From e8474bba3ea29abb45ab39a7cdfd2420bc370ce2 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Wed, 6 May 2026 23:31:25 -0400 Subject: [PATCH] Fixed equicontinuous formulation. --- src/topology/uniform/equicontinuous.tex | 12 ++++++------ 1 file changed, 6 insertions(+), 6 deletions(-) diff --git a/src/topology/uniform/equicontinuous.tex b/src/topology/uniform/equicontinuous.tex index a4529c1..b8e638c 100644 --- a/src/topology/uniform/equicontinuous.tex +++ b/src/topology/uniform/equicontinuous.tex @@ -14,13 +14,13 @@ \begin{enumerate} \item $\cf$ is equicontinuous at $x$. \item For $\angles{x_\alpha}_{\alpha \in A} \subset X$ with $x_\alpha \to x$, $\angles{f_\alpha}_{\alpha \in A} \subset \cf$, and $U \in \fU$, there exists $\alpha_0 \in A$ such that $(f_\alpha(x_\alpha), f_\alpha(x)) \in U$ for all $\alpha \ge \alpha_0$. - \item For any upward-directed set $A$ with $|A| \le |\cn_X(x)|$, $\angles{x_\alpha}_{\alpha \in A} \subset X$ with $x_\alpha \to x$, $\angles{f_\alpha}_{\alpha \in A} \subset \cf$, and $U \in \fU$, there exists $\alpha_0 \in A$ such that $(f_\alpha(x_\alpha), f_\alpha(x)) \in U$ for all $\alpha \ge \alpha_0$. + \item There exists a fundamental system of neighbourhoods $\fB \subset \cn_X(x)$ at $x$ such that for any $\angles{x_V}_{V \in \fB} \subset X$ with $x_\alpha \to x$, $\angles{f_V}_{V \in \fB} \subset \cf$, and $U \in \fU$, there exists $V_0 \in \fB$ such that $(f_V(x_V), f_V(x)) \in U$ for all $V \subset V_0$. \end{enumerate} \end{proposition} \begin{proof} (1) $\Rightarrow$ (2): Since $\cf$ is equicontinuous at $x$, there exists $V \in \cn_X(x)$ such that $(f_\alpha(y), f_\alpha(x)) \in U$ for all $y \in V$ and $\alpha \in A$. Given that $x_\alpha \to x$, there exists $\alpha_0 \in A$ such that $x_\alpha \in V$ for all $\alpha \ge \alpha_0$, so $(f_\alpha(x_\alpha), f_\alpha(x)) \in U$ for all $\alpha \ge \alpha_0$. - $\neg (1) \Rightarrow \neg (3)$: Direct $\cn_X(x)$ under reverse inclusion. If $\cf$ is not equicontinuous at $x$, then there exists $U \in \fU$ such that for every $V \in \cn_X(x)$, there exists $f_V \in \cf$ and $x_V \in V$ with $(f_V(x_V), f_V(x)) \not\in U$. In which case, $x_V \to x$ but $(f_V(x_V), f_V(x)) \not\in U$ for all $V \in \cn_X(x)$. + $\neg (1) \Rightarrow \neg (3)$: If $\cf$ is not equicontinuous at $x$, then there exists $U \in \fU$ such that for every $V \in \fB$, there exists $f_V \in \cf$ and $x_V \in V$ with $(f_V(x_V), f_V(x)) \not\in U$. In which case, $x_V \to x$ but $(f_V(x_V), f_V(x)) \not\in U$ for all $V \in \cn_X(x)$. \end{proof} @@ -38,7 +38,7 @@ then \begin{enumerate}[label=(C\arabic*)] - \item The product uniformity and the compact uniformity on $\cf$ coincide. + \item The uniform structures of pointwise and compact convergence on $\cf$ coincide. \item The closure of $\cf$ in $Y^X$ with respect to the product topology is equicontinuous. \end{enumerate} @@ -49,7 +49,7 @@ then \begin{enumerate}[label=(C\arabic*), start=2] - \item $\cf$ is a precompact subset of $C(X; Y)$ with respect to the compact uniformity. + \item $\cf$ is a precompact subset of $C(X; Y)$ with respect to the uniform structure of compact convergence. \end{enumerate} Conversely, if $X$ is a LCH space, then (C3) implies (E1) + (E2). @@ -67,7 +67,7 @@ \bigcap_{j = 1}^n E(\bracs{x_j}, U) \subset E(K, U \circ U \circ U) \] - so the product uniformity and the compact uniformity coincide. + so the uniform structures of pointwise and compact convergence coincide. (E1) $\Rightarrow$ (C2): Let $\cf'$ be the closure of $\cf$ in $Y^X$ with respect to the product topology. Let $x \in X$ and entourage $U$ of $Y$. Using \autoref{proposition:goodentourages}, assume without loss of generality that $U$ is closed. Since $\cf$ is equicontinuous, there exists $V \in \cn_X(x)$ such that $(f(x), f(y)) \in U$ for all $f \in \cf$ and $y \in V$. For any element $g \in \cf'$, $(g(x), g(y)) \in \ol U = U$ for all $y \in V$. Therefore $\cf'$ is also equicontinuous. @@ -80,7 +80,7 @@ Therefore $g(W) \subset (U \circ U \circ U)(g(x))$, and $\cf$ is equicontinuous. - (C3) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the compact uniformity, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. + (C3) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the uniform structure of compact convergence, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. \end{proof}