Housekeeping.

src/dyn/top/hales-jewett.tex

@@ -0,0 +1,114 @@
+\section{The Hales-Jewett Theorem}
+\label{section:hales-jewett}
+
+
+\begin{definition}[Variable Words]
+\label{definition:variable-words}
+    Let $A$ be a finite alphabet, $S = A^+$ be the free semigroup of all non-empty words in $A$, $T = (A \sqcup \bracsn{*})^+$ be the free semigroup of all non-empty words in $A \sqcup \bracsn{*}$, then the set of \textbf{variable words} in $A$ is the two-sided ideal $V := T \setminus S$ of all words in $A \sqcup \bracsn{*}$ containing $*$. 
+
+    For each $a \in A$, the semigroup homomorphism
+    \[
+    \pi_a: T \to S \quad b \mapsto \begin{cases}
+        b &b \in A \\
+        a &b = *
+    \end{cases}
+    \]
+
+    is the \textbf{substitution map} from $*$ to $a$, with $\pi_a|_S = \text{Id}_S$. 
+\end{definition}
+
+\begin{definition}[Combinatorial Line]
+\label{definition:combinatorial-line}
+    Let $A$ be a finite alphabet, $S$ be the free semigroup of non-empty words in $A$, $V$ be the set of variable words in $A$, and $v \in V$, then
+    \[
+    L(v) = \bracsn{\pi_a(v)|a \in A} \subset S
+    \]
+
+    is the \textbf{combinatorial line}\footnote{Line in the sense that this is a straight line in the hypercube $A^{|v|}$. } generated by $v$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:hales-jewett-minimal-idempotent}
+    Let $A$ be a finite alphabet, $S$ be the free semigroup of non-empty words in $A$, $T = (A \sqcup \bracsn{*})^*$ be the free semigroup of all non-empty words in $A \sqcup \bracsn{*}$, and $V$ be the set of variable words in $A$, then:
+    \begin{enumerate}
+        \item The inclusions $S \to T$ and $V \to T$ admit extensions into embeddings $\beta S \to T$ and $\beta V \to T$, which allows identifying $\beta T = \beta S \bigsqcup \beta V$. Moreover, $\beta S$ may be identified as a subgroup of $T$. In particular, every idempotent in $\beta S$ is idempotent in $\beta T$. 
+        \item For any minimal idempotent $p \in \beta S \setminus S$, $p$ is not minimal in $\beta T$. 
+        \item For any minimal idempotent $p \in \beta S \setminus S$, there exists a minimal idempotent $q \in \beta S \setminus \beta T$ such that $pq = qp = q$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): By (U) of the \hyperref[Stone-Čech compactification]{definition:stone-cech}.
+
+    (2): Suppose for contradiction that $p$ is a minimal idempotent in $\beta T$, then $L = (\beta T)p$ is a minimal left ideal in $\beta T$. Since $(\beta T)*p \subset \beta V \cap (\beta T)p$, $(\beta T)p \cap \beta V =  (\beta T)p$ by minimality, and $(\beta T)p \subset \beta V$. As such, $(\beta S)p \subset \beta S \cap (\beta T)p \cap \beta V = \emptyset$, which is impossible. 
+
+    (3): By \autoref{theorem:ellis-semigroup}, $S \setminus \beta S$ admits a minimal idempotent $p$, which is idempotent in $\beta T$. By the \hyperref[idempotent refinement lemma]{lemma:idempotent-refinement}, there exists a minimial idempotent $q \in \beta T$ such that $pq = qp = q$. 
+
+    Suppose again for contradiction that $q \in \beta S$, then $pq = q \in (\beta S)p$, which is minimal in $\beta S$. As such, $(\beta S)q \subset (\beta S)p$, and $(\beta S)q = (\beta S)p \subset \beta S$, so $q$ is a minimal idempotent in $\beta S$. By (2), $q$ is not minimal in $\beta T$, which is impossible. 
+\end{proof}
+
+\begin{theorem}[Hales-Jewett]
+\label{theorem:hales-jewett}
+    Let $A$ be a finite alphabet, $S$ be the free semigroup of non-empty words in $A$, $V$ be the set of variable words in $A$, then:
+    \begin{enumerate}
+        \item[(IF)] Every finite colouring of $S$ contains a monochromatic combinatorial line. 
+        \item[(F)] For every $r \ge 1$, then there exists $N = N(r) \ge 1$ such that every $r$-colouring of $A^N$ contains a monochromatic combinatorial line. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (IF): Let $T = (A \sqcup \bracsn{*})^+$ be the free semigroup of all non-empty words in $A \sqcup \bracsn{*}$.
+    
+    For each $a \in A$, let $\pi_a: T \to S$ be the corresponding substitution map, and $\beta \pi_a: \beta T \to \beta S$ be its continuous extension. Let $p \in \beta S \setminus S$ be a minimal idempotent. By \autoref{lemma:hales-jewett-minimal-idempotent}, there exists a minimal idempotent $q \in \beta T$ be a minimal idempotent with $pq = qp = q$ and $q \in \beta T \setminus \beta S = \beta V$.
+
+    As $\pi_a|_S = \text{Id}_S$, $\beta \pi_a|_{\beta S} = \text{Id}_{\beta S}$, so $\beta \pi_a(p) = p$. Since $q$ is idempotent and $\beta \pi_a$ is a semigroup homomorphism, $\beta \pi_a(q) \in \beta S$ is also idempotent. Further, 
+    \[
+    \beta \pi_a(q)p = \beta \pi_a(q)\beta \pi_a(p) = \beta\pi_a(qp) = \beta \pi_a(q)
+    \]
+
+    and
+    \[
+    p\beta \pi_a(q) = \beta \pi_a(p)\beta \pi_a(q) = \beta\pi_a(qp) = \beta \pi_a(q)
+    \]
+
+    so $\beta \pi_a(q)p = p\beta \pi_a(q) = \beta \pi_a(q)$. In particular, $\beta \pi_a(q) = \beta \pi_a(q)p \in (\beta S)p$, where $(\beta S)p$ is a minimal left ideal in $\beta S$. By minimality, $(\beta S)\beta \pi_a(q) = (\beta S)p$, and there exists $u \in \beta S$ with $p = u\beta \pi_a(q)$. As such, 
+    \[
+    \beta \pi_a(q) = p\beta \pi_a(q) = u\beta \pi_a(q)^2 = u\beta \pi_a(q) = p
+    \]
+
+    Let $S = \bigsqcup_{j = 1}^r C_r$ be a finite colouring of $S$, then there exists a unique $1 \le i \le r$ such that $C_i \in p$. Since $\beta \pi_a(q) = p$ for all $a \in A$, $\bigcap_{a \in A}\pi_a^{-1}(C_i) \in q$. As $q \in \beta T \setminus \beta S = \beta V$, $V \in q$, and $V \cap \bigcap_{a \in A}\pi_a^{-1}(C_i) \ne \emptyset$. Therefore any element of $V \cap \bigcap_{a \in A}\pi_a^{-1}(C_i)$ generates a combinatorial line, contained in $C_i$. 
+
+    (F): Let $r \ge 1$ and suppose for contradiction that for every $N \in \natp$, there exists an $r$-colouring $c_N: A^N \to \bracsn{1, \cdots, r}$ such that $c_N$ contains no monochromatic combinatorial line. Define
+    \[
+    c: S \to \bracsn{1, \cdots, r} \quad w \mapsto c_{|w|}(w)
+    \]
+
+    then $c$ is a $r$-colouring of $S$ containing no monochromatic combinatorial lines, which contradicts (IF).
+\end{proof}
+
+
+\begin{theorem}[van der Waerden]
+\label{theorem:van-der-waerden}
+    For every $k \in \natp$ and finite colouring $c: \integer \to \bracsn{1, \cdots, r}$, there exists $a \in \integer$ and $n \ge 1$ such that $\bracsn{a + jn|0 \le j \le k}$ is monochromatic. In other words, there exists monochromatic arithmetic progressions of arbitrary length. 
+\end{theorem}
+\begin{proof}
+    Let $A = \bracsn{1, \cdots, k}$ and
+    \[
+    \tilde c: A^N \to \bracsn{1, \cdots, r} \quad x_1 \cdots x_N \mapsto c(x_1 + \cdots + x_N)
+    \]
+
+    By the \hyperref[Hales-Jewett Theorem]{theorem:hales-jewett}, there exists $N = N(r)$ such that $\tilde c$ contains a monochromatic combinatorial line. In other words, there exists a variable word $v \in (A \sqcup \bracsn{*})^N \setminus A^N$ such that the set
+    \[
+    L(v) = \bracsn{\pi_j(v)|1 \le j \le k}
+    \]
+
+    is monochromatic. Let $J = \bracsn{1 \le j \le N|v_j = *}$, then as $v \not\in A^N$, $J$ is non-empty. Take $n = |J|$ and $a = \sum_{j \not\in J}v_j$, then for each $1 \le i \le k$, 
+    \[
+    \tilde c(\pi_i(v)) = c\paren{i\sum_{j \in J}j + \sum_{j \not\in J}v_j} = c\paren{in +a}
+    \]
+
+    In particular, for each $1 \le i, i' \le k$, 
+    \[
+    c\paren{in +a} = \tilde c(\pi_i(v)) = \tilde c(\pi_{i'}(v)) = c(i'n +a)
+    \]
+
+    Therefore the arithmetic progression $\bracsn{a + in|1 \le i \le k}$ is monochromatic.
+\end{proof}

src/dyn/top/index.tex

@@ -5,3 +5,4 @@
 \input{./minimal.tex}
 \input{./recurrence.tex}
 \input{./stone-cech.tex}
+\input{./hales-jewett.tex}

src/dyn/top/stone-cech.tex

@@ -128,21 +128,114 @@
 
 \begin{definition}[Universal Minimal Flow]
 \label{definition:universal-minimal-flow}
-    Let $G$ be a group, then there exists a flow $(M, \alpha)$ such that:
+    Let $G$ be a group, then there exists a flow $L$ such that:
     \begin{enumerate}
-        \item $(M, \alpha)$ is a minimal $G$-flow. 
+        \item $G \curvearrowright L$ is a minimal $G$-flow. 
-        \item[(U)] For any minimal $G$-flow $(X, \beta)$, $x_0 \in X$, and $m_0 \in M$, there exists a unique continuous map $\Phi \in C(M; X)$ such that $\Phi(m_0) = x_0$ and the following diagram commutes
+        \item[(U)] For any minimal $G$-flow $X$, there exists a continuous map $\Phi \in C(M; X)$ such that $\Phi(m_0) = x_0$ and the following diagram commutes
         \[
         \xymatrix{
-        M \ar@{->}[r]^{\Phi} \ar@{->}[d]_{\alpha(g)} & X \ar@{->}[d]^{\beta(g)} \\
+        L \ar@{->}[r]^{\Phi} \ar@{->}[d]_{g} & X \ar@{->}[d]^{g} \\
-        M \ar@{->}[r]_{\Phi} & X
+        L \ar@{->}[r]_{\Phi} & X
         }
         \]
 
         for all $g \in G$.
     \end{enumerate}
+
+    The flow $(M, \alpha)$ is the \textbf{universal minimal $G$-flow}.
+\end{definition}
+\begin{proof}
+    (1): Using \autoref{corollary:stone-cech-minimal-ideal-existence}, let $L \subset \beta G$ be a minimal left ideal, then $G \curvearrowright L$ is a minimal $G$-flow. 
+
+    (U): Let $G \curvearrowright X$ be a minimal $G$-flow, $x_0 \in X$, and $m_0 \in M$. By (U) of the \hyperref[universal flow]{definition:stone-cech-flow}, there exists a unique continuous map $\phi \in C(\beta G; X)$ such that $\phi(1_G) = x_0$ and the following diagram commutes
+    \[
+    \xymatrix{
+    \beta G \ar@{->}[r]^{\Phi} \ar@{->}[d]_{g} & X \ar@{->}[d]^{g} \\
+    \beta G \ar@{->}[r]_{\Phi} & X
+    }
+    \]
+
+    for all $g \in G$. 
+
+    Let $\Phi = \phi|_{L} \in C(L; X)$, then as $L$ is $G$-invariant, $\Phi(L) \supset G\phi(x_0)$. By \autoref{proposition:compact-extensions}, $\Phi(L) = \ol{G\phi(x_0)} = X$.  
+\end{proof}
+
+
+\begin{definition}[Minimal Idempotent]
+\label{definition:tds-minimal-idempotent}
+    Let $S$ be a compact Hausdorff right topological semigroup and $e \in S$ be an idempotent, then $e$ is \textbf{minimal} if there exists a minimal left ideal $I \subset G$ such that $e \in I$. 
 \end{definition}
 
+\begin{theorem}[Ellis]
+\label{theorem:ellis-semigroup}
+    Let $S$ be a non-empty compact Hausdorff right topological semigroup and $I \subset S$ be a closed left ideal, then there exists a minimal left ideal $I \subset S$ and a minimal idempotent $e \in I$. 
+\end{theorem}
+\begin{proof}
+    By Zorn's Lemma, $I$ admits a minimal non-empty closed right topological subsemigroup $K \subset S$. For each $a \in K$, $Ka$ is closed by continuity of right multiplication and \autoref{proposition:compact-extensions}. By minimality of $K$, $K = Ka$, so there exists $e \in K$ such that $a = ea$. Let $E = \bracsn{b \in K|ba = a}$, then $E$ is also a closed and right topological subsemigroup. By minimality of $K$, $E = K$, and in particular $a \in E$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:stone-cech-minimal-idempotent}
+    Let $G$ be an infinite group, then $\beta G$ admits a minimal idempotent in $\beta G \setminus G$. 
+\end{corollary}
+\begin{proof}
+    Let $L \subset (\beta G \setminus G)$ be a minimal left ideal, then $L$ is a non-empty compact Hausdorff right topological semigroup. By \hyperref[Ellis' Theorem]{theorem:ellis-semigroup}, there exists $e \in L$ with $e^2 = e$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:stone-cech-flow-minimal-idempotent}
+    Let $G$ be an infinite group, $G \curvearrowright X$ be a minimal $G$-flow, then there exists a minimal idempotent $\fU \in \beta G \setminus G$ such that $\fU x = x$. 
+\end{proposition}
+\begin{proof}
+    Let $L \subset \beta G$ be a minimal left ideal, then the mapping $\fU \mapsto \fU x$ is a continuous surjection. As $x \in X = \bracsn{\fU x|x \in L}$, there exists $\fU \in L$ with $\fU x = x$. Let $F = \bracsn{\fU \in L|\fU x = x}$, then $F$ is a non-empty closed right topological subsemigroup of $L$. By \hyperref[Ellis' Theorem]{theorem:ellis-semigroup}, $F$ contains an idempotent. 
+\end{proof}
+
+
+\begin{theorem}[Infinite Ramsey Theorem for Pairs]
+\label{theorem:ramsey-pairs}
+    Let
+    \[
+    [\natp]^2 = \bracsn{\bracsn{m, n} \in \nat^2|m, n \in \natp, m \ne n}
+    \]
+    
+    be the set of unordered pairs of distinct elements of $\natp$ and $c: [\natp]^2 \to \bracsn{1, \cdots, r}$, then there exists $A \subset \natp$ infinite such that for any $m, n, m', n' \in A$, $c(\bracsn{m, n}) = c(\bracsn{m', n'})$. 
+\end{theorem}
+\begin{proof}
+    Let $\fU \subset 2^{\natp}$ be a free ultrafilter. For each $n \in \natp$ and $1 \le i \le r$, let $A(n, i) = \bracsn{m > n| c(\bracsn{m,n}) = i}$, then $\natp = \bracsn{1, \cdots, n} \bigsqcup_{i = 1}^r A(n, i)$. As $\fU$ is a free ultrafilter, there exists a unique $1 \le i(n) \le r$ such that $A(n, i(n)) \in \fU$. 
+
+    For each $1 \le i \le r$, let $B_i = \bracsn{n \in \natp|i(n) = i}$, then $\natp = \bigsqcup_{i = 1}^r B_i$. As $\fU$ is an ultrafilter, there exists a unique $1 \le i \le r$ with $B_i \in \fU$. 
+    
+    Let $n_1 \in B_i$, so $i(n_1) = i$. Assume inductively that $\bracsn{n_j}_1^N \subset \natp$ has been defined such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $1 \le j \le N - 1$, $n_j < n_{j+1}$. 
+        \item For each $1 \le j \le N$, $i(n_j) = i$. 
+        \item For each $1 \le j < k \le N$, $c(\bracsn{n_j, n_k}) = i$. 
+    \end{enumerate}
+
+    Let $n_{N+1} \in B_i \cap \bigcap_{j = 1}^N A(n_j, i)$, then:
+    \begin{enumerate}
+        \item As $n_{N+1} \in A(n_N, i)$, $n_{N+1} > n_N$. 
+        \item Since $n_{N+1} \in B_i$, $i(n_{N+1}) = i$. 
+        \item For each $1 \le j \le N$, $n_{N+1} \in A(n_j, i) = A(n_j, i(n))$, so $c(\bracsn{n_j, n_{N+1}}) = i$. 
+    \end{enumerate}
+
+    By induction, there exists $\seq{n_j} \subset \natp$ such that $c(\bracsn{n_j, n_k}) = i$ for all $j, k \in \natp$ with $j \ne k$. 
+\end{proof}
+
+
+\begin{lemma}[Idempotent Refinement]
+\label{lemma:idempotent-refinement}
+    Let $S$ be a semigroup and $f \in \beta S$ be idempotent, then there exists a minimal idempotent $e \in \beta S$ such that $ef = fe = e$. 
+\end{lemma}
+\begin{proof}
+    By \autoref{proposition:compact-extensions}, $(\beta S)f$ is a closed left ideal in $S$. By \hyperref[Ellis' Theorem]{theorem:ellis-semigroup}, $(\beta S)f$ contains a minimal left ideal $L$, and a minimal idempotent $r \in L$. As $L \subset (\beta S)f$, there exists $t \in \beta S$ such that $r = tf$. 
+
+    Let $e = fr \in L$, then $e^2 = frfr = fr^2 = fr = e$. Moreover, 
+    \[
+    ef = frf = ftf^2 = ftf = fr = e = fr = ffr = fe
+    \]
+\end{proof}
+