula $\sigma_{1}\ast\sigma_{2}$
+\end_inset
+
+, a la aplicación 
+\begin_inset Formula $\tau:[0,1]\rightarrow X$
+\end_inset
+
+ dada por
+\begin_inset Formula 
+\[
+\tau(t)=\begin{cases}
+\sigma_{1}(2t) & \text{si }t\in[0,\frac{1}{2}]\\
+\sigma_{2}(2t-1) & \text{si }t\in[\frac{1}{2},1]
+\end{cases}
+\]
+
+\end_inset
+
+que es un arco que une 
+\begin_inset Formula $x$
+\end_inset
+
+ con 
+\begin_inset Formula $z$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Decimos que un subconjunto 
+\begin_inset Formula $S\subseteq\mathbb{R}^{n}$
+\end_inset
+
+ es 
+\series bold
+estrellado
+\series default
+ en 
+\begin_inset Formula $p\in S$
+\end_inset
+
+ si 
+\begin_inset Formula $\forall x\in S,L_{px}\subseteq S$
+\end_inset
+
+.
+ Un espacio topológico 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+ es conexo por arcos si y sólo si existe 
+\begin_inset Formula $p\in X$
+\end_inset
+
+ tal que cualquier 
+\begin_inset Formula $q\in X$
+\end_inset
+
+ se pueda unir con 
+\begin_inset Formula $p$
+\end_inset
+
+ por un arco en 
+\begin_inset Formula $(X,{\cal T})$
+\end_inset
+
+, y en particular los subconjuntos estrellados son conexos por arcos.
+\end_layout
+
+\begin_layout Standard
+Todo subconjunto abierto y conexo de 
+\begin_inset Formula $(\mathbb{R}^{n},{\cal T}_{u})$
+\end_inset
+
+ es conexo por arcos.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $U$
+\end_inset
+
+ un abierto conexo de 
+\begin_inset Formula $(\mathbb{R}^{n},{\cal T}_{u})$
+\end_inset
+
+, 
+\begin_inset Formula $p\in U$
+\end_inset
+
+ y 
+\begin_inset Formula $A$
+\end_inset
+
+ el subconjunto de los puntos de 
+\begin_inset Formula $U$
+\end_inset
+
+ que se pueden unir con 
+\begin_inset Formula $p$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $y\in A$
+\end_inset
+
+, como 
+\begin_inset Formula $U$
+\end_inset
+
+ es abierto, existe 
+\begin_inset Formula $r>0$
+\end_inset
+
+ con 
+\begin_inset Formula $B(y;r)\subseteq U$
+\end_inset
+
+ y si 
+\begin_inset Formula $z\in B(y;r)$
+\end_inset
+
+, la unión del arco que une 
+\begin_inset Formula $p$
+\end_inset
+
+ con 
+\begin_inset Formula $y$
+\end_inset
+
+ y el radio que une 
+\begin_inset Formula $y$
+\end_inset
+
+ con 
+\begin_inset Formula $z$
+\end_inset
+
+ es un arco que une 
+\begin_inset Formula $p$
+\end_inset
+
+ con 
+\begin_inset Formula $z$
+\end_inset
+
+, luego 
+\begin_inset Formula $B(y;r)\subseteq A$
+\end_inset
+
+ y, como 
+\begin_inset Formula $y$
+\end_inset
+
+ es arbitrario, 
+\begin_inset Formula $A$
+\end_inset
+
+ es abierto.
+ Ahora bien, sea 
+\begin_inset Formula $y\in U\backslash A$
+\end_inset
+
+, existe 
+\begin_inset Formula $r>0$
+\end_inset
+
+ con 
+\begin_inset Formula $B(y;r)\subseteq U$
+\end_inset
+
+.
+ Pero si existiera 
+\begin_inset Formula $z\in B(y;r)$
+\end_inset
+
+ con 
+\begin_inset Formula $z\in A$
+\end_inset
+
+, la unión del arco que une 
+\begin_inset Formula $p$
+\end_inset
+
+ con 
+\begin_inset Formula $z$
+\end_inset
+
+ y el radio que une 
+\begin_inset Formula $z$
+\end_inset
+
+ con 
+\begin_inset Formula $y$
+\end_inset
+
+ es un arco que une 
+\begin_inset Formula $p$
+\end_inset
+
+ con 
+\begin_inset Formula $y$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $y\in A\#$
+\end_inset
+
+, luego 
+\begin_inset Formula $B(y;r)\subseteq U\backslash A$
+\end_inset
+
+, y como 
+\begin_inset Formula $y$
+\end_inset
+
+ es arbitrario, 
+\begin_inset Formula $U\backslash A$
+\end_inset
+
+ es abierto y 
+\begin_inset Formula $A$
+\end_inset
+
+ es cerrado.
+ Como 
+\begin_inset Formula $A$
+\end_inset
+
+ es abierto y cerrado en un espacio conexo y 
+\begin_inset Formula $A\neq\emptyset$
+\end_inset
+
+ porque 
+\begin_inset Formula $p\in A$
+\end_inset
+
+, entonces 
+\begin_inset Formula $A=U$
+\end_inset
+
+ y 
+\begin_inset Formula $U$
+\end_inset
+
+ es conexo por arcos.
+\end_layout
+
+\end_body
+\end_document
-- 
cgit v1.2.3