From ce52ac4b63af6b9abe0cf7e092300fe54986f9bf Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sat, 27 Jun 2026 22:35:32 -0400 Subject: [PATCH] Edited the open preimage functions. --- src/topology/main/preimage.tex | 33 +++++++++++++++++++++------------ 1 file changed, 21 insertions(+), 12 deletions(-) diff --git a/src/topology/main/preimage.tex b/src/topology/main/preimage.tex index f458cb2..d8c30e4 100644 --- a/src/topology/main/preimage.tex +++ b/src/topology/main/preimage.tex @@ -31,16 +31,17 @@ \begin{definition}[Basic Preimage Function] \label{definition:basic-preimage-function} - Let $X$ be a set, $Y$ be a topological space, $\mathcal{B}$ be a base for $Y$ with $\emptyset \in \mathcal{B}$, and $P: \mathcal{B} \to 2^X$, then $P$ is a \textbf{basic preimage function} if: + Let $X$ be a set, $Y$ be a topological space, $\mathcal{B}$ be a base for $Y$ with $\emptyset \in \mathcal{B}$, and $p: \mathcal{B} \to 2^X$, then $p$ is a \textbf{basic preimage function} if: \begin{enumerate}[label=(PF\arabic*)] \item $P(\emptyset) = \emptyset$. - \item[(PF2')] For each $\mathcal{U} \subset \mathcal{B}$ and $V \in \mathcal{B}$ with $V \subset \bigcup_{U \in \mathcal{U}}U$, $P(V) \subset \bigcup_{U \in \mathcal{U}}P(U)$. + \item[(PF2')] For each $\mathcal{U} \subset \mathcal{B}$ and $V \in \mathcal{B}$ with $V \subset \bigcup_{U \in \mathcal{U}}U$, $p(V) \subset \bigcup_{U \in \mathcal{U}}p(U)$. + \item[(PF3')] For each $U, V \subset \mathcal{B}$, $p(U) \cap p(V) \subset \bigcup_{W \in \mathcal{B}, W \subset U \cap V}p(W)$. \end{enumerate} \end{definition} \begin{proposition} \label{proposition:basic-preimage-function} - Let $X$ be a set and $(Y, \topo)$ be a topological space with base $\mathcal{B}$, then: + Let $X$ be a set, $(Y, \topo)$ be a topological space, and $\mathcal{B}$ be a base for $Y$ with $\emptyset \in \mathcal{B}$, then: \begin{enumerate} \item For any open preimage function $P: \topo \to 2^X$, $P|_{\mathcal{B}}$ is a basic preimage function. \item For any basic preimage function $p: \mathcal{B} \to 2^X$, there exists a unique open preimage function $P: \topo \to 2^X$ such that $p = P|_{\mathcal{B}}$. @@ -52,7 +53,7 @@ P: \topo \to 2^X \quad U \mapsto \bigcup_{V \in \mathcal{B}(U)}p(V) \] - then the only element in $\mathcal{B}$ that can be contained in $\emptyset$ is $\emptyset$ itself, so $P(\emptyset) = \emptyset$. + then $P(\emptyset) = \emptyset$. For any $U \in\topo$, $U = \bigcup_{V \in \mathcal{B}(U)}V$, so if $P$ is an extension of $p$ to $\topo$ as an open preimage function, then $P$ must be unique by (PF2'). Let $\mathcal{U} \subset \topo$. Since $\bigcup_{U \in \mathcal{U}}\mathcal{B}(U) \subset \mathcal{B}\paren{\bigcup_{U \in \mathcal{U}}U}$, $\bigcup_{U \in \mathcal{U}}P(U) \subset P\paren{\bigcup_{U \in \mathcal{U}}U}$. On the other hand, for each $V \in \mathcal{B}\paren{\bigcup_{U \in \mathcal{U}}U}$, $V \subset \bigcup_{U \in \mathcal{U}}\bigcup_{W \in \mathcal{B}(U)}W$. By (PF2'), \[ @@ -64,8 +65,21 @@ P\paren{\bigcup_{U \in \mathcal{U}}U} = \bigcup_{V \in \mathcal{B}(\bigcup_{U \in \mathcal{U}}U)}p(V) \subset \bigcup_{U \in \mathcal{U}}P(U) \] - Finally, let $U, V \in \topo$, then since $\mathcal{B}(U \cap V) = \mathcal{B}(U) \cap \mathcal{B}(V)$, $P(U \cap V) = P(U) \cap P(V)$. Therefore $P$ is an open preimage function. + Finally, let $U, V \in \topo$, then since $\mathcal{B}(U \cap V) = \mathcal{B}(U) \cap \mathcal{B}(V)$, + \begin{align*} + P(U \cap V) &= \bigcup_{W \in \mathcal{B}(U \cap V)}p(W) \\ + &\subset \bigcup_{W \in \mathcal{B}(U)}p(W) \cap \bigcup_{W \in \mathcal{B}(V)}p(W) = P(U) \cap P(V) + \end{align*} + + On the other hand, by (PF3'), + \begin{align*} + P(U) \cap P(V) &= \bigcup_{W \in \mathcal{B}(U)}\bigcup_{W' \in \mathcal{B}(V)}p(W) \cap p(W') \\ + &\subset \bigcup_{W \in \mathcal{B}(U)}\bigcup_{W' \in \mathcal{B}(V)}\bigcup_{S \in \mathcal{B}(W \cap W')}p(S) \\ + &\subset \bigcup_{W \in \mathcal{B}(U \cap V)} p(W) = P(U \cap V) + \end{align*} + + so $P(U \cap V) = P(U) \cap P(V)$. Therefore $P$ is an open preimage function. \end{proof} \begin{theorem} @@ -80,17 +94,12 @@ \begin{proof} Since $P$ is an open preimage function and $Y$ is Hausdorff, \autoref{proposition:open-preimage-function-gymnastics} implies that such a function is unique if it exists, so it is sufficient to demonstrate existence. - For each $x \in X$, let $\fF(x) = \bracs{U \in \topo|x \in P(U)}$, then by (PF1) and (PF3'), $\fF(x)$ is a filter base in $(Y, \fU)$. Moreover, (S) implies that $\fF(x)$ is a Cauchy filter base. Since $Y$ is complete and Hausdorff, there exists a unique $y \in Y$ such that $\fF(x) \to y$. Thus the function in question must be + For each $x \in X$, let $\fF(x) = \bracs{U \in \topo|x \in P(U)}$, then by (S), $\fF(x)$ is non-empty. By (PF1) and (PF3'), $\fF(x)$ is a filter base in $(Y, \fU)$. Moreover, (S) implies that $\fF(x)$ is a Cauchy filter base. Since $Y$ is complete and Hausdorff, there exists a unique $y \in Y$ such that $\fF(x) \to y$. Thus the function in question must be \[ f: X \to Y \quad x \mapsto \lim_{y, \fF(x)}y \] - It remains to verify that $P$ is the open preimage function of $f$. To this end, let $U \in \topo$ and $x \in X$ such that $f(x) \in U$. Since $U \in \cn_Y(f(x))$, there exists a symmetric entourage $V \in \fU$ such that $(V \circ V)(f(x)) \subset U$. By (S), there exists a $V$-small set $W \in \fF(x)$. As $f(x) \in \ol W \subset V \circ W$ and $W$ is $V$-small, $W \subset (V \circ V)(f(x))$. Thus - \[ - x \in P(W) \subset P((V \circ V)(f(x))) \subset P(U) - \] - - and $f^{-1}(U) \subset P(U)$. + It remains to verify that $P$ is the open preimage function of $f$. To this end, let $U \in \topo$ and $x \in X$ such that $f(x) \in U$. Since $U \in \cn_Y(f(x))$, there exists a symmetric entourage $V \in \fU$ such that $(V \circ V)(f(x)) \subset U$. By (S), there exists a $V$-small set $W \in \fF(x)$. As $f(x) \in \ol W \subset V \circ W$ and $W$ is $V$-small, $W \subset (V \circ V)(f(x)) \subset U$. Thus $x \in P(W) \subset P(U)$, and $f^{-1}(U) \subset P(U)$. On the other hand, let $x \in P(U)$. Suppose for contradiction that $f(x) \not\in U$. For each $y \in U$, there exists $V_y \in \cn_Y^o(y)$ and $W_y \in \cn_Y^o(f(x))$ such that $V_y \subset U$ and $V_y \cap W_y = \emptyset$. Since $x \in f^{-1}(W_y) \subset P(W_y)$ and $P(V_y) \cap P(W_y) = P(V_y \cap W_y) = \emptyset$, $x \not\in P(V_y)$. By (PF2'), $\bigcup_{y \in U}P(V_y) = P(U)$, so $x \not\in P(U)$, which is a contradiction. Therefore $f^{-1}(U) = P(U)$ for all $U \in \topo$. \end{proof}