Changed layout for more manageable volumes

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+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\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_roman_osf false
+\font_sans_osf false
+\font_typewriter_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
+\float_placement class
+\float_alignment class
+\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_formatted_ref 0
+\use_minted 0
+\use_lineno 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 english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO 1,
+ 3,
+ 6 (2pp.,
+ 0:46)
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc1[M10]
+\end_layout
+
+\end_inset
+
+The text refers to a set 
+\begin_inset Formula $S$
+\end_inset
+
+ containing finite sequences of positive integers,
+ and states that this set is 
+\begin_inset Quotes eld
+\end_inset
+
+essentially an oriented tree.
+\begin_inset Quotes erd
+\end_inset
+
+ What is the root of this oriented tree,
+ and what are the arcs?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The vertices are the elements of 
+\begin_inset Formula $S$
+\end_inset
+
+,
+ the root is 
+\begin_inset Formula $\emptyset$
+\end_inset
+
+,
+ and the arcs go from 
+\begin_inset Formula $(x_{1},\dots,x_{n})$
+\end_inset
+
+ to 
+\begin_inset Formula $(x_{1},\dots,x_{n-1})$
+\end_inset
+
+,
+ for every 
+\begin_inset Formula $(x_{1},\dots,x_{n})\in S\setminus\{\emptyset\}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc3[M23]
+\end_layout
+
+\end_inset
+
+If it is possible to tile the upper right quadrant of the plane when given an 
+\emph on
+infinite
+\emph default
+ set of tetrad types,
+ is it always possible to tile the whole plane?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+No.
+ For example,
+ our tiles might be of the form
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{center}
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{tikzpicture}
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+draw (0,0) -- (0,2) -- (2,2) -- (2,0) -- (0,0) -- (2,2)
+\end_layout
+
+\begin_layout Plain Layout
+
+      (0,2) -- (2,0)
+\end_layout
+
+\begin_layout Plain Layout
+
+      (0.5,1) node{$n$} (1,0.5) node{$n$}
+\end_layout
+
+\begin_layout Plain Layout
+
+      (1.5,1) node{$n+1$} (1,1.5) node{$n+1$};
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{tikzpicture}
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{center}
+\end_layout
+
+\end_inset
+
+for 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+.
+ Then in the 
+\begin_inset Formula $(x,y)$
+\end_inset
+
+ square we might place the tile with 
+\begin_inset Formula $n=\max\{x,y\}$
+\end_inset
+
+ and this would tile the upper right quadrant,
+ but there's obviously no way to tile the whole plane.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc6[M23]
+\end_layout
+
+\end_inset
+
+(Otto Schreier.) In a famous paper,
+ B.
+ L.
+ van der Waerden proved the following theorem:
+\end_layout
+
+\begin_layout Quote
+
+\shape slanted
+If 
+\begin_inset Formula $k$
+\end_inset
+
+ and 
+\begin_inset Formula $m$
+\end_inset
+
+ are positive integers,
+ and if we have 
+\begin_inset Formula $k$
+\end_inset
+
+ sets 
+\begin_inset Formula $S_{1},\dots,S_{k}$
+\end_inset
+
+ of positive integers with every positive integer included in at least one of these sets,
+ then at least of the sets 
+\begin_inset Formula $S_{j}$
+\end_inset
+
+ contains an arithmetic progression of length 
+\begin_inset Formula $m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+(The latter statement means there exist integers 
+\begin_inset Formula $a$
+\end_inset
+
+ and 
+\begin_inset Formula $\delta>0$
+\end_inset
+
+ such that 
+\begin_inset Formula $a+\delta$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a+2\delta$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $a+m\delta$
+\end_inset
+
+ are all in 
+\begin_inset Formula $S_{j}$
+\end_inset
+
+.) If possible,
+ use this result and the infinity lemma to prove the following stronger statement:
+\end_layout
+
+\begin_layout Quote
+
+\shape slanted
+If 
+\begin_inset Formula $k$
+\end_inset
+
+ and 
+\begin_inset Formula $m$
+\end_inset
+
+ are positive integers,
+ there is a number 
+\begin_inset Formula $N$
+\end_inset
+
+ such that if we have 
+\begin_inset Formula $k$
+\end_inset
+
+ sets 
+\begin_inset Formula $S_{1},\dots,S_{k}$
+\end_inset
+
+ of integers with every integer between 1 and 
+\begin_inset Formula $N$
+\end_inset
+
+ included in at least one of these sets,
+ then at least one of the sets 
+\begin_inset Formula $S_{j}$
+\end_inset
+
+ contains an arithmetic progression of length 
+\begin_inset Formula $m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+It is enough to prove it when the set that contains every integer between 1 and 
+\begin_inset Formula $N$
+\end_inset
+
+ is 
+\begin_inset Formula $S_{1}$
+\end_inset
+
+.
+ We define a tree whose vertices are the tuples 
+\begin_inset Formula $(S_{1},\dots,S_{k})$
+\end_inset
+
+ of finite sets of integers such that,
+ if 
+\begin_inset Formula $n=\max\bigcup_{i=1}^{k}S_{i}$
+\end_inset
+
+,
+ every positive integer up to 
+\begin_inset Formula $n$
+\end_inset
+
+ is in one of the sets,
+ and such that no set contains an arithmetic progression of length 
+\begin_inset Formula $m$
+\end_inset
+
+.
+ The edges go from one such tuple to 
+\begin_inset Formula $(S_{1}\setminus\{n\},\dots,S_{k}\setminus\{n\})$
+\end_inset
+
+,
+ so the root is 
+\begin_inset Formula $(\emptyset,\dots,\emptyset)$
+\end_inset
+
+ and 
+\begin_inset Formula $n$
+\end_inset
+
+ is the height of the node in the tree.
+\end_layout
+
+\begin_layout Standard
+Since each vertex contains a finite number of children (
+\begin_inset Formula $2^{k}-1$
+\end_inset
+
+,
+ corresponding to the ways of adding the next number to one or more of the sets),
+ it follows that if this tree were infinite,
+ there would be an infinite path.
+ By van der Waerden theorem,
+ the component-wise union of this path would give us a tuple 
+\begin_inset Formula $(S_{1},\dots,S_{k})$
+\end_inset
+
+ such that some 
+\begin_inset Formula $S_{j}$
+\end_inset
+
+ contains an arithmetic progression of length 
+\begin_inset Formula $m$
+\end_inset
+
+,
+ so by construction one of the nodes in the path would follow this condition.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Thus the tree is finite and we just need to take 
+\begin_inset Formula $N$
+\end_inset
+
+ as one plus the depth of the tree.
+\end_layout
+
+\end_body
+\end_document