From 3ba29569efe0a6cec30c5a9946b84333b599429a Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Thu, 6 Aug 2026 15:59:15 -0400 Subject: [PATCH] Added universality of the zero-dimensional spaces. --- src/fa/lc/convex.tex | 2 +- src/measure/sets/borel.tex | 26 ++++++++ src/topology/dst/analytic.tex | 34 ++++++++++ src/topology/dst/index.tex | 2 + src/topology/dst/zero.tex | 113 ++++++++++++++++++++++++++++++++++ src/topology/index.tex | 1 + src/topology/notation.tex | 3 + 7 files changed, 180 insertions(+), 1 deletion(-) create mode 100644 src/topology/dst/analytic.tex create mode 100644 src/topology/dst/zero.tex diff --git a/src/fa/lc/convex.tex b/src/fa/lc/convex.tex index d4327ab..bd33b25 100644 --- a/src/fa/lc/convex.tex +++ b/src/fa/lc/convex.tex @@ -135,7 +135,7 @@ \label{definition:seminorm-topology} Let $E$ be a vector space over $K \in \RC$ and $\seqi{[\cdot]}$ be seminorms, then: \begin{enumerate} - \item For each $i \in I$, $d_i: E \times E \to [0, \infty)$ defined by $(x, y) \mapsto [x - y]_i$ is a pseudo-metric. + \item For each $i \in I$, $d_i: E \times E \to [0, \infty)$ defined by $(x, y) \mapsto [x - y]_i$ is a pseudometric. \item The topology induced by $\seqi{d}$ makes $E$ a topological vector space. \item For each $i \in I$, $[\cdot]_i: E \to [0, \infty)$ is continuous. \end{enumerate} diff --git a/src/measure/sets/borel.tex b/src/measure/sets/borel.tex index 8383811..aa1cff7 100644 --- a/src/measure/sets/borel.tex +++ b/src/measure/sets/borel.tex @@ -8,6 +8,32 @@ \end{definition} + +\begin{lemma} +\label{lemma:monotone-borel-characterisation} + Let $X$ be a topological space, then $\cb_X$ is the smallest subset of $2^X$ that: + \begin{enumerate} + \item contains open and closed subsets of $X$. + \item is closed under countable intersections. + \item is closed under countable disjoint unions. + \end{enumerate} +\end{lemma} +\begin{proof}[Proof, {{\cite[Lemma 8.2.4]{CohnMeasure}}}. ] + Let $\cf \subset 2^X$ be the smallest subset of $2^X$ satisfying the lemma. Since $\cb_X$ satisfies the lemma, $\cf \subset \cb_X$. On the other hand, let + \[ + \cf_0 = \bracs{A \subset X| A \in \cf, A^c \in \cf} + \] + + then by definition, $\cf_0$ is closed under complements. Let $\seq{A_n} \subset \cf_0$, then + \[ + \bigcup_{n \in \natp}A_n = \bigsqcup_{n \in \natp}A_n \setminus \bigcup_{k = 1}^{n-1} A_k + = \bigsqcup_{n \in \natp}A_n \cap \bigcap_{k = 1}^{n - 1}A_k^c + \] + + Since $\cf_0 \subset \cf$ is closed under complements, $\seq{A_n^c} \subset \cf$ as well. By (2) and (3), $\bigcup_{n \in \natp}A_n \in \cf$ and $\bigcap_{n \in \natp}A_n^c \in \cf$. Thus $\bigcup_{n \in \natp}A_n \in \cf$ as well. By (1), $\cf_0$ is a $\sigma$-algebra that contains all open subsets of $X$, so $\cf \supset \cf_0 \supset \cb_X$. +\end{proof} + + \begin{definition}[Borel $\sigma$-Algebra on $\ol{\real}$] \label{definition:borel-sigma-algebra-extended} The family diff --git a/src/topology/dst/analytic.tex b/src/topology/dst/analytic.tex new file mode 100644 index 0000000..3ac17c0 --- /dev/null +++ b/src/topology/dst/analytic.tex @@ -0,0 +1,34 @@ +\section{Analytic Sets} +\label{section:analytic-sets} + +\begin{definition}[Analytic Set] +\label{definition:analytic-set} + Let $X$ be a Polish space and $A \subset X$, then $A$ is \textbf{analytic} if there exists a Polish space $Z$ and $f \in C(Z; X)$ such that $f(Z) = A$. +\end{definition} + + +\begin{proposition} +\label{proposition:analytic-sets} + Let $X$ be a Polish space, then: + \begin{enumerate} + \item For each family $\seq{A_n} \subset X$ of analytic sets, $\bigcup_{n \in \natp}A_n$ is analytic. + \item For each family $\seq{A_n} \subset X$ of analytic sets, $\bigcap_{n \in \natp}A_n$ is analytic. + \item For any $A \in \cb_X$, $A$ is analytic. + \end{enumerate} +\end{proposition} +\begin{proof}[Proof, {{\cite[Proposition 8.2.1-8.2.3]{CohnMeasure}}}. ] + For each $n \in \natp$, let $Z_n$ be a Polish space and $f_n \in C(Z_n; X)$ such that $A_n = f_n(Z_n)$. + + (1): By \autoref{proposition:polish-space-extension}, $Z = \bigsqcup_{n \in \natp}Z_n$ is a Polish space. Let $f \in C(Z; X)$ be the gluing of $\seq{f_n}$, then $\bigcup_{n \in \natp}A_n = f(Z)$. + + (2): By \autoref{proposition:polish-space-extension}, $Z = \prod_{n \in \natp}Z_n$ is a Polish space. For each $m, n \in \natp$, $\bracs{f_m \circ \pi_m = f_n \circ \pi_n}$ is closed. Thus + \[ + \Delta := \bigcap_{m \in \natp}\bigcap_{n \in \natp}\bracs{f_m \circ \pi_m = f_n \circ \pi_n} + \] + + is closed, and Polish by \autoref{proposition:polish-space-extension}. Hence $\bigcap_{n \in \natp}A_n = f_1 \circ \pi_1(\Delta)$ is also analytic. + + (3): By \autoref{proposition:polish-space-extension}, every open and closed subset of $X$ is analytic. Thus (1), (2), and \autoref{lemma:monotone-borel-characterisation} imply that every element of $\cb_X$ is analytic. +\end{proof} + + diff --git a/src/topology/dst/index.tex b/src/topology/dst/index.tex index 2a11c8d..d83906f 100644 --- a/src/topology/dst/index.tex +++ b/src/topology/dst/index.tex @@ -2,3 +2,5 @@ \label{chap:polish-spaces} \input{./polish.tex} +\input{./analytic.tex} +\input{./zero.tex} diff --git a/src/topology/dst/zero.tex b/src/topology/dst/zero.tex new file mode 100644 index 0000000..914ce6f --- /dev/null +++ b/src/topology/dst/zero.tex @@ -0,0 +1,113 @@ +\section{Zero Dimensional Spaces} +\label{section:zero-dimensional} + +\begin{definition}[Zero-Dimensional] +\label{definition:zero-dimentional} + Let $X$ be a topological space, then $X$ is \textbf{zero-dimensional} if $X$ admits a base consisting of clopen sets. +\end{definition} + +\begin{proposition} +\label{proposition:zero-dimensional-extension} + Let $\seqi{X}$ and $X$ be zero-dimensional spaces, then: + \begin{enumerate} + \item For any $A \subset X$, $A$ is zero-dimensional. + \item $\prod_{i \in I}X_i$ is zero-dimensional. + \item $\bigsqcup_{i \in I}X_i$ is zero-dimensional. + \end{enumerate} +\end{proposition} +% Proof omitted. + +\begin{definition}[Cantor Space] +\label{definition:cantor-space} + Let $2 = \bracs{0, 1}$ be equipped with the discrete topology, then $2^{\natp}$ is the \textbf{Cantor space}. +\end{definition} + +\begin{definition}[Baire Space] +\label{definition:the-baire-space} + Let $\natp$ be equipped with the discrete topology, then $\mathscr{N} = (\natp)^{\natp}$ is the \textbf{Baire space}. +\end{definition} + +\begin{proposition} +\label{proposition:baire-universality} + Let $X$ be a non-empty Polish space, then there exists a surjective mapping $f \in C(\mathscr{N}; X)$. +\end{proposition} +\begin{proof}[Proof, {{\cite[Proposition 8.2.7]{CohnMeasure}}}. ] + Let $d: X^2 \to [0, \infty)$ be a complete metric on $X$. To construct the desired map, it is sufficient to construct $\bracsn{C(n_1, \cdots, n_N)| \bracsn{n_k}_1^N \subset \natp, N \in \natp} \subset 2^X$ such that: + \begin{enumerate}[label=(\roman*)] + \item For each $N \in \natp$, $\bracsn{n_k}_1^N \subset \natp$, $C(n_1, \cdots, n_N) \subset X$ is closed and non-empty. + \item For each $N \in \natp$, $\bracsn{n_k}_1^N \subset \natp$, $\text{diam}(C(n_1, \cdots, n_N)) \le 1/N$. + \item For each $N \in \natp$ and $\bracsn{n_k}_1^{N} \subset \natp$, + \[ + C(n_1, \cdots, n_{N}) = \bigcup_{n_{N+1} \in \natp}C(n_1, \cdots, n_{N+1}) + \] + \item $X = \bigcup_{n_1 \in \natp}C(n_1)$. + \end{enumerate} + + Since $X$ is Polish, there exists a countable dense subset $\seq{x_k} \subset X$. For each $n_1 \in \natp$, let $C(n_1) = \ol{B(x_{n_1}, 1/2)}$, then $\bracsn{C(n_1)|n_1 \in \natp}$ satisfies (i) and (ii) by definition. As $\seq{x_k}$ is dense in $X$, $X = \bigcup_{n_1 \in \natp}C(n_1)$, and (iv) is also satisfied. + + Let $N \in \natp$ and suppose inductively that $\bracsn{C(n_1, \cdots, n_k)| \bracsn{n_j}_1^k \subset \natp,1 \le k \le N}$ has been constructed to satisfy (i)-(iv) for each $1 \le k \le N$. + + Fix $\bracsn{n_k}_1^N \subset \natp$. By \autoref{proposition:separable-metric-space}, there exists a countable dense subset $\seq{y_k} \subset C(n_1, \cdots, n_N)$. For each $n_{N+1} \in \natp$, let $C(n_1, \cdots, n_{N+1}) = \ol{B(y_{n_{N+1}}, 1/[2(N+1)])} \cap C(n_1, \cdots, n_N)$, then $\bracsn{C(n_1, \cdots, n_{N+1})|n_{N+1} \in \natp}$ satisfies (i) and (ii) by definition. Since $\seq{y_k}$ is dense in $C(n_1, \cdots, n_N)$, + \[ + C(n_1, \cdots, n_{N}) = \bigcup_{n_{N+1} \in \natp}C(n_1, \cdots, n_{N+1}) + \] + + so (iii) is also satisfied. + + Finally, suppose that $\bracsn{C(n_1, \cdots, n_N)| \bracsn{n_k}_1^N \subset \natp, N \in \natp} \subset 2^X$ has been constructed to satisfy (i)-(iv). Since $X$ is complete, for each $\seq{n_k} \in \mathscr{N}$, there exists a unique $f(\seq{n_k}) \in X$ such that $\bracs{f(\seq{n_k})} = \bigcap_{N \in \natp}C(n_1, \cdots, n_N)$. For each $x \in X$, (iii) and (iv) imply that there exists $\seq{n_k} \in \mathscr{N}$ such that $x = f(\seq{n_k})$. As such, $f: \mathscr{N} \to X$ is a surjective mapping. For any $\seq{m_k}, \seq{n_k} \in \mathscr{N}$ and $N \in \natp$ with $m_k = n_k$ for each $1 \le k \le N$, $d(f(\seq{m_k}), f(\seq{n_k})) \le 1/N$. Therefore $f \in C(\mathscr{N}; X)$. +\end{proof} + +\begin{proposition} +\label{proposition:cantor-universality} + Let $X$ be a non-empty compact metrisable space, then there exists a surjective mapping $f \in C(2^{\natp}; X)$. +\end{proposition} +\begin{proof} + Let $d: X \times X \to [0, \infty)$ be a metric on $X$. Since $X$ is compact, for each $N \in \natp$, there exists $K_N \in \natp$ and $\seqf{x_{N, k}|1 \le k \le K_N} \subset X$ such that $X = \bigcup_{k = 1}^{K_N}B(x_{N, k}, 1/N)$. + + To construct the desired map, it is sufficient to construct + \[ + \bracs{C(n_1, \cdots, n_N) \bigg | \bracsn{n_k}_1^N \in \prod_{n = 1}^{N}\bracs{1, \cdots, K_n}, N \in \natp} \subset 2^X + \] + + + such that: + \begin{enumerate}[label=(\roman*)] + \item For each $N \in \natp$ and $\bracsn{n_k}_1^N$, $C(n_1, \cdots, n_N) \subset X$ is closed and non-empty. + \item For each $N \in \natp$ and $\bracsn{n_k}_1^N$, $\text{diam}(C(n_1, \cdots, n_N)) \le 4/N$. + \item For each $N \in \natp$ and $\bracsn{n_k}_1^{N}$, + \[ + C(n_1, \cdots, n_{N}) = \bigcup_{n_{N+1} = 1}^{K_{N+1}}C(n_1, \cdots, n_{N+1}) + \] + \item $X = \bigcup_{n_1 = 1}^{K_1}C(n_1)$. + \end{enumerate} + + For each $1 \le n_1 \le K_1$, let $C(n_1) = \ol{B(x_{1, n_1}, 1)}$, then $\bracsn{C(n_1)|1 \le n_1 \le K_1}$ satisfies (i), (ii), and (iv) by definition. + + Let $N \in \natp$ and suppose inductively that + \[ + \bracs{C(n_1, \cdots, n_K) \bigg | \bracsn{n_k}_1^K \in \prod_{n = 1}^{K}\bracs{1, \cdots, K_n}, 1 \le K \le N} \subset 2^X + \] + + + has been constructed to satisfy (i)-(iv) for each $1 \le K \le N$. Fix $\bracsn{n_k}_1^N \in \prod_{n = 1}^{N}\bracs{1, \cdots, K_n}$. Since $X = \bigcup_{k = 1}^{K_{N+1}}B(x_{N+1, k}, 1/(N+1))$, there exists $\bracsn{y_k}_1^{K_{N+1}} \subset C(n_1, \cdots, n_N)$ such that $C(n_1, \cdots, n_N) \subset \bigcup_{k = 1}^{K_{N+1}}B(y_k, 2/(N+1))$. For each $1 \le n_{N+1} \le K_{N+1}$, let $C(n_1, \cdots, n_{N+1}) = C(n_1, \cdots, n_N) \cap \ol{B(y_{n_{N+1}}, 2/(N+1))}$, then $\bracsn{C(n_1, \cdots, n_{N+1})|1 \le n_{N+1} \le K_{N+1}}$ satisfies (i)-(iii). + + Now, suppose that + \[ + \bracs{C(n_1, \cdots, n_N) \bigg | \bracsn{n_k}_1^N \in \prod_{n = 1}^{N}\bracs{1, \cdots, K_n}, N \in \natp} \subset 2^X + \] + + has been constructed to satisfy (i)-(iv). Since $X$ is complete, for each $\seq{n_k} \in \prod_{n \in \natp}\bracs{1, \cdots, K_n}$, there exists a unique $f(\seq{n_k}) \in X$ such that $\bracs{f(\seq{n_k})} = \bigcap_{N \in \natp}C(n_1, \cdots, n_N)$. For each $x \in X$, (iii) and (iv) imply that there exists $\seq{n_k} \in \prod_{n \in \natp}\bracs{1, \cdots, K_n}$ such that $x = f(\seq{n_k})$. As such, $f: \prod_{n \in \natp}\bracs{1, \cdots, K_n} \to X$ is a surjective mapping. For any $\seq{m_k}, \seq{n_k} \in \prod_{n \in \natp}\bracs{1, \cdots, K_n}$ and $N \in \natp$ with $m_k = n_k$ for each $1 \le k \le N$, $d(f(\seq{m_k}), f(\seq{n_k})) \le 4/N$. Therefore $f \in C(\prod_{n \in \natp}\bracs{1, \cdots, K_n}; X)$. + + Finally, for each $n \in \natp$, there exists $L_n \in \natp$ and a surjective mapping $g_n: 2^{L_n} \to \bracs{1, \cdots, K_n}$. Let + \[ + g = \prod_{n \in \natp}g_n: \prod_{n \in \natp}2^{L_n} \to \prod_{n \in \natp}\bracs{1, \cdots, K_n} + \] + + be the product of $\seq{g_n}$, then $g \in C(\prod_{n \in \natp}2^{L_n}; \prod_{n \in \natp}\bracs{1, \cdots, K_n})$, and + \[ + f \circ g: 2^{\natp} \iso \prod_{n \in \natp}2^{L_n} \to X + \] + + is the desired continuous surjection. +\end{proof} + diff --git a/src/topology/index.tex b/src/topology/index.tex index b82bad1..f1b37f3 100644 --- a/src/topology/index.tex +++ b/src/topology/index.tex @@ -6,4 +6,5 @@ \input{./functions/index.tex} \input{./metric/index.tex} \input{./groups/index.tex} +\input{./dst/index.tex} \input{./notation.tex} diff --git a/src/topology/notation.tex b/src/topology/notation.tex index 903a2bd..6b3b0ed 100644 --- a/src/topology/notation.tex +++ b/src/topology/notation.tex @@ -26,4 +26,7 @@ $f \prec U$ & $f \in C_c(X; [0,1])$ with $\mathrm{supp}(f) \subset U$. & \autoref{definition:compactly-supported-01} \\ $C_0(X; E)$ & Continuous functions vanishing at infinity. & \autoref{definition:vanish-at-infinity} \\ $BC(X; E)$ & Bounded continuous functions $X \to E$. & \autoref{definition:bounded-continuous-function-space} \\ + + % DST + $\mathscr{N}$ & The Baire space. & \autoref{definition:the-baire-space} \\ \end{tabular}