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Bokuan Li
2026-01-17 23:11:19 -05:00
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1
src/topology/functions/index.tex
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@@ -2,3 +2,4 @@
\label{chap:function-spaces}
\input{./src/topology/functions/set-systems.tex}
\input{./src/topology/functions/uniform.tex}

9
src/topology/main/interiorclosureboundary.tex
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@@ -78,6 +78,15 @@
Let $x \in \prod_{i \in I}X_i$ and $U \in \cn(x)$, then there exists $J \subset I$ finite and $\seqf{U_j}$ such that $U_j \in \cn(\pi_j(x))$ for each $j \in J$, and $x \in \bigcap_{j \in J}\pi_j^{-1}(U_j) \subset U$. By density of each $A_j$, there exists $y \in X$ such that $y_j \in U_j$ for each $j \in J$. Thus $y \in U$, and $A$ is dense.
\end{proof}
\begin{proposition}
\label{proposition:separable-product}
Let $\seqf{X_j}$ be separable topological spaces, then $\prod_{j = 1}^n X_j$ is also separable.
\end{proposition}
\begin{proof}
Let $\seq{A_j}$ such that $A_j \subset X_j$ is countable and dense for each $1 \le j \le n$. By \ref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$.
\end{proof}
\begin{lemma}
\label{lemma:closurecomplement}
Let $X$ be a topological space and $A \subset X$, then $(A^o)^c = \overline{A^c}$.

50
src/topology/main/metric.tex
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@@ -1,50 +0,0 @@
\section{Metric Spaces}
\label{section:metric}
\begin{definition}[Metric]
\label{definition:metric}
Let $X$ be a set, a \textbf{pseudo-metric} is a mapping $d: X \times X \to [0, \infty)$ such that:
\begin{enumerate}
\item For any $x \in X$, $d(x, x) = 0$.
\item For any $x, y \in X$, $d(x, y) = d(y, x)$.
\item For any $x, y, z \in X$, $d(x, z) \le d(x, y) + d(y, z)$.
\end{enumerate}
and the pair $(X, d)$ is a \textbf{pseudo-metric space}. If in addition,
\begin{enumerate}
\item[(4)] For any $x, y \in X$, $d(x, y) = 0$ if and only if $x = y$.
\end{enumerate}
then $d$ is a \textbf{metric}.
\end{definition}
\begin{definition}[Induced Uniformity]
\label{definition:metric-uniformity}
Let $(X, d)$ be a metric space. For each $r > 0$, let
\[
U_r = \bracs{(x, y) \in X \times Y| d(x, y) < r}
\]
then there exists a uniformity $\fU$ on $X$ such that the family $\fB = \bracs{U_r| r > 0}$ is a fundamental system of entourages for $\fU$, known as the \textbf{uniformity induced by} $d$.
Let $\seqi{d}$ be pseudo-metrics on $X$. For each $i \in I$, let $X_i = X$ be equipped with the uniformity induced by $d_i$, then the \textbf{uniformity induced by} $\seqi{d}$ is the initial uniformity generated by the identity $\text{Id}_X: X \to X_i$.
\end{definition}
\begin{lemma}
\label{lemma:uniform-first-countable}
Let $X$ be a uniform space with a countable fundamental system of entourages, then there exists a countable fundamental system of symmetric entourages $\fB = \bracs{U_{1/2^n}|n \in \nat^+}$ such that $U_{1/2^{n+1}} \circ U_{1/2^{n+1}} \subset U_{1/2^n}$ for all $n \in \nat^+$.
\end{lemma}
\begin{proof}
Let $\seq{V}$ be a countable fundamental system of entourages. By \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $U_{1/2} \subset V_1$. Suppose inductively that $U_{1/2^n}$ has been constructed, then by (U2) and \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $U_{1/2^{n+1}} \subset U_{1/2^n} \cap V$. Thus $\fB = \bracs{U_{1/2^n}|n \in \nat^+}$ is the desired family.
\end{proof}
% Work in progress: pseudo-metrisability of uniform spaces.
\begin{lemma}
\label{lemma:uniform-urysohn}
Let $X$ be a uniform space with a countable fundamental system of entourages, $\mathbb{D}$ be the dyadic rational numbers in $[0, 1]$, then there exists a fundamental system of entourages $\fB = \bracs{}$
\end{lemma}
\begin{proposition}
\label{proposition:uniform-pseudometric}
Let
\end{proposition}

1
src/topology/main/neighbourhoods.tex
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@@ -22,7 +22,6 @@
\label{lemma:openneighbourhood}
Let $(X, \topo)$ be a topological space, then $U \subset X$ is open if and only if $U \in \cn_\topo(x)$ for all $x \in U$.
\end{lemma}
\begin{proof}
Suppose that $U \in \cn_\topo(x)$ for all $x \in U$. For each $x \in U$, there exists $V_x \in \topo$ with $x \in V_x \subset U$.
Thus $U = \bigcup_{x \in U}V_x \in \topo$.

2
src/topology/main/normal.tex
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@@ -82,6 +82,6 @@
Since $\real$ is complete, so is $BC(X; \real)$ by \ref{proposition:set-uniform-complete}. Using successive approximations (\ref{theorem:successive-approximation}), for every $f \in BC(A; \real)$, there exists $F \in BC(X; \real)$ such that $RF = F|_A = f$ and
\[
\norm{F}_u \le \frac{1}{3} \cdot \frac{1}{1 - 2/3} \cdot \norm{f}_u = 1
\norm{F}_u \le \frac{1}{3} \cdot \frac{1}{1 - 2/3} \cdot \norm{f}_u = \norm{f}_u
\]
\end{proof}

98
src/topology/uniform/cauchy.tex
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@@ -0,0 +1,98 @@
\section{Cauchy Filters}
\label{section:cauchy-filter}
\begin{definition}[Small]
\label{definition:small}
Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A \subset X$, then $A$ is \textbf{$V$-small} if $A \times A \subset V$.
\end{definition}
\begin{lemma}[{{\cite[Proposition 2.3.1]{Bourbaki}}}]
\label{lemma:small-intersect}
Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A, B \subset X$ such that:
\begin{enumerate}
\item $A, B$ are $V$-small.
\item $A \cap B \ne \emptyset$.
\end{enumerate}
then $A \cup B$ is $V \circ V$-small.
\end{lemma}
\begin{proof}
Since $V \circ V \supset V$, it is sufficient to consider the case where $x \in A$ and $y \in B$. By assumption (2), there exists $z \in A \cap B$. By assumption (1), $(x, z), (z, y) \in V$, so $(x, y) \in V \circ V$.
\end{proof}
\begin{definition}[Cauchy Filter]
\label{definition:cauchyfilter}
Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a filter on $X$, then $\fF$ is \textbf{Cauchy} if for every $V \in \fU$, there exists $E \in \fF$ such that $E$ is $V$-small.
\end{definition}
\begin{proposition}[Cauchy Criterion]
\label{proposition:cauchycriterion}
Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a convergent filter, then $\fF$ is Cauchy.
\end{proposition}
\begin{proof}
Let $x \in X$ such that $\cn(x) \subset \fF$ and $V \in \fU$. By \ref{lemma:symmetricfundamentalentourage}, assume without loss of generality that $V \in \fU$. Since $\cn(x) \subset \fF$, then $V(x) \in \fU$ and $V(x) \times V(x) \subset V$.
\end{proof}
\begin{definition}[Cauchy Continuous]
\label{definition:cauchy-continuous}
Let $X, Y$ be uniform spaces and $f: X \to Y$, then $f$ is \textbf{Cauchy continuous} if for any Cauchy filter base $\fB \subset 2^X$, $f(\fB) \subset 2^Y$ is Cauchy.
\end{definition}
\begin{proposition}[{{\cite[Proposition 2.3.3]{Bourbaki}}}]
\label{proposition:imagecauchy}
Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f \in UC(X; Y)$, then $f$ is Cauchy continuous.
\end{proposition}
\begin{proof}
Let $V \in \mathfrak{V}$, then there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$. For any Cauchy filter base $\fB \subset 2^X$, there exists $E \in \fB$ such that $E \times E \subset V'$, so $f(E) \times f(E) \subset V$.
\end{proof}
\begin{definition}[Minimal Cauchy Filter]
\label{definition:minimalcauchy}
Let $X$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then $\fF$ is \textbf{minimal} if it is minimal with respect to inclusion.
\end{definition}
\begin{proposition}[{{\cite[Proposition 2.3.5]{Bourbaki}}}]
\label{proposition:minimalcauchyexistence}
Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then:
\begin{enumerate}
\item There exists a unique minimal Cauchy filter $\fF_0$ such that $\fF_0 \subset \fF$.
\item If $\fB \subset \fF$ is a base of $\fF$, and $\fV \subset \fU$ is a fundamental system of symmetric entourages, then $\mathfrak{M} = \bracs{V(M): V \in \fV, M \in \fB}$ is a base for $\fF_0$.
\end{enumerate}
\end{proposition}
\begin{proof}
Since $\fV$ and $\fB$ are both non-empty, $\mathfrak{M}$ is non-empty as well. Let $V_1, V_2 \in \fV$ and $M_1, M_2 \in \fB$. By (F2) of $\fV$ and $\fB$, there exists $V_0 \in \fV$ and $M_0 \in \fB$ such that $V_0 \subset V_1 \cap V_2$ and $M_0 \subset M_1 \cap M_2$. So $V_0(M_0) \subset V_1(M_1) \cap V_2(M_2)$, and $\mathfrak{M}$ satisfies (FB1). Given that $\emptyset \not\in \fV, \fB$, $\emptyset \not\in \mathfrak{M}$, so $\mathfrak{M}$ satisfies (FB2). By \ref{proposition:filterbasecriterion}, $\mathfrak{M}$ is a filter base for
\[
\fF_0 = \bracs{E \subset X| \exists V \in \fV, M \in \fB: V(M) \subset E} \subset \fF
\]
To see that $\mathfrak{M}$ is a Cauchy filter base, let $V \in \fV$. By (U2), there exists $W \in \fV$ such that $W \circ W \circ W \subset V$. Since $\fB$ is a Cauchy filter base, there exists $M \in \fB$ such that $M \times M \subset W$. Let $(x, y) \in W(M) \times W(M)$, then there exists $(p, q) \in M \times M$ such that $(p, x), (q, y) \in W$. As $M \times M$ \subset $W$, $(x, y) \in W \circ W \circ W$ by symmetry of $W$ and \ref{lemma:compositiongymnastics}. Therefore $W(M) \times W(M) \subset V$ and $\mathfrak{M}$ is a Cauchy filter base.
Finally, let $\fF' \subset \fF$ be a Cauchy filter. Let $M \in \fB$ and $V \in \fV$, then there exists $N \in \fF'$ such that $N \times N \subset V$. Since $N \in \fF$, $M \cap N \ne \emptyset$, and there exists $y \in M \cap N$. For any $x \in N$, $N \times N \subset V$ implies that $x \in V(y)$. Hence $N \subset V(M)$ and $\fF' \supset \fF_0$.
\end{proof}
\begin{proposition}[{{\cite[Corollary 2.3.4]{Bourbaki}}}]
\label{proposition:cauchyinterior}
Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a minimal Cauchy filter, then $\fF$ admits a base consisting of open sets.
\end{proposition}
\begin{proof}
Let $\fV \subset \fU$ be the set of all symmetric, open entourages. By \ref{proposition:goodentourages}, $\fV$ is a fundamental system of entourages. By \ref{proposition:minimalcauchyexistence}, $\mathfrak{M} = \bracs{U(M)| U \in \fV, M \in \fF}$ is a base for $\fF$.
Let $V \in \fU$, then there exists $U \in \fV$ such that $U \subset V$. For any $M \in \fF$, $U(M)$ is open by \ref{lemma:openentourageneighbourhoods}. Thus $\mathfrak{M}$ consists of open sets.
\end{proof}
\begin{proposition}[{{\cite[Corollary 2.3.1, 2.3.2]{Bourbaki}}}]
\label{proposition:cauchyfilterlimit}
Let $(X, \fU)$ be a uniform space, then:
\begin{enumerate}
\item For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter.
\item For each Cauchy filter $\fF \subset 2^X$ and $x \in X$, $x$ is an accumulation point of $\fF$ if and only if $\fF$ converges to $x$.
\item For any $x \in X$, filter $\fF \subset 2^X$ converging to $x$, and Cauchy filter $\mathfrak{G} \subset \fF$, $\mathfrak{G}$ converges to $x$.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $\fF = \bracs{E \subset X: x \in E}$, then $\fF$ is a filter generated by the filter base $\fB = \bracs{\bracs{x}}$. By \ref{proposition:minimalcauchyexistence}, $\bracs{U(x): U \in \fU} = \cn(x)$ is the unique minimal filter contained in $\fF$.
(2): Let $x \in X$ be an accumulation point of $\fF$, then $U(x) \cap E \ne \emptyset$ for all $U \in \fU$ and $E \in \fF$. By \ref{definition:generatedfilter}, there exists a filter $\fF' \supset \fF \cup \cn(x)$. Since $\fF$ is a Cauchy filter, so is $\fF'$. By minimality of $\cn(x)$, $\fF \supset \cn(x)$.
(3): Since $\fF$ converges to $x$, $x$ is a cluster point of $\fF$ and $\mathfrak{G}$. By (2), $x$ is also a limit point of $\mathfrak{G}$.
\end{proof}

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src/topology/uniform/complete.tex
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@@ -1,103 +1,6 @@
\section{Completeness}
\label{section:completeness}
\begin{definition}[Small]
\label{definition:small}
Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A \subset X$, then $A$ is \textbf{$V$-small} if $A \times A \subset V$.
\end{definition}
\begin{lemma}[{{\cite[Proposition 2.3.1]{Bourbaki}}}]
\label{lemma:small-intersect}
Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A, B \subset X$ such that:
\begin{enumerate}
\item $A, B$ are $V$-small.
\item $A \cap B \ne \emptyset$.
\end{enumerate}
then $A \cup B$ is $V \circ V$-small.
\end{lemma}
\begin{proof}
Since $V \circ V \supset V$, it is sufficient to consider the case where $x \in A$ and $y \in B$. By assumption (2), there exists $z \in A \cap B$. By assumption (1), $(x, z), (z, y) \in V$, so $(x, y) \in V \circ V$.
\end{proof}
\begin{definition}[Cauchy Filter]
\label{definition:cauchyfilter}
Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a filter on $X$, then $\fF$ is \textbf{Cauchy} if for every $V \in \fU$, there exists $E \in \fF$ such that $E$ is $V$-small.
\end{definition}
\begin{proposition}[Cauchy Criterion]
\label{proposition:cauchycriterion}
Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a convergent filter, then $\fF$ is Cauchy.
\end{proposition}
\begin{proof}
Let $x \in X$ such that $\cn(x) \subset \fF$ and $V \in \fU$. By \ref{lemma:symmetricfundamentalentourage}, assume without loss of generality that $V \in \fU$. Since $\cn(x) \subset \fF$, then $V(x) \in \fU$ and $V(x) \times V(x) \subset V$.
\end{proof}
\begin{definition}[Cauchy Continuous]
\label{definition:cauchy-continuous}
Let $X, Y$ be uniform spaces and $f: X \to Y$, then $f$ is \textbf{Cauchy continuous} if for any Cauchy filter base $\fB \subset 2^X$, $f(\fB) \subset 2^Y$ is Cauchy.
\end{definition}
\begin{proposition}[{{\cite[Proposition 2.3.3]{Bourbaki}}}]
\label{proposition:imagecauchy}
Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f \in UC(X; Y)$, then $f$ is Cauchy continuous.
\end{proposition}
\begin{proof}
Let $V \in \mathfrak{V}$, then there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$. For any Cauchy filter base $\fB \subset 2^X$, there exists $E \in \fB$ such that $E \times E \subset V'$, so $f(E) \times f(E) \subset V$.
\end{proof}
\begin{definition}[Minimal Cauchy Filter]
\label{definition:minimalcauchy}
Let $X$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then $\fF$ is \textbf{minimal} if it is minimal with respect to inclusion.
\end{definition}
\begin{proposition}[{{\cite[Proposition 2.3.5]{Bourbaki}}}]
\label{proposition:minimalcauchyexistence}
Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then:
\begin{enumerate}
\item There exists a unique minimal Cauchy filter $\fF_0$ such that $\fF_0 \subset \fF$.
\item If $\fB \subset \fF$ is a base of $\fF$, and $\fV \subset \fU$ is a fundamental system of symmetric entourages, then $\mathfrak{M} = \bracs{V(M): V \in \fV, M \in \fB}$ is a base for $\fF_0$.
\end{enumerate}
\end{proposition}
\begin{proof}
Since $\fV$ and $\fB$ are both non-empty, $\mathfrak{M}$ is non-empty as well. Let $V_1, V_2 \in \fV$ and $M_1, M_2 \in \fB$. By (F2) of $\fV$ and $\fB$, there exists $V_0 \in \fV$ and $M_0 \in \fB$ such that $V_0 \subset V_1 \cap V_2$ and $M_0 \subset M_1 \cap M_2$. So $V_0(M_0) \subset V_1(M_1) \cap V_2(M_2)$, and $\mathfrak{M}$ satisfies (FB1). Given that $\emptyset \not\in \fV, \fB$, $\emptyset \not\in \mathfrak{M}$, so $\mathfrak{M}$ satisfies (FB2). By \ref{proposition:filterbasecriterion}, $\mathfrak{M}$ is a filter base for
\[
\fF_0 = \bracs{E \subset X| \exists V \in \fV, M \in \fB: V(M) \subset E} \subset \fF
\]
To see that $\mathfrak{M}$ is a Cauchy filter base, let $V \in \fV$. By (U2), there exists $W \in \fV$ such that $W \circ W \circ W \subset V$. Since $\fB$ is a Cauchy filter base, there exists $M \in \fB$ such that $M \times M \subset W$. Let $(x, y) \in W(M) \times W(M)$, then there exists $(p, q) \in M \times M$ such that $(p, x), (q, y) \in W$. As $M \times M$ \subset $W$, $(x, y) \in W \circ W \circ W$ by symmetry of $W$ and \ref{lemma:compositiongymnastics}. Therefore $W(M) \times W(M) \subset V$ and $\mathfrak{M}$ is a Cauchy filter base.
Finally, let $\fF' \subset \fF$ be a Cauchy filter. Let $M \in \fB$ and $V \in \fV$, then there exists $N \in \fF'$ such that $N \times N \subset V$. Since $N \in \fF$, $M \cap N \ne \emptyset$, and there exists $y \in M \cap N$. For any $x \in N$, $N \times N \subset V$ implies that $x \in V(y)$. Hence $N \subset V(M)$ and $\fF' \supset \fF_0$.
\end{proof}
\begin{proposition}[{{\cite[Corollary 2.3.4]{Bourbaki}}}]
\label{proposition:cauchyinterior}
Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a minimal Cauchy filter, then $\fF$ admits a base consisting of open sets.
\end{proposition}
\begin{proof}
Let $\fV \subset \fU$ be the set of all symmetric, open entourages. By \ref{proposition:goodentourages}, $\fV$ is a fundamental system of entourages. By \ref{proposition:minimalcauchyexistence}, $\mathfrak{M} = \bracs{U(M)| U \in \fV, M \in \fF}$ is a base for $\fF$.
Let $V \in \fU$, then there exists $U \in \fV$ such that $U \subset V$. For any $M \in \fF$, $U(M)$ is open by \ref{lemma:openentourageneighbourhoods}. Thus $\mathfrak{M}$ consists of open sets.
\end{proof}
\begin{proposition}[{{\cite[Corollary 2.3.1, 2.3.2]{Bourbaki}}}]
\label{proposition:cauchyfilterlimit}
Let $(X, \fU)$ be a uniform space, then:
\begin{enumerate}
\item For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter.
\item For each Cauchy filter $\fF \subset 2^X$ and $x \in X$, $x$ is an accumulation point of $\fF$ if and only if $\fF$ converges to $x$.
\item For any $x \in X$, filter $\fF \subset 2^X$ converging to $x$, and Cauchy filter $\mathfrak{G} \subset \fF$, $\mathfrak{G}$ converges to $x$.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $\fF = \bracs{E \subset X: x \in E}$, then $\fF$ is a filter generated by the filter base $\fB = \bracs{\bracs{x}}$. By \ref{proposition:minimalcauchyexistence}, $\bracs{U(x): U \in \fU} = \cn(x)$ is the unique minimal filter contained in $\fF$.
(2): Let $x \in X$ be an accumulation point of $\fF$, then $U(x) \cap E \ne \emptyset$ for all $U \in \fU$ and $E \in \fF$. By \ref{definition:generatedfilter}, there exists a filter $\fF' \supset \fF \cup \cn(x)$. Since $\fF$ is a Cauchy filter, so is $\fF'$. By minimality of $\cn(x)$, $\fF \supset \cn(x)$.
(3): Since $\fF$ converges to $x$, $x$ is a cluster point of $\fF$ and $\mathfrak{G}$. By (2), $x$ is also a limit point of $\mathfrak{G}$.
\end{proof}
\begin{definition}[Complete Space]
\label{definition:completespace}
Let $(X, \fU)$ be a uniform space, then $X$ is \textbf{complete} if every Cauchy filter converges to at least one point.

1
src/topology/uniform/index.tex
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@@ -3,6 +3,7 @@
\input{./src/topology/uniform/definition.tex}
\input{./src/topology/uniform/uc.tex}
\input{./src/topology/uniform/metric.tex}
\input{./src/topology/uniform/cauchy.tex}
\input{./src/topology/uniform/complete.tex}
\input{./src/topology/uniform/completion.tex}

245
src/topology/uniform/metric.tex Normal file
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@@ -0,0 +1,245 @@
\section{Pseudometrics}
\label{section:pseudometric}
The axioms of uniform spaces strongly resembles working in a metric space. In fact, any uniform space may arise from a family of uniformly continuous pseudometrics. This allows understanding uniform spaces in a more familiar languagge.
\begin{definition}[Pseudometric]
\label{definition:pseudometric}
Let $X$ be a set, a \textbf{pseudometric} is a mapping $d: X \times X \to [0, \infty)$ such that:
\begin{enumerate}
\item[(PM1)] For any $x \in X$, $d(x, x) = 0$.
\item[(PM2)] For any $x, y \in X$, $d(x, y) = d(y, x)$.
\item[(PM3)] For any $x, y, z \in X$, $d(x, z) \le d(x, y) + d(y, z)$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma:pseudometric-continuous}
Let $(X, \fU)$ be a uniform space and $d: X \times X \to [0, \infty)$ be a pseudometric, then the following are equivalent:
\begin{enumerate}
\item $d \in UC(X \times X; [0, \infty))$.
\item For each $r > 0$, $U_r = \bracs{(x, y) \in X \times X| d(x, y) < r} \in \fU$.
\end{enumerate}
\end{lemma}
\begin{proof}
(1) $\Rightarrow$ (2): Let $r > 0$, then there exists $V \in \fU$ symmetric such that for any $(x, x'), (y, y') \in V$, $\abs{d(x, y) - d(x', y')} < r$. In particular, for any $(x, y) \in V$, $(x, x), (x, y) \in V$. Thus $d(x, y) < d(x, x) + r = r$. Therefore $V \subset U_r$, and $U_r \in \fU$.
(2) $\Rightarrow$ (1): Let $r > 0$, then for any $(x, x'), (y, y') \in U_{r/2}$, $\abs{d(x, y) - d(x', y')} < r$ by the triangle inequality.
\end{proof}
\begin{definition}[Pseudometric Uniformity]
\label{definition:pseudometric-uniformity}
Let $X$ be a set and $\seqi{d}$ be pseudometric on $X$. For each $i \in I$, $r > 0$, and $x \in X$, let
\[
B_i(x, r) = \bracs{y \in X| d(x, y) < r}
\]
and
\[
U_{i, r} = \bracs{(x, y) \in X \times X| d(x, y) < r}
\]
then there exists a uniformity $\fU$ on $X$ such that:
\begin{enumerate}
\item The family
\[
\fB = \bracs{\bigcap_{j \in J}U_{j, r} \bigg | J \subset I \text{ finite}, r > 0}
\]
forms a fundamental system of entourages consisting of symmetric open sets.
\item For any $x \in X$,
\[
\cb(x) = \bracs{\bigcap_{j \in J}B_j(x, r) \bigg | J \subset I \text{ finite}, r > 0}
\]
is a fundamental system of neighbourhoods at $x$.
\item For each $U \subset X$, $U$ is open if and only if for every $x \in U$, there exists $J \subset I$ finite and $r > 0$ such that $\bigcap_{j \in J}B_j(x, r) \subset U$.
\item For each $i \in I$, $d_i \in UC(X \times X; [0, \infty))$.
\item[(U)] For any other uniformity $\mathfrak{V}$ satisfying (4), $\mathfrak{U} \subset \mathfrak{V}$.
\end{enumerate}
The uniformity $\fU$ is the \textbf{pseudometric uniformity} induced by $\seqi{d}$, and the topology induced by $\fU$ is the \textbf{pseudometric topology} on $X$ induced by $\seqi{d}$.
\end{definition}
\begin{proof}
(1, fundamental system): To see that $\fB$ is a fundamental system of entourages for a uniformity on $X$, it is sufficient to verify the conditions of \ref{proposition:fundamental-entourage-criterion}.
\begin{enumerate}
\item[(FB1)] For any $J, J' \subset I$ finite and $r, r' > 0$,
\[
\bigcap_{j \in J \cup J'}U_{j, \min(r, r')} \subset \paren{\bigcap_{j \in J}U_{j, r}} \cap \paren{\bigcap_{j \in J'}U_{j, r'}}
\]
\item[(UB1)] For each $i \in I$, $d(x, x) = 0$ for all $x \in X$. Thus for any $i \in I$ and $r > 0$, $U_{i, r}$ contains the diagonal.
\item[(UB2)] For each $J \subset I$ finite and $r > 0$,
\[
\paren{\bigcap_{j \in J}U_{j, r/2}} \circ \paren{\bigcap_{j \in J}U_{j, r}} \subset \bigcap_{j \in J}U_{j, r/2} \circ U_{j, r/2} \subset \bigcap_{j \in J}U_{j, r}
\]
by the triangle inequality.
\end{enumerate}
(2): Since $\fB$ is a fundamental system of entourages for $\fU$,
\[
\cb(x) = \bracs{U(x)| U \in \fB}
\]
is a fundamental system of neighbourhoods at $x$.
(3): By definition of the uniform topology, for any $U \subset X$, $U$ is open if and only if for any $x \in U$, there exists $V \in \fU$ such that $x \in V(x) \subset U$. As $\fB$ is a fundamental system of entourages for $\fU$, this is equivalent to the existence of $J \subset I$ finite and $r > 0$ such that
\[
x \in \paren{\bigcap_{j \in J}U_{j, r}}(x) = \bigcap_{j \in J}B_j(x, r) \subset U
\]
(1, symmetric open): Let $i \in I$ and $r > 0$. Since $d_i$ is symmetric, so is $U_{i, r}$. For any $(x, y) \in U_{i, r}$, let $s = r - d_i(x, y)$, then for any $(x', y') \in X \times X$ with $d_i(x, x') < s/2$ and $d_i(y, y') < s/2$, $d(x', y') < s + d_i(x, y) = r$. Thus $B_i(x, s/2) \times B_i(y, s/2) \subset U_{i, r}$.
By (3), $B_i(x, s/2) \in \cn(x)$ and $B_i(y, s/2) \in \cn(y)$, so $B_i(x, s/2) \times B_i(y, s/2) \in \cn((x, y))$. As such, $U_{i, r}$ is open by \ref{lemma:openneighbourhood}.
(4): By \ref{lemma:pseudometric-continuous}.
(5): By \ref{lemma:pseudometric-continuous}, $U_{i, r} \in \mathfrak{V}$ for all $i \in I$ and $r > 0$, so $\mathfrak{V} \supset \mathfrak{B}$ by (F2), and $\mathfrak{V} \supset \mathfrak{U}$.
\end{proof}
\begin{lemma}
\label{lemma:uniform-sequence-pseudometric}
Let $(X, \fU)$ be a uniform space and $\bracsn{U_n}_0^\infty \subset \fU$ such that:
\begin{enumerate}
\item[(a)] $U_{0} = X \times X$.
\item[(b)] For each $n \in \natz$, $U_n$ is symmetric.
\item[(c)] For each $n \in \natz$, $U_{n + 1} \circ U_{n+1} \subset U_n$.
\end{enumerate}
then there exists a pseudometric $d: X \times X \to [0, 1]$ such that
\[
U_{n+1} \subset \bracs{(x, y)| d(x, y) < 2^{-n}} \subset U_{n-1}
\]
for each $n \in \natp$.
\end{lemma}
\begin{proof}
Let
\[
\rho: X \times X \to [0, 1] \quad (x, y) \mapsto \inf\bracs{2^{-n}| n \in \natz, (x, y) \in U_n}
\]
and $d: X \times X \to [0, 1]$ by
\[
d(x, y) = \inf\bracs{\sum_{j = 1}^n\rho(x_{j-1}, x_j) \bigg | \seqf{x_j} \subset X, x_0 = x, x_1 = y, n \in \natp}
\]
then
\begin{enumerate}
\item[(PM1)] For any $x \in X$, $x \in \bigcap_{n \in \natp}U_n$. Thus $\rho(x, x) = 0$ and $d(x, x) \le \rho(x, x) = 0$.
\item[(PM2)] Let $x, y \in X$. By assumption (b), $\rho(x, y) = \rho(y, x)$. Thus $d(x, y) = d(y, x)$ as well.
\item[(PM3)] Let $x, y, z \in X$, then for any $\seqf{x_j}$ and $\seqf[m]{y_j}$ with $x_0 = x$, $x_n = y = y_0$, and $y_m = z$,
\[
d(x, z) \le \sum_{j = 1}^n \rho(x_{j - 1}, x_j) + \sum_{j = 1}^m \rho(y_{j - 1}, y_j)
\]
As this holds for all such $\seqf{x_j}$ and $\seqf[m]{y_j}$, $d(x, z) \le d(x, y) + d(y, z)$.
\end{enumerate}
so $d$ is a pseudometric.
For any $(x, y) \in U_{n+1}$, $d(x, y) \le \rho(x, y) < 2^{-n}$, so $U_{n+1} \subset \bracs{(x, y)| d(x, y) \le 2^{-n}}$.
Let $x, y \in X$ with $d(x, y) < 2^{-n}$. If $\rho(x, y) < 2^{-n}$, then the claim holds directly. Assume inductively that for any $x, y \in X$ with $d(x, y) < 2^{-n}$, if there exists $\seqf[m]{x_j} \subset X$ such that $x_0 = x$, $x_m = y$ and $\sum_{j = 1}^m \rho(x_{j - 1}, x_j) < 2^{-n}$, then $(x, y) \in U_{n - 1}$.
Let $x, y \in X$ and $\seqf[m+1]{x_j} \subset X$ such that $x_0 = x$, $x_{m+1} = y$, and $\sum_{j = 1}^{m+1} \rho(x_{j - 1}, x_j) < 2^{-n}$. Let $1 \le k < m+1$ such that $\sum_{j = 1}^{k}\rho(x_{j - 1}, x_j) < 2^{-n-1}$ and $\sum_{j = 1}^{k+1}\rho(x_{j-1}, x_j) \ge 2^{-n-1}$, then $\sum_{j = k + 1}^{m+1}\rho(x_{j-1}, x_j) < 2^{-n-1}$ as well. By the inductive hypothesis, $(x, x_k), (x_{k+1}, y) \in U_{n+1} \subset U_n$. Given that $\rho(x_k, x_{k+1}) < 2^{-n}$, $(x_k, x_{k+1}) \in U_{n+1}$ too. Thus
\[
(x, y) \subset U_{n+1} \circ U_{n + 1} \circ U_{n + 1} \subset U_n \circ U_n \subset U_{n-1}
\]
\end{proof}
\begin{remark}
Unfortunately, it was hard to access sources for this proof online, so I could not provide a specific citation. As such, I followed an online PDF given by the link: https://krex.k-state.edu/server/api/core/bitstreams/1bdf2b14-3b5a-4962-a589-93ee1998950c/content
\end{remark}
\begin{remark}
It may be tempting to construst the level sets of the pseudometric on the dyadic rational numbers by composing these sets, then proceed to construct the pseudometric as in Urysohn's lemma. However, this approach has a major shortcoming in that the composition of symmetric entourages are not necessarily symmetric. As such, it is difficult to construct symmetric level sets for the desired pseudometric.
\end{remark}
\begin{theorem}
\label{theorem:uniform-pseudometric}
Let $(X, \fU)$ be a uniform space, then $\fU$ is the pseudometric uniformity induced by the family of all uniformly continuous pseudometrics on $X$.
\end{theorem}
\begin{proof}
Let $\seqi{d}$ be the family of all uniformly continuous pseudometrics on $X$, and $\mathfrak{V}$ be the pseudometric uniformity induced by $\seqi{d}$. By (U) of \ref{definition:pseudometric-uniformity}, $\fU \supset \mathfrak{V}$.
On the other hand, let $U_1 \in \mathfrak{U}$. By (U3), there exists $\seq{U_n} \subset \mathfrak{U}$ such that $U_{n + 1} \circ U_{n + 1} \subset U_n$ for all $n \in \natp$. Let $U_0 = X \times X$, then $\bracsn{U_n}_0^\infty \subset \fU$ satisfies the hypothesis of \ref{lemma:uniform-sequence-pseudometric}. Thus there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for all $n \in \natp$,
\[
U_{n + 1} \subset \bracs{(x, y) \in X \times X|d(x, y) < 2^{-n}} \subset U_{n-1}
\]
By \ref{definition:pseudometric-uniformity}, $d$ is a uniformly continuous pseudometric on $X$. Since $\bracs{(x, y) \in X \times X|d(x, y) < 1/4} \subset U_1$, $U_1 \in \mathfrak{V}$. Therefore $\fU = \mathfrak{V}$.
\end{proof}
\begin{definition}[Equivalent Pseudometrics]
\label{definition:equivalent-pseudometrics}
Let $X$ be a set and $\seqi{d}, \seqj{d}$ be pseudometrics on $X$, then $\seqi{d}$ and $\seqj{d}$ are \textbf{equivalent} if their induced uniformities coincide.
\end{definition}
\begin{lemma}
\label{lemma:pseudometric-clamp}
Let $X$ be a set and $d: X \times X \to [0, \infty)$ be a pseudometric, then the pseudometric
\[
\td d: X \times X \to [0, \infty) \quad (x, y) \mapsto d(x, y) \wedge 1
\]
is equivalent to $d$.
\end{lemma}
\begin{proof}
For any $r \in (0, 1]$,
\[
\bracsn{(x, y) \in X \times X| d(x, y) < r} = \bracsn{(x, y) \in X \times X| \td d(x, y) < r}
\]
Since sets of the above form generate the uniformity induced by $d$ and $\td d$, their induced uniformities coincide.
\end{proof}
\begin{proposition}
\label{proposition:countable-equivalent-pseudometrics}
Let $X$ be a set and $\seq{d_n}$ be pseudometrics on $X$, then there exists a pseudometric $d: X \times X \to [0, \infty)$ equivalent to $\seq{d_n}$.
\end{proposition}
\begin{proof}
Using \ref{lemma:pseudometric-clamp}, assume without loss of generality that for each $n \in \natp$, $d_n$ takes values in $[0, 1]$. Let
\[
d(x, y) = \sum_{n \in \natp}\frac{d_n(x, y)}{2^n}
\]
then $d$ is a well-defined a pseudometric.
For each $n \in \nat$ and $r > 0$, denote
\begin{align*}
U_{n, r} &= \bracs{(x, y) \in X \times X| d_n(x, y)< r} \\
U_{r} &= \bracs{(x, y) \in X \times X|d(x, y) < r}
\end{align*}
Let $r > 0$, then there exists $n \in \natp$ such that $2^{-n} < r$. Take $s = r - 2^{-n}$, then $\bigcap_{k = 1}^nU_{k, s} \subset U_{r}$. On the other hand, for any $n \in \natp$ and $r > 0$, $U_{r/2^n} \subset \bigcap_{k = 1}^n U_{k, r}$. Therefore $\seq{d_n}$ and $d$ are equivalent.
\end{proof}
\begin{theorem}[Metrisability of Uniform Spaces]
\label{theorem:uniform-metrisable}
Let $(X, \fU)$ be a uniform space, then the following are equivalent:
\begin{enumerate}
\item There exists a countable fundamental system of entourages for $X$.
\item There exists a pseudometric $d: X \times X \to [0, \infty)$ that induces the uniformity on $X$.
\item There exists a countable family $\seq{d_n}$ of pseudometrics on $X$ that induce the uniformity on $X$.
\end{enumerate}
\end{theorem}
\begin{proof}
(1) $\Rightarrow$ (2): Let $\seq{U_n} \subset \fU$ be a fundamental system of entourages for $X$ and $V_1 = U_n$. Assume inductively that $\bracs{V_k|1 \le k \le n} \subset \fU$ has been constructed such that
\begin{enumerate}
\item[(a)] For each $1 \le k \le n$, $V_k$ is symmetric.
\item[(b)] For each $1 \le k \le n$, $V_k \subset U_k$.
\item[(c)] For each $1 \le k < n$, $V_{k+1} \circ V_{k+1} \subset V_{k}$.
\end{enumerate}
Let $W = V_n \cap U_{n+1}$, then by \ref{lemma:symmetricfundamentalentourage}, there exists $V_{n+1} \in \fU$ symmetric such that $V_{n+1} \circ V_{n+1} \subset W$. Thus $\bracs{V_k|1 \le k \le n + 1} \subset \fU$ satisfies (a), (b), and (c) for $n + 1$.
Let $V_0 = X \times X$, then by \ref{lemma:uniform-sequence-pseudometric}, there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for each $n \in \natp$,
\[
V_{n+1} \subset \bracs{(x, y) \in X \times X| d(x, y) < 2^{-n}} \subset V_{n-1}
\]
For any $U \in \fU$, there exists $n \in \nat$ such that
\[
U \supset U_n \supset V_n \supset \bracs{(x, y) \in X \times X| d(x, y) < 2^{-n-1}}
\]
so $d$ induces the uniformity on $\fU$.
(3) $\Rightarrow$ (1): By (1) of \ref{definition:pseudometric-uniformity},
\[
\fB = \bracs{\bigcap_{j \in J}U_{j, r} \bigg | J \subset \nat \text{ finite}, r > 0}
\]
where
\[
U_{j, r} = \bracs{(x, y) \in X \times X| d_n(x, y) < r}
\]
is a fundamental system of entourages for $\fU$. Since for any $r > 0$, there exists $q \in \rational \cap (0, r)$,
\[
\bracs{\bigcap_{j \in J}U_{j, r} \bigg | J \subset \nat \text{ finite}, r \in \rational, r > 0}
\]
is a countable fundamental system of entourages for $\fU$.
\end{proof}