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\part{General Topology}
\label{part:-part-topology}
\input{./src/topology/main/index.tex}
\input{./src/topology/uniform/index.tex}

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\section{Baire Spaces}
\label{section:baire}
\begin{definition}[Baire Space]
\label{definition:baire}
Let $X$ be a topological space, then the following are equivalent:
\begin{enumerate}
\item For any $\seq{A_n}$ nowhere dense, $\bigcup_{n \in \nat}A_n \subsetneq X$.
\item For any $\seq{A_n}$ closed with empty interior, $\bigcup_{n \in \nat}A_n \subsetneq X$.
\item For any $\seq{A_n}$ closed with $\bigcup_{n \in \nat}A_n = X$, there exists $N \in \nat$ such that $\bigcup_{n \le N}A_n$ has non-empty interior.
\item For any $\seq{U_n}$ open and dense, $\bigcap_{n \in \nat}U_n$ is dense.
\end{enumerate}
If the above holds, then $X$ is a \textbf{Baire space}.
\end{definition}
\begin{theorem}[Baire Category Theorem]
\label{theorem:baire}
Let $X$ be a topological space, then the following are sufficient conditions for $X$ to be Baire:
\begin{enumerate}
\item $X$ is completely metrisable.
\item $X$ is locally compact.
\end{enumerate}
\end{theorem}

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\section{Continuous Maps}
\label{section:continuity}
\begin{definition}[Continuity]
\label{definition:continuity}
Let $X$ and $Y$ be topological spaces, $f: X \to Y$ be a function, and $x \in X$, then the following are equivalent:
\begin{enumerate}
\item For each $V \in \cn(f(x))$, $f^{-1}(V) \in \cn(x)$.
\item For each filter base $\fB \subset 2^X$ converging to $x$, $f(\fB)$ converges to $f(x)$.
\end{enumerate}
If the above holds, then $f$ is \textbf{continuous at} $x \in X$.
The following are also equivalent:
\begin{enumerate}
\item For each $U \subset Y$ open, $f^{-1}(U)$ is open in $X$.
\item $f$ is continuous at every $x \in X$.
\item For each convergent filter base $\fB \subset 2^X$, $f(\fB)$ is convergent.
\end{enumerate}
If the above holds, then $f$ is \textbf{continuous}.
The collection $C(X; Y)$ is the space of all continuous functions from $X$ to $Y$.
\end{definition}
\begin{proof}
Local continuity, $(1) \Rightarrow (2)$: Let $V \in \cn(f(x))$, then $f^{-1}(V) \in \cn(x)$. Since $\fB$ converges to $x$, there exists $B \in \fB$ with $B \subset f^{-1}(V)$. In which case, $f(B) \subset V$ and $V \in \fF(\fB)$ by (F1).
Local continuity, $(2) \Rightarrow (1)$: Since $f(\cn(x))$ converges to $f(x)$, for each $U \in \cn(f(x))$, there exists $V \in \cn(x)$ such that $f(V) \subset V$. Thus $V \subset f^{-1}(U) \in \cn(x)$ by (F1).
Global continuity, $(1) \Rightarrow (2)$: Let $U \subset Y$ be open, then $U \in \cn(f(x))$ for all $x \in f^{-1}(U)$. Hence $f^{-1}(U) \in \cn(x)$ for all $x \in f^{-1}(U)$. Thus $U$ is open by \ref{lemma:openneighbourhood}.
Global continuity, $(2) \Rightarrow (1)$: Let $x \in X$ and $V \in \cn(f(x))$, then there exists $U \in \cn(f(x))$ open, so $f^{-1}(V) \supset f^{-1}(U)$ contains an open set that contains $x$. Therefore $f^{-1}(V) \in \cn(x)$.
Global continuity, $(2) \Leftrightarrow (3)$: Follows from local continuity.
\end{proof}

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\section{Definitions}
\label{section:top-definitions}
\begin{definition}[Topological Space]
\label{definition:topspace}
Let $X$ be a non-empty set. A \textbf{topology} over $X$ is a family $\topo \subset 2^X$ such that
\begin{enumerate}
\item[(O1)] $\emptyset \in \topo$ and $X \in \topo$.
\item[(O2)] For any $U, V \in \topo$, $U \cap V \in \topo$.
\item[(O3)] For any $\seqi{U} \subset \topo$, $\bigcup_{i \in I}U_i \in \topo$.
\end{enumerate}
The elements of $\topo$ are known as \textbf{open sets}, and the pair $(X, \topo)$ is known as a \textbf{topological space}.
\end{definition}
\begin{definition}[Closed Set]
\label{definition:closedset}
Let $(X, \topo)$ be a topological space, then $A \subset X$ is \textbf{closed} if $A^c \in \topo$.
\end{definition}
\begin{definition}[Separation Axioms]
\label{definition:separation}
Let $(X, \topo)$ be a topological space, then $X$ may satisfy the following \textbf{separation axioms}:
\begin{enumerate}
\item[(T0)] For any $x, y \in X$ with $x \ne y$, there exists $U \in \topo$ with $x \in U$ and $y \not\in U$, or there exists $U \in \topo$ with $x \not\in U$ and $y \in U$.
\item[(T1)] For any $x, y \in X$ with $x \ne y$, there exists $U \in \topo$ with $x \in U$ and $y \not\in U$.
\item[(T2)] For any $x, y \in X$ with $x \ne y$, there exists $U, V \in \topo$ with $x \in U$, $y \in V$, and $U \cap V = \emptyset$.
\item[(T3)] $X$ is (T1), and for any $x \in X$ and $A \subset X$ closed with $x \not\in A$, there exists $U, V \in \topo$ such that $x \in U$, $A \subset V$, and $U \cap V = \emptyset$.
\item[(T4)] $X$ is (T1), and for any $A, B \subset X$ closed with $A \cap B = \emptyset$, there exists $U, V \in \topo$ such that $A \subset U$, $B \subset V$, and $U \cap V = \emptyset$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma:t1}
Let $X$ be a topological space, then $X$ satisfies (T1) if and only if $\bracs{x}$ is closed for all $x \in X$.
\end{lemma}
\begin{proof}
$(\Rightarrow)$: Let $x \in X$, then for each $y \in X \setminus {x}$, there exists $U_y \subset X$ open such that $x \not\in U_y$. Thus $U^c = \bigcup_{y \in X \setminus \bracs{x}}U_y$ is open.
$(\Leftarrow)$: Let $x, y \in X$ with $x \ne y$, then $y \in \bracs{x}^c$, $x \not\in \bracs{x}^c$, and $\bracs{x}^c$ is open.
\end{proof}
\begin{definition}[Base]
\label{definition:base}
Let $(X, \topo)$ be a topological space, then a \textbf{base} for $\topo$ is a family $\cb \subset \topo$ such that:
\begin{enumerate}
\item For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
\item For every $x \in X$ and $U \subset X$ open with $x \in U$, there exists $V \in \cb$ such that $x \in V \subset U$.
\end{enumerate}
In which case,
\[
\topo = \topo(\cb) = \bracs{\bigcup_{i \in I}U_i \bigg | \seqi{U} \subset \cb, I \text{ index set}}
\]
Conversely, if $\cb \subset 2^X$ is a family such that:
\begin{enumerate}
\item[(TB1)] For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
\item[(TB2)] For every $x \in X$ and $U, V \in \cb$ such that $x \in U \cap$, there exists $W \in \cb$ such that $x \in W \subset U \cap V$.
\end{enumerate}
then $\topo(\cb)$ is a topology on $X$, and $\cb$ is a base for $\topo(\cb)$.
\end{definition}
\begin{proof}
Let $U \in \topo$. If $U = \emptyset$, then $U$ is a union over an empty index set. Otherwise, for each $x \in U$, there exists $V_x \in \cb$ such that $x \in V_x \subset U$. In which case, $U = \bigcup_{x \in U}V_x \in \topo$ and $\topo \subset \topo(\cb)$. On the other hand, since $\cb \subset \topo$, $\topo \supset \topo(\cb)$ by (O3).
For the converse, $\emptyset \in \topo(\cb)$ as a union over an empty index set and $X = \bigcup_{U \in \cb}U \in \topo(\cb)$ by (TB1). For any $\seqi{U}, \seqj{V} \subset \cb$,
\[
\paren{\bigcup_{i \in I}U_i} \cap \paren{\bigcup_{j \in J}V_j} = \bigcup_{i \in I}\bigcup_{j \in J}U_i \cap V_j
\]
Thus to show that $\topo(\cb)$ satisfies (O2), it is sufficient to show that $U \cap V \in \topo(\cb)$ for all $U, V \in \cb$. To this end, for every $x \in U \cap V$, there exists $W_x \in \cb$ such that $x \in W_x \subset U \cap V$. Therefore $U \cap V = \bigcup_{x \in U \cap V}W_x \in \topo(\cb)$, and $\topo(\cb)$ satisfies (O2).
By definition, $\topo(\cb)$ satisfies (O3).
\end{proof}
\begin{definition}[Generated Topology]
\label{definition:generated-topology}
Let $X$ be a set and $\ce \subset 2^X$ such that $\bigcup_{U \in \ce}U = X$, then the smallest topology on $X$ containing $\ce$ is given by
\[
\topo(\ce) = \bracs{\bigcup_{i \in I}U_i \bigg | U_i \in \cb(\ce)}
\]
where
\[
\cb(\ce) = \bracs{\bigcap_{j = 1}^n U_j \bigg | \seqf{U_j} \subset \ce, n \in \nat^+}
\]
is a base for $\topo(\ce)$. The topology $\topo(\ce)$ is known as the topology \textbf{generated by} $\ce$.
\end{definition}
\begin{proof}
Since $\cb(\ce) \supset \ce$, $\cb(\ce)$ satisfies (TB1). In addition, $\cb(\ce)$ is closed under finite intersections, so it satisfies (TB2). By \ref{definition:base}, $\cb(\ce)$ is a base for $\topo(\ce)$.
\end{proof}
\begin{definition}[Initial Topology]
\label{definition:initial-topology}
Let $X$ be a set, $\bracsn{(Y_j, \topo_i)}$ be a family of topological spaces, and $\seqi{f}$ be a family of maps such that $f_i: X \to Y_i$ for each $i \in I$, then there exists a topology $\topo$ on $X$ such that:
\begin{enumerate}
\item For each $i \in I$, $f_i \in C(X; Y_i)$.
\item[(U)] If $\mathcal{S}$ is a topology on $X$ satisfying $(1)$, then $\mathcal{S} \supset \topo$.
\item The family
\[
\mathcal{B} = \bracs{\bigcap_{j \in J}f_j^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \topo_j}
\]
is a base for $\topo$.
\end{enumerate}
The topology $\topo$ is known a the \textbf{initial/weak topology} generated by the maps $\seqi{f}$.
\end{definition}
\begin{proof}
Let $\topo$ be the topology genereated by sets of the form $\ce = \bracs{f_i^{-1}(U_i)| i \in I, U_i \in \topo_i}$. Let $\topo$ be the topology generated by $\ce$, then
\begin{enumerate}
\item For each $i \in I$, $\topo \supset \bracs{f_i^{-1}(U)|U \in \topo_i}$, so $f_i \in C(\topo; Y_i)$.
\item If $\mathcal{S}$ is a topology such that $f_i \in C(X, \mathcal{S}; Y_i)$, then $\bracs{f_i^{-1}(U)|U \in \topo_i} \subset \mathcal{S}$. Thus $\ce \subset \mathcal{S}$ and $\mathcal{S} \supset \topo$.
\item By \ref{definition:generated-topology}, $\cb$ is a base for $\topo$.
\end{enumerate}
\end{proof}

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\section{Filters}
\label{section:filters}
\begin{definition}[Filter]
\label{definition:filter}
Let $X$ be a set, a \textbf{filter} $\fF \subset X^2$ is a non-empty family of sets such that:
\begin{enumerate}
\item[(F1)] For any $E \in \fF$ and $X \supset F \supset E$, $F \in \fF$.
\item[(F2)] For any $E, F \in \fF$, $E \cap F \in \fF$.
\item[(F3)] $\emptyset \not\in \fF$
\end{enumerate}
\end{definition}
\begin{definition}[Filter Base]
\label{definition:filterbase}
Let $X$ be a set, $\fF \subset 2^X$ be a filter, and $\fB \subset \fF$, then $\fB$ is a \textbf{filter base} for $\fF$ if for every $F \in \fF$, there exists $E \in \fB$ such that $E \subset F$.
\end{definition}
\begin{proposition}
\label{proposition:filterbasecriterion}
Let $X$ be a set, $\fF \subset 2^X$ be a filter, and $\fB \subset \fF$ be a filter base, then:
\begin{enumerate}
\item[(FB1)] For any $E, F \in \fB$, there exists $G \in \fB$ such that $G \subset E \cap F$.
\item[(FB2)] $\emptyset \not\in \fB$.
\end{enumerate}
Conversely, if $\fB \subset 2^X$ is a non-empty collection that satisfies (FB1) and (FB2), then $\fB$ is a base for the filter
\[
\fF = \bracs{F \subset X| \exists E \in \fB: E \subset F}
\]
\end{proposition}
\begin{proof}
Filter Base $\Rightarrow$ (FB1): Let $E, F \in \fB$, then $E \cap F \in \fF$. Thus there exists $G \in \fB$ such that $G \subset E \cap F$.
Filter Base $\Rightarrow$ (FB2): By (F3), $\emptyset \not\in \fB \subset \fU$.
(FB1) + (FB2) $\Rightarrow$ (F1): Let $E \in \fF$, then there exists $E_0 \in \fB$ such that $E_0 \subset E$. Thus for any $X \supset F \supset E$, $E_0 \subset F \in \fF$.
(FB1) + (FB2) $\Rightarrow$ (F2): Let $E, F \in \fF$, then there exists $E_0, F_0 \in \fB$ such that $E_0 \subset E$ and $F_0 \subset F$. By (FB1), there exists $G \in \fB$ such that $G \subset E_0 \cap F_0$, then $G \subset E \cap F \in \fF$.
(FB1) + (FB2) $\Rightarrow$ (F3): Let $E \in \fF$, then there exists $E_0 \in \fB$ such that $E_0 \subset E$. By (FB2), $E_0 \ne \emptyset$, so $E \ne \emptyset$ as well.
Finally, $\fB$ is a filter base for $\fF$ by definition.
\end{proof}
\begin{proposition}
\label{proposition:imagefilterbase}
Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^X$ be a filter base, then $f(\fB) = \bracs{f(E)| E \in \fB}$ is also a filter base.
\end{proposition}
\begin{proof}
(FB1): Let $E, F \in \fB$, then there exists $G \in \fB$ such that $G \subset E \cap F$. In which case, $f(G) \subset f(E \cap F) \subset f(E) \cap f(F)$.
(FB2): For any $f(E) \in f(\fB)$, $E \ne \emptyset$ implies that $f(E) \ne \emptyset$.
\end{proof}
\begin{proposition}
\label{proposition:preimage-filterbase}
Let $X, Y$ be sets, $f: X \to Y$ be a function, and $\fB \subset 2^X$ be a filter base. If $\emptyset \not\in f^{-1}(\fB) = \bracs{f^{-1}(E)| E \in \fB}$, then $f^{-1}(\fB)$ is also a filter base.
\end{proposition}
\begin{proof}
(FB1): Let $E, F \in \fB$, then there exists $G \in \fB$ such that $G \subset E \cap F$. In which case, $f^{-1}(G) \subset f^{-1}(E \cap F) = f^{-1}(E) \cap f^{-1}(F)$.
(FB2): By assumption, $\emptyset \not \in f^{-1}(\fB)$.
\end{proof}
\begin{definition}[Filter Subbase]
\label{definition:filtersubbase}
Let $X$ be a set and $\fB_0 \subset 2^X$ be a non-empty collection, then $\fB_0$ is a \textbf{filter subbase} if for any $\seqf{E_j} \subset \fB_0$, $\bigcap_{j = 1}^n E_j \ne \emptyset$.
\end{definition}
\begin{definition}[Generated Filter]
\label{definition:generatedfilter}
Let $X$ be a set and $\fB_0 \subset 2^X$ be a filter subbase, then there exists a filter containing $\fB_0$.
The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \textbf{generated by} $\fB_0$, which is given by $\fF(\fB_0) = \bracs{E \subset X| \exists F \in \fB: F \subset E}$, where
\[
\fB = \bracs{\bigcap_{j = 1}^n E_j \bigg | \seqf{E_j} \subset \fB_0, n \in \nat^+}
\]
\end{definition}
\begin{proof}
For any $\seqf{E_j}, \bracsn{F_j}_1^m \subset \fB_0$,
\[
G = \paren{\bigcap_{j = 1}^n E_j} \cap \paren{\bigcap_{j = 1}^m F_j} \in \fB
\]
Thus $\fB$ satisfies (FB1). Since $\bigcap_{j = 1}^n E_j \ne \emptyset$, $\emptyset \not\in \fB$, and $\fB$ satisfies (FB2).
By \ref{proposition:filterbasecriterion}, $\fB$ is a base for the filter $\fF = \bracs{E \subset X| \exists F \in \fB: F \subset E}$.
If $\fF' \supset \fB_0$ is a filter, then $\fF \supset \fB$ by (F2), and $\fF' \supset \fF$ by (F1). Thus $\fF$ is the smallest filtetr containing $\fB$.
\end{proof}
\begin{definition}[Convergence]
\label{definition:filterconvergence}
Let $X$ be a topological space, $\fB \subset 2^X$ be a filter base, and $x \in X$, then \textbf{$\fB$ converges} to $x$ if $\cn(x) \subset \fF(\fB)$.
If $A \subset X$ and $\fB \subset 2^A$, then $\fB$ converges to $x$ if $\fF(\fB) \supset \bracsn{U \cap A| U \in \cn(x)}$.
\end{definition}
\begin{definition}[Accumulation Point]
\label{definition:accumulationpoint}
Let $X$ be a topological space, $\fB \subset 2^X$ be a filter base, and $x \in X$, then $x$ is an \textbf{accumulation point} of $\fB$ if $x \in \bigcap_{E \in \fB}\ol{E}$.
\end{definition}
\begin{definition}[Limit]
\label{definition:limit}
Let $X, Y$ be topological spaces, $A \subset X$, and $f: A \to Y$ be a function. For any filter base $\fB \subset 2^A$, if $f(\fB)$ converges to $y \in Y$, then $y = \lim_{x, \fB}f(x)$ is a \textbf{limit} of $f$ with respect to $\fB$.
For any $x_0 \in \overline{A}$, let $\fF = \bracs{U \cap A| U \in \cn(x_0)}$ be the trace of $\cn(x_0)$ on $A$, then $\fF \subset 2^A$ is a filter. If $f(\fF)$ converges to $y \in Y$, then
\[
y = \lim_{x \to x_0 \\ x \in A}f(x)
\]
is a \textbf{limit} of $f$ at $y$ with respect to $A$. If $A = X$, then $x \in A$ may be omitted.
\end{definition}

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\section{Hausdorff Spaces}
\label{section:hausdorff}
\begin{definition}[Hausdorff]
\label{definition:hausdorff}
Let $X$ be a topological space, then the following are equivalent:
\begin{enumerate}
\item For any $x, y \in X$ with $x \ne y$, there exists $U \in \cn(x)$ and $V \in \cn(y)$ such that $U \cap V = \emptyset$.
\item For every $x \in X$, $\bracs{x} = \bigcap_{U \in \cn(x)}\overline{U}$.
\item Every convergent filter in $X$ has exactly one cluster point.
\item Every filter in $X$ converges to at most one point.
\item For any index set $I$, the diagonal $\Delta$ is closed in $X^I$.
\item The diagonal $\Delta$ is closed in $X \times X$.
\end{enumerate}
If the above holds, then $X$ is a \textbf{T2/Hausdorff} space.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: Let $y \in X \setminus \bracs{x}$, then there exists $U \in \cn(x)$ and $V \in \cn(y)$ such that $U \cap V = \emptyset$. By (2) of \ref{definition:closure}, $y \not\in \overline{U} \subset \bigcap_{U \in \cn(x)}\overline{U}$.
$(2) \Rightarrow (3)$: Let $\fF \subset 2^X$ be a filter and $x \in X$ such that $\cn(x) \subset \fF$, then
\[
\bracs{x} = \bigcap_{U \in \cn(x)}\ol{U} \supset \bigcap_{U \in \cn(x)}\ol{U} \supset \bracs{x}
\]
so $x$ is the only cluster point of $\fF$.
$(3) \Rightarrow (4)$: Let $\fF \subset 2^X$ be a filter. If $\fF$ converges to $x \in X$, then $x$ is a cluster point of $\fF$. As $\fF$ admits only one cluster point, $x$ is the only limit point of $\fF$.
$(4) \Rightarrow (5)$: Let $x \in \overline{\Delta}$, then by (4) of \ref{definition:closure}, there exists $\fF \subset 2^\Delta$ converging to $x$. Let $i, j \in I$, then for any $y \in \Delta$, $\pi_i(y) = \pi_j(y)$. Thus $\pi_i(\fF) = \pi_j(\fF)$. By \ref{proposition:productfilterconvergence}, $\pi_i(\fF)$ converges to $\pi_i(x)$ and $\pi_j(\fF)$ converges to $\pi_j(x)$. By assumption, $\pi_i(x) = \pi_j(x)$. Since this holds for all pairs $i, j \in I$, $x \in \Delta$.
$(5) \Rightarrow (6)$: Take $I = \bracs{1, 2}$.
$(6) \Rightarrow (1)$: Let $x, y \in X$ with $x \ne y$, then $\Delta^c \in \cn(x, y)$, and there exists $U \in \cn(x)$ and $V \in \cn(y)$ such that $(x, y) \in \pi_1^{-1}(U) \cap \pi_2^{-1}(V) \subset \Delta^c$. Thus $U \cap V = \emptyset$ contain the desired neighbourhoods.
\end{proof}
\begin{proposition}
\label{proposition:continuousextensionunique}
Let $X$ be a topological space, $Y$ be a Hausdorff space, $A \subset X$ be a dense subset, and $F, G \in C(X; Y)$. If $F|_A = G|_A$, then $F = G$.
\end{proposition}
\begin{proof}
Let $x \in X$. By (4) \ref{definition:closure}, there exists a filter base $\fB \subset 2^A$ that converges to $x$. By (2) of local continuity, $F(\fB)$ converges to $F(x)$ and $G(\fB)$ converges to $G(x)$. Since $F(\fB) = G(\fB)$, $F(x) = G(x)$ by (4) of \ref{definition:hausdorff}.
\end{proof}
\begin{proposition}
\label{proposition:product-hausdorff}
Let $\seqi{X}$ be Hausdorff spaces, then $\prod_{i \in I}X_i$ is Hausdorff.
\end{proposition}
\begin{proof}
Let $x, y \in \prod_{i \in I}X_i$ with $x \ne y$, then there exists $i \in I$ such that $\pi_i(x) \ne \pi_i(y)$. In which case, there exists $U \in \cn(\pi_i(x))$ and $V \in \cn(\pi_i(y))$ with $U \cap V = \emptyset$. Thus $\pi_i^{-1}(U) \in \cn(x)$, $\pi_i^{-1}(V) \in \cn(y)$, and $\pi_i^{-1}(U) \cap \pi_i^{-1}(V) = \emptyset$.
\end{proof}

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\chapter{Topological Spaces}
\label{chap:topological-spaces}
\input{./src/topology/main/definition.tex}
\input{./src/topology/main/filters.tex}
\input{./src/topology/main/nets.tex}
\input{./src/topology/main/neighbourhoods.tex}
\input{./src/topology/main/interiorclosureboundary.tex}
\input{./src/topology/main/continuity.tex}
\input{./src/topology/main/product.tex}
\input{./src/topology/main/hausdorff.tex}
\input{./src/topology/main/regular.tex}
\input{./src/topology/main/metric.tex}
\input{./src/topology/main/baire.tex}

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\section{Interior, Closure, Boundary}
\label{section:icb}
\begin{definition}[Interior]
\label{definition:interior}
Let $X$ be a topological space, $A \subset X$, $x \in X$, and $\fB \subset \cn(x)$ be a fundamental system of neighbourhoods, then the following are equivalent:
\begin{enumerate}
\item $A \in \cn(x)$.
\item There exists $U \in \fB$ with $U \subset A$.
\item There exists $U \in \cn(x)$ with $U \subset A$.
\end{enumerate}
The set of all points satisfying the above is the \textbf{interior} $A^o$ of $A$, which is the largest open set contained in $A$.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: Since $\fB$ is a fundamental system of neighbourhoods, there exists $U \in \fB$ with $U \subset A$.
$(2) \Rightarrow (3)$: $\fB \subset \cn(x)$.
$(3) \Rightarrow (1)$: By (F1) of $\cn(x)$, $A \in \cn(x)$.
Let $U \subset A$ be open, then $U \in \cn(x)$ for all $x \in U$ by \ref{lemma:openneighbourhood}. By (3), $U \subset A^o$, so $A^o$ contains every open subset of $A$. On the other hand, for any $x \in A^o$, (2) implies that there exists $U \in \cn^o(x)$ with $U \subset A$. In which case, $U \subset A^o$, and $A^o \in \cn(x)$. Therefore $A^o$ itself is open by \ref{lemma:openneighbourhood}.
\end{proof}
\begin{definition}[Closure]
\label{definition:closure}
Let $X$ be a topological space, $A \subset X$, $x \in X$, and $\fB \subset \cn(x)$ be a fundamental system of neighbourhoods, then the following are equivalent:
\begin{enumerate}
\item For every $B \supset A$ closed, $x \in B$.
\item For every $U \in \cn(x)$, $U \cap A \ne \emptyset$.
\item For every $U \in \fB$, $U \cap A \ne \emptyset$.
\item There exists a filter $\fF \subset 2^A$ that converges to $\fF$.
\end{enumerate}
The set $\ol{A}$ of all points satisfying the above is the \textbf{closure} of $A$ in $X$. By (1), it is the smallest closed set containing $A$.
\end{definition}
\begin{proof}
$\neg (2) \Rightarrow \neg (1)$: Let $U \in \cn(x)$ such that $U \cap A = \emptyset$ and $V \in \cn^o(x)$ with $V \subset U$. Since $x \in V$ by (V1), $V^c \supset A$ is closed with $x \not\in V^c$.
$(2) \Rightarrow (3)$: $\cn(x) \supset \fB$.
$(3) \Rightarrow (4)$: Let $\fF = \bracs{V \cap A| V \in \cn(x)}$. For any $V \cap A \in \fF$, there exists $W \in \fB$ such that $W \subset V$. Since $W \cap A \ne \emptyset$, $V \cap A \ne \emptyset$ as well. Thus $\emptyset \not\in \fF$, and $\fF$ is a filter in $A$, and $\fF$ converges to $x$ by definition.
$\neg (1) \Rightarrow \neg (4)$: Let $B \supset A$ such that $x\not\in B$, then $B^c \in \cn^o(x)$ with $B^c \cap A = \emptyset$. Thus there exists no filter on $A$ containing $B^c \cap A$, and no filter on $A$ converging to $x$.
\end{proof}
\begin{proposition}
\label{proposition:closure-of-image}
Let $X, Y$ be topological spaces, $A \subset X$, and $f: X \to Y$ be continuous, then $f(\ol{A}) \subset \ol{f(A)}$.
\end{proposition}
\begin{proof}
Since $f$ is continuous, $f^{-1}(\ol{f(A)})$ is closed and contains $A$.
\end{proof}
\begin{definition}[Dense]
\label{definition:dense}
Let $X$ be a topologicial space and $A \subset X$, then the following are equivalent:
\begin{enumerate}
\item $\ol{A} = X$.
\item For every $\emptyset \ne U \subset X$ open, $A \cap U \ne \emptyset$.
\item For every $U \subset X$ open, $\overline{A \cap U} \supset U$.
\end{enumerate}
If the above holds, then $A$ is a \textbf{dense} subset of $X$.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: Since $U \ne \emptyset$, there exists $x \in U$, so $U \in \cn(x)$. By (2) of \ref{definition:closure}, $U \cap A \ne \emptyset$.
$(2) \Rightarrow (3)$: Let $\emptyset \ne U \subset X$ open. For each $x \in U$ and $V \in \cn(x)$, $U \cap V \in \cn(x)$ by (F2). Thus there exists $y \in A \cap U \cap V$. By (3) of \ref{definition:closure}, $x \in \overline{A \cap U}$.
$(3) \Rightarrow (1)$: $X$ is open.
\end{proof}
\begin{proposition}
\label{proposition:dense-product}
Let $\seqi{X}$ be topological spaces and $\seqi{A}$ such that each $A_i \subset X_i$ is dense, then $\prod_{i \in I}A_i$ is dense in $\prod_{i \in I}X_i$.
\end{proposition}
\begin{proof}
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{lemma}
\label{lemma:closurecomplement}
Let $X$ be a topological space and $A \subset X$, then $(A^o)^c = \overline{A^c}$.
\end{lemma}
\begin{proof}
Let $x \in X$. By (3) of \ref{definition:interior}, $x \in (A^o)^c$ if and only if there exists no $U \in \cn(x)$ such that $U \subset A$. Thus $x \in (A^o)^c$ if and only if $U \cap A^c \ne \emptyset$ for all $U \in \cn(x)$. By (2) of \ref{definition:closure}, this is equivalent to $x \in \ol{A^c}$.
\end{proof}
\begin{definition}[Boundary]
\label{definition:boundary}
Let $X$ be a topological space, $A \subset X$, $x \in X$, and $\fB \subset \cn(x)$ be a fundamental system of neighbourhoods, then the following are equivalent:
\begin{enumerate}
\item For every $V \in \cn(x)$, $V \cap A \ne \emptyset$ and $V \cap A^c \ne \emptyset$.
\item For every $V \in \fB$, $V \cap A \ne \emptyset$ and $V \cap A^c \ne \emptyset$.
\item $x \in \overline{A} \setminus A^o$.
\item $x \in \overline{A} \cap \overline{A^c}$.
\end{enumerate}
The set $\partial A$ of all points satisfying the above is the \textbf{boundary} of $A$.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: $\cn(x) \supset \fB$.
$(2) \Rightarrow (3)$: By (2) of \ref{definition:closure}, $x \in \overline{A}$. Since $V \cap A^c \ne \emptyset$ for all $V \in \fB$, there exists no open set $U \subset A$ with $x \in A$. By (2) of \ref{definition:interior}, $x \not\in A^o$.
$(3) \Rightarrow (4)$: By \ref{lemma:closurecomplement}.
$(4) \Rightarrow (1)$: By (2) of \ref{definition:closure}.
\end{proof}

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\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}

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\section{Neighbourhoods}
\label{section:neighbourhoods}
\begin{definition}[Neighbourhood]
\label{definition:neighbourhood}
Let $(X, \topo)$ be a topological space and $A \subset X$.
A \textbf{neighbourhood} of $A$ is a set $V \supset X$ where there exists $U \in \topo$ such that $A \subset U \subset V \subset X$.
The set $\cn_{X, \topo}(A) = \cn_X(A) = \cn_\topo(A) = \cn(A)$ denotes the collection of all neighbourhoods of $A$, and $\cn^o(A)$ denotes the set of \textit{open} neighbourhoods of $A$.
\textit{If $A = \bracs{x}$ is a single point, the above definition and notation applies to $x$ directly}.
\end{definition}
\begin{definition}[Fundamental System of Neighbourhoods]
\label{definition:fundamentalneighbourhoods}
Let $X$ be a topological space and $A \subset X$. A family $\fB \subset \cn(A)$ is a \textbf{fundamental system of neighbourhoods/neighbourhood base} at $A$ if for every $U \in \cn(A)$, there exists $V \in \fB$ such that $V \subset U$.
\end{definition}
\begin{lemma}[{{\cite[Proposition 1.2.1]{Bourbaki}}}]
\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 $U = \bigcup_{x \in U}V_x \in \topo$.
\end{proof}
\begin{proposition}[{{\cite[Proposition 1.2.2]{Bourbaki}}}]
\label{proposition:neighbourhoodcharacteristic}
Let $(X, \topo)$ be a topological space, then for each $x \in X$,
\begin{enumerate}
\item[(F1)] For every $V \in \cn_\topo(x)$, $W \in \cn_\topo(x)$ for all $X \supset W \supset V$.
\item[(F2)] For any $A, B \in \cn_\topo(x)$, $A \cap B \in \cn_\topo(x)$.
\item[(V1)] For every $A \in \cn_\topo(x)$, $x \in A$.
\item[(V2)] For every $V \in \cn_\topo(x)$, there exists $W \in \cn_\topo(x)$ such that $V \in \cn_\topo(y)$ for all $y \in W$.
\end{enumerate}
Conversely, if $\cn: X \to 2^X$ is a mapping such that
\begin{enumerate}
\item $\cn(x) \ne \emptyset$ for all $x \in X$.
\item $\cn(x)$ satisfies (F1), (F2), (V1), and (V2).
\end{enumerate}
then there exists a unique topology $\topo \subset 2^X$ such that $\cn = \cn_\topo$.
\end{proposition}
\begin{proof}
\textit{Existence}: Suppose that $\cn: X \to 2^X$ is a mapping such that (F1), (F2), (V1), and (V2) holds for all $x \in X$. Let
\[
\topo = \bracs{U \subset X| U \in \cn(x) \forall x \in U}
\]
Firstly, $\emptyset$ satisfies the condition vacuously, so $\emptyset \in \topo$.
For any $x \in X$, $\cn(x)$ is non-empty and there exists $V \in \cn(x)$. Since $X \supset V$, $X \in \topo$ by (F1).
Let $U, V \in \topo$. If $U \cap V \ne \emptyset$, then there exists $x \in U \cap V$, and $U, V \in \cn(x)$. For any $y \in U \cap V$, $U, V \in \cn(y)$ by (F2). Hence $U \cap V \in \cn(y)$ as well. Thus $U \cap V \in \topo$.
Now, let $\seqi{U} \subset \topo$ and $U = \bigcup_{i \in I}U_i$. For any $x \in U$, there exists $i \in I$ such that $x \in U_i$. In which case, $U \in \cn(x)$ by (F1), and $U \in \cn(x)$. Therefore $U \in \topo$, and $\topo$ forms a topology on $X$.
\textit{Agreement with $\cn_\topo$: } Fix $x \in X$. Let $V \in \cn_\topo(x)$, then there exists $U \in \topo$ such that $x \in U \subset V$. Since $U \in \topo$ and $x \in U$, $U \in \cn(x)$. By (F1), $V \in \cn(x)$ as well, so $\cn_\topo(x) \subset \cn(x)$.
Conversely, let $V \in \cn(x)$. By (V4), there exists $U_0 \in \cn(x)$ such that $V \in \cn(y)$ for all $y \in U_0$. Define
\[
U = \bracs{y \in V: V \in \cn(y)}
\]
then $U \supset U_0$ and $U \in \cn(x)$ by (V1). Let $y \in U$, then (V4) implies that there exists $W \in \cn(y)$ such that $V \in \cn(z)$ for all $z \in W$. Thus $W \subset U$ and $U \in \cn(y)$ by (F1).
\textit{Uniqueness}: Let $\mathcal{R}$ be a topology such that $\cn_{\mathcal{R}} = \cn$, then by \ref{lemma:openneighbourhood},
\[
\mathcal{R} = \bracs{U \subset X| U \in \cn_{\mathcal{R}}(x) \forall x \in U} = \bracs{U \subset X| U \in \cn(x) \forall x \in U} = \topo
\]
so $\topo$ is unique.
\end{proof}

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\section{Nets}
\label{section:nets}
\begin{definition}[Net]
\label{definition:net}
Let $X$ be a set and $A$ be an upward-directed set, then a \textbf{net} in $X$ is a mapping $A \to X$, denoted $\net{x} \subset X$.
\end{definition}
\begin{definition}[Eventually]
\label{definition:eventually}
Let $X$ be a set, $\net{x} \subset X$, and $E \subset X$, then $\net{x}$ is \textbf{eventually in} $E$ if there exists $\alpha_0 \in A$ such that $x_\alpha \in E$ for all $\alpha \gtrsim \alpha_0$.
\end{definition}
\begin{definition}[Frequently]
\label{definition:frequently}
Let $X$ be a set, $\net{x} \subset X$, and $E \subset X$, then $\net{x}$ is \textbf{frequently in} $E$ if for all $\alpha_0 \in A$, there exists $\alpha \gtrsim \alpha_0$ such that $x_\alpha \in E$.
\end{definition}
\begin{definition}[Convergence]
\label{definition:net-convergence}
Let $X$ be a topological space and $\net{x} \subset X$, then $\net{x}$ \textbf{converges} to $x$ if for every $U \in \cn(x)$, $\net{x}$ is eventually in $U$, denoted $x_\alpha \to x$.
\end{definition}

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\section{Product Spaces}
\label{section:producttopology}
\begin{definition}[Product Topology]
\label{definition:product-topology}
Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces, then the \textbf{product} topology $\topo$ is the initial topology generated by the projections $\seqi{\pi}$, and:
\begin{enumerate}
\item The family
\[
\cb(\ce) = \bracs{\bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_{k}) \bigg | U_{k} \in \topo_{i_k}, \seqf{i_k} \subset I, n \in \nat^+}
\]
is a base for $\topo$.
\item[(U)] For any topological space space $Y$ and $\seqi{f}$ where $f_i \in C(Y; X_i)$ for all $i \in I$, there exists a unique $f \in C(Y; X)$ such that the following diagram commutes
\[
\xymatrix{
Y \ar@{->}[d]_{f} \ar@{->}[rd]^{f_i} & \\
X \ar@{->}[r]_{\pi_i} & X_i
}
\]
for all $i \in I$.
\end{enumerate}
\end{definition}
\begin{proof}
(1): By (3) of \ref{definition:initial-topology}.
(U): Let $f \in \prod_{i \in I}f_i$, then $f$ is the unique function such that the diagrams commute. For each $\bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k) \in \topo$,
\[
f^{-1}\paren{\bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k)} = \bigcap_{k = 1}^n f_{i_k}^{-1}(U_k)
\]
which is open in $Y$ by (O2).
\end{proof}
\begin{proposition}
\label{proposition:productfilterconvergence}
Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\B$ be a filter base on $\prod_{i \in I}X_i$, then $\fB$ converges to $x \in \prod_{i \in I}X_i$ if and only if $\pi_i(\fB)$ converges to $\pi_i(x)$ for all $i \in I$.
\end{proposition}
\begin{proof}
$(\Rightarrow)$: Let $i \in I$ and $U \in \cn(\pi_i(x))$, then $\pi_i^{-1}(U) \in \cn(x)$. Since $\fB$ converges to $x$, there exists $B \in \fB$ with $B \subset \pi_i^{-1}(U)$. In which case, $\pi_i(B) \subset U$ and $\pi_i(\fB)$ converges to $\pi_i(x)$.
$(\Leftarrow)$: Let $U \in \cn(x)$, then there exists $\seqf{i_k}$ and open sets $\seqf{U_k}$ such that $x \in \bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k) \subset U$. For each $1 \le k \le n$, since $\pi_{i_k}(\fB)$ converges to $\pi_{i_k}(x)$ for each $1 \le k \le n$, there exists $B_k \in \fB$ such that $B_k \subset \pi_{i_k}^{-1}(U_k)$. As $\fB$ is a filter base, there exists $B \in \fB$ with
\[
B \subset \bigcap_{k = 1}^n B_k \subset \bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k) \subset U
\]
Therefore $\fB$ converges to $x$.
\end{proof}
\begin{proposition}
\label{proposition:initial-product-topology}
Let $X$ be a topological space, $\seqi{Y}$ be topological spaces, $\seqi{f}$ where $f_i \in C(X; Y_i)$ for each $i \in I$, then
\begin{enumerate}
\item There exists a unique $f \in C(X; \prod_{i \in I}Y_i)$ such that the following diagram commutes
\[
\xymatrix{
& \prod_{i \in I} Y_i \ar@{->}[rd]^{\pi_i} & \\
X \ar@{->}[rr]_{f_i} \ar@{->}[ru]^{f} & & Y_i
}
\]
for all $i \in I$.
\item If $X$ is T0 and equipped with the initial topology induced by $\seqi{f}$, then $f$ is an embedding.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): By (U) of \ref{definition:product-topology}.
(2): Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U \in \cn(x)$ such that $y \not\in U$. Let $J \subset I$ finite, $\seqj{U}$, and $\seqj{V}$ such that
\[
U = \bigcap_{j \in J}f_j^{-1}(U_j) = f^{-1}\paren{\bigcap_{j \in J}\pi_j^{-1}(U_j)}
\]
then $f(x) \in \bigcap_{j \in J}\pi_j^{-1}(U_j)$ but $f(y) \not\in \bigcap_{j \in J}\pi_j^{-1}(U_j)$. Thus $f$ is injective.
Let $U \subset X$ be open. Assume without loss of generality that there exists $J \subset I$ finite and $\bracs{U_j}_{j \in J}$ such that $U = \bigcap_{j \in J}f_j^{-1}(U_j)$. In which case, $\bigcap_{j \in J}\pi_j^{-1}(U_j)$ is open in $\prod_{i \in I}Y_i$ and
\[
f(U) = f(X) \cap \bigcap_{j \in J}\pi_j^{-1}(U_j)
\]
is a relatively open set.
\end{proof}
\begin{proposition}
\label{proposition:product-topology-embedding}
Let $\seqi{X}$, $\seqi{Y}$ be topological/uniform spaces and $\seqi{f}$ such that $f_i \in C(X_i; Y_i)$ is an embedding for all $i \in I$, then
\begin{enumerate}
\item There exists a unique $f \in C(\prod_{i \in I}X_i; \prod_{i \in I}Y_i)$/$f \in UC(\prod_{i \in I}X_i; \prod_{i \in I}Y_i)$ such that the following diagram commutes
\[
\xymatrix{
\prod_{i \in I}Y_i \ar@{->}[r]^{\pi_i} & Y_i \\
\prod_{i \in I} X_i \ar@{->}[r]_{\pi_i} \ar@{->}[u]^{f} & X_i \ar@{^{(}->}[u]_{f_i}
}
\]
for all $i \in I$.
\item $f$ is an embedding.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): By (U) of \ref{definition:product-topology}/\ref{definition:product-uniform}.
(2): Consider the following diagram
\[
% https://darknmt.github.io/res/xypic-editor/#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
\xymatrix{
\prod_{i \in I}Y_i \ar@{->}[rd]^{\pi_i} & \\
\iota_P(\prod_{i \in I}X_i) \ar@{->}[u] & Y_i \\
\prod_{i \in I} X_i \ar@{->}[r]_{\pi_i} \ar@{->}[u]^{f} & X_i \ar@{^{(}->}[u]_{f_i}
}
\]
Since each $X_i \to Y_i$ is an embedding, the composition
\[
\xymatrix{
\iota_P(\prod_{i \in I}X_i) \ar@{->}[r]^{\pi_i} & Y_i \ar@{->}[r]^{f_i^{-1}} & X_i
}
\]
is continuous/uniformly continuous. By (U) of \ref{definition:product-topology}/\ref{definition:product-uniform}, $f$ is an embedding.
\end{proof}

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\section{Regular Spaces}
\label{section:regularspaces}
\begin{definition}[Regular Space, {{\cite[Proposition 1.4.11]{Bourbaki}}}]
\label{definition:regular}
Let $X$ be a topological space, then the following are equivalent:
\begin{enumerate}
\item For each $x \in X$ and $A \subset X$ closed with $x \not\in A$, there exists $U \in \cn(x)$ and $V \in \cn(A)$ such that $U \cap V = \emptyset$.
\item For each $x \in X$, the closed neighbourhoods of $x$ forms a fundamantal system of neighbourhoods at $x$.
\end{enumerate}
If $X$ is a T1 space such that the above holds, then $X$ is \textbf{regular}.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: Let $U \in \cn^o(x)$, then $U^c$ is closed with $x \not\in U^c$ by (V1). Thus there exists $V \in \cn^o(U^c)$ such that $x \not\in V$, so $V^c \in \cn(x)$ is closed with $V^c \subset U$.
$(2) \Rightarrow (1)$: Since $X$ is Hausdorff, $X$ is T1 as well. Let $x \in X$ and $A \subset X$ closed such that $x \not\in A$, then $A^c \in \cn(x)$. So there exists $K \in \cn(x)$ closed such that $x \in K \subset A^c$. Thus $K \in \cn(x)$ and $K^c \in \cn(A)$ are the desired neighbourhoods.
\end{proof}
\begin{theorem}[{{\cite[Theorem 1.8.1]{Bourbaki}}}]
\label{theorem:regularextension}
Let $X$ be a topological space, $Y$ be a regular space, $A \subset X$, and $f \in C(A; Y)$, then the following are equivalent:
\begin{enumerate}
\item There exists $F \in C(X; Y)$ such that $F|_A = f$.
\item For each $x \in X$, $\lim_{y \to x, y \in A}f(y)$ exists.
\end{enumerate}
\end{theorem}
\begin{proof}
$(\Rightarrow)$: Let $x \in X$ and $\fF = \bracs{U \cap A| U \in \cn(x)}$, then $f(\fF) = F(\fF)$. By continuity of $F$, $F(\fF)$ converges to $F(x)$, so $\lim_{y \to x, x \in A}f(x)$ exists.
$(\Leftarrow)$: Let $F(x) = \lim_{y \to x, y \in A}f(y)$, then $F$ is well-defined and $F|_A = f$ by (4) of \ref{definition:hausdorff}.
Let $x \in X$ and $V \in \cn(F(x))$. Using (2) of \ref{definition:regular}, assume without loss of generality that $V$ is closed. By assumption, there exists $U \in \cn^o(x)$ such that $f(U \cap A) \subset V$. In which case, \ref{lemma:openneighbourhood} implies that $U \in \cn^o(y)$ for all $y \in U$. Since every limit point of a filter is a cluster point,
\[
F(y) = \lim_{\substack{z \to y \\ z \in A}}f(z) \in \ol{f(U \cap A)} \subset V
\]
as $V$ is closed. Therefore $F(U) \subset V$, and $F$ is continuous.
\end{proof}

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\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{proposition}[{{\cite[Proposition 2.3.3]{Bourbaki}}}]
\label{proposition:imagecauchy}
Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces, $f \in UC(X; Y)$, and $\fB \subset 2^X$ be a Cauchy filter base, then $f(\fB) \subset 2^Y$ is a Cauchy filter base.
\end{proposition}
\begin{proof}
Let $V \in \mathfrak{V}$, then there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$. Given that $\fB$ is a Cauchy filter base, 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.
\end{definition}
\begin{proposition}
\label{proposition:product-complete}
Let $\seqi{X}$ be complete uniform spaces, then $\prod_{i \in I}X_i$ is complete.
\end{proposition}
\begin{proof}
For any Cauchy filter $\fF$ in $\prod_{i \in I}\wh X_i$, $\pi_i(\fF)$ is a Cauchy filter base by (1) of \ref{definition:product-uniform} and \ref{proposition:imagecauchy}. For each $i \in I$, $\wh X_i$ is complete, so there exists $x_i \in \wh X_i$ such that $\pi_i(\fF)$ converges to $x_i$. Let $x \in X$ such that $\pi_i(x) = x_i$ for all $i \in I$, then $\fF$ converges to $x$ by \ref{proposition:productfilterconvergence}. Therefore $\prod_{i \in I}X_i$ is complete.
\end{proof}
\begin{proposition}
\label{proposition:complete-closed}
Let $X$ be a uniform space and $A \subset X$ be closed, then $A$ is complete.
\end{proposition}
\begin{proof}
Let $\fF$ be a Cauchy filter on $A$, then there exists $x \in X$ such that $\fF$ converges to $x$. Since $A$ is closed, $x \in X$ by (4) of \ref{definition:closure}.
\end{proof}
\begin{lemma}
\label{lemma:complete-dense}
Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $X$ is \textbf{complete} if and only if every Cauchy filter on $A$ converges to at least one point in $X$.
\end{lemma}
\begin{proof}
Let $\fF \subset X$ be a Cauchy filter. Using \ref{proposition:minimalcauchyexistence}, assume without loss of generality that $\fF$ is minimal. By \ref{proposition:cauchyinterior}, $\fF$ admits a base consisting of open sets. Hence $E \cap A \ne \emptyset$ for all $E \in \fF$. Let $\fF_A = \bracs{E \cap A| E \in \fF}$, then $\fF_A$ is a filter base on $A$.
By assumption, there exists $x \in X$ such that $\fF_A$ converges to $x$. Since $\fF$ is a Cauchy filter contained in the filter generated by $\fF_A$ in $X$, $x$ is also a limit point of $\fF$ by \ref{proposition:cauchyfilterlimit}.
\end{proof}
\begin{proposition}[{{\cite[Proposition 2.3.11]{Bourbaki}}}]
\label{proposition:uniformextension}
Let $X$ be a topological space, $Y$ be a complete Hausdorff uniform space, $A \subset X$ be a dense subset, and $f \in C(A; Y)$ be a continuous function, then the following are equivalent:
\begin{enumerate}
\item There exists a continuous function $F \in C(X; Y)$ such that $F|_A = f$.
\item For each $x \in X$, $f(\bracs{U \cap A| U \in \cn(x)})$ is a Cauchy filter base.
\end{enumerate}
\end{proposition}
\begin{proof}
By \ref{proposition:uniform-neighbourhoods}, $Y$ is regular. Since $Y$ is complete, (2) is equivalent to (2) of \ref{theorem:regularextension}. Therefore the proposition follows from \ref{theorem:regularextension}.
\end{proof}
\begin{theorem}[Extension of Uniformly Continuous Functions, {{\cite[Theorem 1.3.2]{Bourbaki}}}]
\label{theorem:uniform-continuous-extension}
Let $(X, \fU)$, $(Y, \fV)$ be uniform spaces, $A \subset Y$ be a dense subset, and $f \in UC(A; Y)$, then:
\begin{enumerate}
\item There exists a unique $F \in UC(X; Y)$ such that $F|_A = f$.
\item $F \in UC(X; Y)$.
\end{enumerate}
\end{theorem}
\begin{proof}
(1): By \ref{proposition:imagecauchy}, $f(\bracs{U \cap A| U \in \cn(x)})$ is a Cauchy filter base for all $x \in X$. Thus the existence and uniqueness of the extension follows from \ref{proposition:uniformextension}.
(2): Let $V \in \fV$. Using \ref{proposition:goodentourages}, assume without loss of generality that $V$ is symmetric and closed. Since $f \in UC(A; Y)$, there exists $U \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in U$. Using \ref{proposition:subspace-entourage}, assume without loss of generality that $U = \overline{U} = \ol{U \cap (A \times A)}$. In which case, since $F$ is continuous,
\[
F(U) = F(\overline{U}) = F(\ol{U \cap (A \times A)}) \subset \overline{F(U \cap (A \times A))} = \overline{f(U \cap (A \times A))} \subset V
\]
by \ref{proposition:closure-of-image}.
\end{proof}

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\section{The Hausdorff Completion}
\label{section:completion}
\begin{definition}[Hausdorff Completion, {{\cite[Theorem 2.3.3]{Bourbaki}}}]
\label{definition:hausdorff-completion}
Let $(X, \fU)$ be a uniform space, then there exists a $(\wh X, \iota)$ such that:
\begin{enumerate}
\item $(\wh X, \wh \fU)$ is a complete Hausdorff uniform space.
\item $\iota \in UC(X; \wh X)$.
\item[(U)] For any pair $(Y, f)$ satisfying (1) and (2), there exists unique $F \in C(\wh X; Y)$ such that the following diagram commutes
\[
\xymatrix{
& \wh X \ar@{->}[rd]^{F} & \\
X \ar@{->}[rr]_{f} \ar@{->}[ru]^{\iota} & & Y
}
\]
and $F \in UC(\wh X; Y)$.
\end{enumerate}
Moreover,
\begin{enumerate}
\item[(4)] For any symmetric entourage $V \in \fU$, let
\[
\wh V = \bracsn{(\fF, \mathfrak{G}) \in \wh X| \exists U \in \fF \cap \mathfrak{G}: U \times U \subset V}
\]
then[(5)] The family $\wh \fB = \bracsn{\wh V| V \in \fU, V \text{ symmetric}}$ forms a fundamental system of entourages for $\wh X$. In particular, $\bracs{\wh V \cap \iota(X)| V \in \fU, V \text{ symmetric}}$ is a fundamental system of entuorages for $\iota(X)$.
\item $\iota(X)$ is dense in $\wh X$.
\end{enumerate}
The pair $(\wh X, \iota)$ is the \textbf{Hausdorff completion} of $X$.
\end{definition}
\begin{proof}
(1, Uniform), (4): Let $\wh X$ be the set of all minimal Cauchy filters on $X$. For each $V \in \fU$, let
\[
\wh V = \bracsn{(\fF, \mathfrak{G}) \in \wh X| \exists U \in \fF \cap \mathfrak{G}: U \times U \subset V}
\]
and $\wh \fB = \bracsn{\wh V| V \in \fU, V \text{ symmetric}}$, then
\begin{enumerate}
\item[(FB1)] Let $\wh U, \wh V \in \wh \fB$. By \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $W \in \fU$ such that $W \subset U \cap V$. In which case, for any $(\fF, \mathfrak{G}) \in \wh W$, there exists $E \in \fF \cap \mathfrak{G}$ with $E \times E \subset W \subset U \cap V$. Thus $\wh W \subset \wh U \cap \wh V$.
\item[(UB1)] Let $\wh U \in \wh \fB$ and $\fF \in \wh X$, then since $\fF$ is Cauchy, there exists $E \in \fF$ such that $E \times E \subset U$, so $(\fF, \fF) \in \wh U$.
\item[(UB2)] Let $\wh U \in \wh \fB$. By \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $W \in \fU$ such that $W \circ W \subset U$. For any $\fF, \mathfrak{G}, \mathfrak{H} \in \wh X$ such that $(\fF, \mathfrak{G}), (\mathfrak{G}, \mathfrak{H}) \in \wh W$, there exists $W$-small sets $E \in \fF \cap \mathfrak{G}$ and $F \in \fF \cap \mathfrak{H}$. Since $\mathfrak{G}$ is a filter, $E \cap F \ne \emptyset$ by (F2) and (F3). By \ref{lemma:small-intersect}, $E \cup H$ is $W \circ W$-small and thus $U$-small. Using (F1), $E \cup H \in \fF \cap \mathfrak{H}$, so $(\fF, \mathfrak{H}) \in \wh U$. Therefore $\wh W \circ \wh W \subset U$.
\end{enumerate}
By \ref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\wh \fU \supset \wh \fB$. Moreover, $\wh \fB$ consists of symmetric entourages by construction.
(1, Hausdorff): It is sufficient to show that $\Delta$ is closed and use (6) of \ref{definition:hausdorff}. Let $(\fF, \mathfrak{G}) \in \ol{\Delta}$, then $(\fF, \mathfrak{G}) \in U$ for all $U \in \fU$ closed. Let $\fB = \bracs{F \cup G| F \in \fF, G \in \mathfrak{G}}$, then
\begin{enumerate}
\item[(FB1)] For any $F \cup G, F' \cup G' \in \fB$, $(F \cup G) \cap (F' \cup G') \supset (F \cap F') \cup (G \cap G') \in \fB$.
\item[(FB2)] By (F3), $\emptyset \not\in \fF \cup \mathfrak{G}$, so $\emptyset \not\in \fB$.
\end{enumerate}
Thus $\fB$ is a filter base by \ref{proposition:filterbasecriterion}, and the filter $\mathfrak{H}$ generated by $\fB$ is contained in $\fF$ and $\mathfrak{G}$. By \ref{proposition:goodentourages}, for every $U \in \fU$, there exists a $U$-small set $E \in \fF \cap \mathfrak{G} \subset \fB \subset \mathfrak{H}$. So $\mathfrak{H} \subset \fF, \mathfrak{G}$ is a Cauchy filter. By minimality of $\fF$ and $\mathfrak{G}$, $\fF = \mathfrak{G} = \mathfrak{H}$.
(2): For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter by (1) of \ref{proposition:cauchyfilterlimit}. Define $\iota: X \to \wh X$ by $x \mapsto \cn(x)$. Let $\wh U \in \wh \fU$ and $(\cn(x), \cn(y)) \in \wh U$, then there exists a $U$-small set $E \in \cn(x) \cap \cn(y)$. By (V1), $(x, y) \in E \times E \in U$.
Conversely, if $(x, y) \in U$, then $E = U(x) \cap U(y) \in \cn(x) \cap \cn(y)$ by (F2), and $E$ is $U$-small by symmetry of $U$.
Thus $(\iota \times \iota)^{-1}(\wh U) = U$, $(\iota \times \iota)^{-1}: \wh \fU \to \fU$ is a bijection, and $\iota \in UC(X; \wh X)$.
(4): Let $\fF \in \wh X$ and $\wh U \in \wh \fU$. Since $\fF$ is a Cauchy filter, there exists a $U$-small set $E \in \fF$. Using \ref{proposition:cauchyinterior}, assume without loss of generality that $E$ is open. Let $x \in E$, then $E \in \cn(x)$, so $(\fF, \iota(x)) \in \wh \fU$, and $X$ is dense in $\wh X$ by (3) of \ref{definition:closure}.
(1, Complete): Using \ref{lemma:complete-dense}, it is sufficient to show that every Cauchy filter in $\iota(X)$ converges in $\wh X$.
Let $\wh \fF \in 2^{\iota(X)}$ be a Cauchy filter, then since $\iota: X \to \iota(X)$ is surjective, $\iota^{-1}(\wh \fF)$ is a filter base by \ref{proposition:preimage-filterbase}.
For any $U \in \fU$, there exists a $\wh U$-small set $\wh E \in \wh \fF$. Since $U = (\iota \times \iota)^{-1}(\wh U)$, $\iota^{-1}(\wh E) \subset U$, so $\iota^{-1}(\wh \fF)$ is a Cauchy filter base.
Let $\mathfrak{G} \subset \iota^{-1}(\wh \fF)$ be a minimal Cauchy filter, then $\iota(\mathfrak{G})$ is a Cauchy filter base by \ref{proposition:imagecauchy}.
Let $U \in \fU$. By \ref{proposition:cauchyinterior}, there exists a $U$-small open set $E \in \mathfrak{G}$. For any $x \in E$, $E \in \cn(x)$ by \ref{lemma:openneighbourhood}, so $(\cn(x), \mathfrak{G}) = (\iota(x), \mathfrak{G}) \in \wh U$, and $\iota(E) \subset \wh U(\mathfrak{G})$. Since $E \in \mathfrak{G}$, $\wh U(\mathfrak{G}) \supset \iota(E) \in \iota(\mathfrak{G})$, so $\iota(\mathfrak{G})$ converges to $\mathfrak{G}$.
Now, given that $\mathfrak{G} \subset \iota^{-1}(\wh \fF)$, $\iota(\mathfrak{G}) \subset \iota(\iota^{-1}(\fF)$, so $\iota(\iota^{-1}(\fF))$ converges to $\mathfrak{G}$ as well. Since $\fF$ is a filter on $\iota(X)$, $\iota(\iota^{-1}(\fF)) = \fF$, thus $\fF$ is convergent.
(U): Let $(Y, \mathfrak{V})$ be a complete Hausdorff uniform space and $f \in UC(X; Y)$. For each $\fF \in \wh X$, let $F(\fF) = \lim_{x, \fF}f(x)$. For any $x \in X$, $F(\iota(x)) = \lim_{y \to x}f(y) = f(x)$, so $f = F \circ \iota$. Since $\fF$ is a Cauchy filer, $f \in UC(X; Y)$, and $Y$ is a complete Hausdorff uniform space, the limit exists and is unique. Thus $F: \iota(X) \to Y$ is well-defined.
Let $U \in \mathfrak{V}$, then there exists a symmetric entourage $V \in \fU$ such that $(f(x), f(y)) \in U$ for all $(x, y) \in V$, then for any $(\iota(x), \iota(y)) \in \wh V \cap (\iota(X) \times \iota(X))$, $(F(\iota(x)), F(\iota(y))) \in U$, so $F$ is uniformly continuous. By \ref{theorem:uniform-continuous-extension}, there exists a unique $\td F \in C(\wh X; Y)$ such that $\td F|_{\iota(X)} = F|_{\iota(X)}$, and any such $\td F$ is uniformly continuous.
\end{proof}
\begin{lemma}
\label{lemma:completion-of-hausdorff}
Let $(X, \fU)$ be a Hausdorff uniform space, then the canonical map $\iota: X \to \wh X$ is an embedding.
\end{lemma}
\begin{proof}
Since the mapping $(\iota \times \iota)^{-1}$ is a bijection between two bases of uniformities of $X$ and $\wh X$, it is sufficient to show that $\iota$ is injective. To this end, observe that for any $x, y \in X$, $\cn(x) = \cn(y)$ if and only if $x = y$ by (4) of \ref{definition:hausdorff}.
\end{proof}
\begin{definition}[Associated Hausdorff Uniform Space, {{\cite[Proposition 2.8.16]{Bourbaki}}}]
\label{definition:associated-hausdorff}
Let $(X, \fU)$ be a uniform space, then there exists $(X', i)$ such that:
\begin{enumerate}
\item $X'$ is a Hausdorff uniform space.
\item $i \in UC(X; X')$.
\item For any pair $(Y, f)$ satisfying (1) and (2), there exists a unique $f' \in UC(X'; Y)$ such that the following diagram commutes
\[
\xymatrix{
X' \ar@{->}[rd]^{F} & \\
X \ar@{->}[r]_{f} \ar@{->}[u]^{i} & Y
}
\]
\end{enumerate}
known as the \textbf{Hausdorff uniform space associated with} $(X, \fU)$.
\end{definition}
\begin{proof}
Let $(\wh X, \iota)$ be the Hausdorff completion of $X$, $X' = \iota(X)$, and $i = \iota$, then $(X', i)$ satisfies (1) and (2).
(U): Let $(\wh Y, \iota)$ be the Hausdorff completion of $Y$. Using \ref{lemma:completion-of-hausdorff}, identify $Y$ as a subspace of $\wh Y$. By (U) of the Hausdorff completion, there exists a unique $\ol F \in UC(\wh X; \wh Y)$ such that the following diagram commutes:
\[
\xymatrix{
\wh X \ar@{->}[r]^{\overline{F}} & \wh Y \\
X' \ar@{->}[u]^{\iota} \ar@{->}[r]^{F} & Y' \ar@{->}[u]_{\iota} \\
X \ar@{->}[r]_{f} \ar@{->}[u]^{i} & Y \ar@{=}[u]
}
\]
Since $\ol F(X') \subset \iota(Y) = Y'$, $F = \ol F|_{X'} \in UC(X'; Y')$ is continuous. By \ref{lemma:completion-of-hausdorff}, $\iota \in UC(Y; Y')$ is a homeomorphism. Upon identifying $Y$ with $Y'$, $F \in UC(X'; Y)$ is the desired map.
\end{proof}
\begin{proposition}
\label{proposition:hausdorff-uniform-factor}
Let $X, Y$ be uniform spaces, $X', Y'$ be their associated Hausdorff uniform spaces, and $\wh X, \wh Y$ be their Hausdorff completions, then there exists a unique $F \in UC(X'; Y')$ and $\ol F \in UC(\wh X; \wh Y)$ such that the following diagram commutes:
\[
\xymatrix{
\wh X \ar@{->}[r]^{\overline{F}} & \wh Y \\
X' \ar@{->}[u] \ar@{->}[r]^{F} & Y' \ar@{->}[u] \\
X \ar@{->}[r]_{f} \ar@{->}[u] & Y \ar@{->}[u]
}
\]
\end{proposition}
\begin{proof}
By (U) of \ref{definition:hausdorff-completion} and \ref{definition:associated-hausdorff}.
\end{proof}
\begin{proposition}
\label{proposition:initial-completion}
Let $X$ be a set, $\seqi{Y}$ be uniform spaces, $\seqi{f}$ with $f_i: X \to Y_i$ for each $i \in I$. Let $\fU$ be the initial uniformity on $X$ induced by $\seqi{f}$, and $(\wh X, \iota_X)$ be the Hausdorff completion of $X$.
\begin{enumerate}
\item For each $i \in I$, let $(\wh Y_i, \iota_i)$ be the Hausdorff completion of $Y_i$, then there exists a unique $F_i \in UC(\wh X; \wh Y_i)$ such that the following diagram commutes
\[
\xymatrix{
\wh X \ar@{->}[r]^{F_i} & \wh Y \\
X \ar@{->}[r]_{f_i} \ar@{->}[u] & Y \ar@{->}[u]
}
\]
\item The uniformity of $\wh X$ is the initial uniformity induced by $\seqi{F}$.
\item There exists a unique $F \in UC(X; \prod_{i \in I}\wh Y_i)$ and $\ol{F} \in UC(\wh X; \prod_{i \in I}\wh Y_i)$ such that the following diagram commutes
\[
\xymatrix{
& \prod_{i \in I} \wh Y_i \ar@{->}[rd]^{\pi_i} & \\
& \wh X \ar@{->}[r]^{F_i} \ar@{->}[u]_{\overline{F}} & \wh Y_i \\
X \ar@{->}[ruu]^{F} \ar@{->}[rr]_{f_i} \ar@{->}[ru]_{\iota_X} & & Y_i \ar@{->}[u]_{\iota_i}
}
\]
Moreover, $\ol{F}(\wh X) = \overline{F(X)}$, and $\ol{F}$ is an embedding.
\end{enumerate}
In particular, by \ref{proposition:dense-product}, there is a natural isomorphism
\[
\prod_{i \in I}\wh X_i \iso \wh{\prod_{i \in I}X_i}
\]
induced by extending the identity on $\pi_{i \in I}X_i$.
\end{proposition}
\begin{proof}
(1): By (U) of \ref{definition:hausdorff-completion}.
(2), (3): By (U) of the product (\ref{definition:product-topology}), there exists $f \in UC(X; \prod_{i \in I}Y_i)$ such that the following diagram commutes:
\[
\xymatrix{
& \prod_{i \in I} Y_i \ar@{->}[rd]^{\pi_i} & \\
X \ar@{->}[rr]_{f_i} \ar@{->}[ru]^{f} & & Y_i
}
\]
First suppose that $X$ and $\seqi{Y}$ are Hausdorff. For each $i \in I$, $\iota_i \circ \pi_i \in UC(\prod_{i \in I} Y_i; \wh Y_i)$, so by (U) of \ref{proposition:product-complete}, there exists a unique $\iota_P \in UC(\prod_{i \in I} Y_i; \prod_{i \in I} \wh Y_i)$ such that the following diagram commutes
\[
\xymatrix{
\prod_{i \in I}\wh Y_i \ar@{->}[r]^{\pi_i} & \wh Y_i \\
\prod_{i \in I}Y_i \ar@{->}[r]^{\pi_i} \ar@{^{(}->}[u]^{\iota_P} & Y_i \ar@{^{(}->}[u]_{\iota_i}
}
\]
for all $i \in I$. By \ref{lemma:completion-of-hausdorff}, each $\iota_i \in UC(Y_i; \wh Y_i)$ is an embedding, so \ref{proposition:product-topology-embedding} implies that $\iota_P$ is an embedding as well.
Since $X$ has the initial topology, $f: X \to \prod_{i \in I}Y_i$ is an embedding by \ref{proposition:initial-product-topology}. Thus the composition $\iota_P \circ f$ is an embedding. As $\prod_{i \in I}\wh Y_i$ is complete by \ref{proposition:product-complete} and \ref{proposition:product-hausdorff}, $\overline{\iota_P \circ f(X)} \subset \prod_{i \in I}\wh Y_i$ is a complete Hausdorff uniform space. Let $Z$ be a complete Hausdorff uniform space and $g \in UC(X; Z)$, then \ref{theorem:uniform-continuous-extension} implies that there exists a unique $G \in UC(\overline{\iota_P \circ f(X)}; Z)$ such that the following diagram commutes
\[
\xymatrix{
\prod_{i \in I}\wh Y_i & \\
\overline{\iota_P \circ f(X)} \ar@{^{(}->}[u] \ar@{->}[rd]^{G} & \\
X \ar@{^{(}->}[u]^{\iota_P\circ f} \ar@{->}[r]_{g} & Z
}
\]
Thus $(\overline{\iota_P \circ f(X)}, \iota \circ f)$ satisfies (1), (2), and (U) of the Hausdorff completion (\ref{definition:hausdorff-completion}). Therefore $\wh X$ may be identified as a subspace of $\prod_{i \in I}\wh Y_i$ as follows:
\[
% https://darknmt.github.io/res/xypic-editor/#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
\xymatrix{
\prod_{i \in I}\wh Y_i \ar@{->}[r]^{\pi_i} & \wh Y_i \\
\overline{\iota_P \circ f(X)} \ar@{^{(}->}[u] \ar@{=}[r] & \wh X \\
X \ar@{^{(}->}[u]^{\iota_P\circ f} \ar@{->}[ru]_{\iota_X} &
}
\]
In which case, $\wh X$ must be equipped with the initial topology induced by the projection maps.
Now assume that $X$ and $Y$ are arbitrary. Let $X'$ and $\seqi{Y'}$ be the Hausdorff spaces associated with $X$ and $\seqi{Y}$, respectively. By \ref{proposition:hausdorff-uniform-factor}, there exists $\seqi{f'}$ such that the following diagram commutes
\[
\xymatrix{
X' \ar@{->}[r]^{f_i'} & Y_i' \\
X \ar@{->}[u] \ar@{->}[r]_{f_i} & Y_i \ar@{->}[u]
}
\]
for all $i \in I$. By (5) of \ref{definition:hausdorff-completion}, there is a correspondence between the uniformities of $X$ and $X'$, and $Y$ and $Y'$. Thus $X'$ is equipped with the initial uniformity genereated by $\seqi{f'}$.
\end{proof}

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\section{Uniform Structures}
\label{section:uniformstructures}
\begin{definition}[Inversion]
\label{definition:inversion}
Let $X$ be a set and $U \subset X \times X$, then the \textbf{inversion} of $U$ is the set
\[
U^{-1} = \bracs{(y, x)| (x, y) \in U}
\]
A set $U \subset X \times X$ is \textbf{symmetric} if $U = U^{-1}$.
\end{definition}
\begin{definition}[Composition]
\label{definition:composition}
Let $X$ be a set and $U, V \subset X \times X$, then the \textbf{composition} of $U$ and $V$ is the set
\[
U \circ V = \bracs{(x, z) \in X \times X| \exists y \in Y: (x, y) \in U, (y, z) \in V}
\]
\end{definition}
\begin{definition}[Slice]
\label{definition:slice}
Let $X, Y$ be sets, $U \subset X \times Y$, and $A \subset X$, then
\[
U(A) = \bracs{y \in Y: (x, y) \in U, x \in A}
\]
is the \textbf{slice} of $U$ at $A$.
\end{definition}
\begin{definition}[Uniformity]
\label{definition:uniformity}
Let $X$ be a set, then a non-empty family $\fU \subset X \times X$ is a \textbf{uniformity} on $X$ if:
\begin{enumerate}
\item[(F1)] For any $U \in \fU$ and $V \supset U$, $V \in \fU$.
\item[(F2)] For any $U, V \in \fU$, $U \cap V \in \fU$.
\item[(U1)] For every $U \in \fU$, $U \supset \Delta = \bracs{(x, x)| x \in X}$.
\item[(U2)] For any $U \in \fU$, $U^{-1} \in \fU$.
\item[(U3)] For any $U \in \fU$, there exists $V \in \fU$ such that $V \circ V \subset U$.
\end{enumerate}
The elements of $\fU$ are called the \textbf{entourages} of $\fU$, and the pair $(X, \fU)$ is a \textbf{uniform space}.
For any $x, y \in X$ and $U \in \fU$, $x$ and $y$ are \textbf{$U$-close} if $(x, y) \in U$.
\end{definition}
\begin{definition}[Subspace Uniformity]
\label{definition:subspaceuniform}
Let $(X, \fU)$ be a uniform space and $A \subset X$, then the family
\[
\fU_A = \bracs{U \cap (A \times A)| U \in \fU}
\]
forms a uniformity on $A$, known as the \textbf{subspace uniformity} induced on $A$.
\end{definition}
\begin{proposition}
\label{proposition:subspace-uniform-compatible}
Let $(X, \fU)$ be a uniform space and $A \subset X$, then the subspace topology of $A$ coincides with the topology induced by the subspace uniformity on $A$.
\end{proposition}
\begin{proof}
Let $x \in A$ and $U \in \fU$, then $U(x) \cap A = [U \cap (A \times A)](x)$. Thus $V \subset X$ is a neighbourhood of $x$ with respect to the subspace topology if and only if it is a neighbourhood of $x$ with respect to the topology on $A$ induced by the subspace uniformity.
\end{proof}
\begin{definition}[Fundamental System of Entourages]
\label{definition:fundamentalentourage}
Let $(X, \fU)$ be a uniform space, then a family $\fB \subset \fU$ is a \textbf{fundamental system of entourages} for $\fU$ if for every $U \in \fU$, there exists $V \in \fB$ such that $V \subset U$.
\end{definition}
\begin{proposition}
\label{proposition:fundamental-entourage-criterion}
Let $X$ be a set and $\fB \subset 2^{X \times X}$ be a non-empty family of sets. If
\begin{enumerate}
\item[(FB1)] For each $U, V \in \fB$, there exists $W \in \fB$ such that $W \subset U \cap V$.
\item[(UB1)] For each $V \in \fB$, $\Delta \subset V$.
\item[(UB2)] For each $V \in \fB$, there exists $W \in \fB$ such that $W \circ W \subset V$.
\end{enumerate}
then there exists a unique uniformity $\fU \subset 2^{X \times X}$, which is given by
\[
\fU = \bracs{U \subset X \times X| \exists V \in \fB: V \subset U}
\]
\end{proposition}
\begin{proof}
(F1): By definition of $\fU$.
(F2): For any $U, V \in \fU$, there exists $U_0, V_0 \in \fB$ such that $U_0 \subset U$ and $V_0 \subset V$. By (FB1), there exists $W \in \fB$ with $W \subset U_0 \cap V_0 \subset U \cap V$. Thus $U \cap V \in \fU$.
(U1) and (U2): For any $U \in \fU$, there exists $U_0 \in \fB$ with $U_0 \subset U$. By (UB1), $\Delta \subset U_0 \subset U$. By (UB2), there exists $V_0 \in \fB \subset \fU$ with $V_0 \circ V_0 \subset U_0 \subset U$.
\end{proof}
\begin{lemma}
\label{lemma:symmetricfundamentalentourage}
Let $(X, \fU)$ be a uniform space, $\fB \subset \fU$ be a fundamental system of entourages, then
\[
\fB_S = \bracsn{U \cap U^{-1}| U \in \fB}
\]
is also a fundamental system of entourages.
\end{lemma}
\begin{proof}
By (F2), $\fB \subset \fU$. For any $U \in \fU$, there exists $V \in \fB$ such that $V \subset U$. In which case,
\[
U \supset V \supset V \cap V^{-1} \in \fB_S
\]
so $\fb_S$ is a fundamental system of entourages.
\end{proof}
\begin{definition}[Topology of a Uniform Space, {{\cite[Proposition 2.1.2]{Bourbaki}}}]
\label{definition:uniformtopology}
Let $(X, \fU)$ be a uniform space and
\[
\cn: X \to 2^X \quad x \mapsto \bracsn{U(x)| U \in \fU}
\]
then there exists a unique topology $\topo \subset 2^X$ such that $\cn_\topo = \cn$, known as the \textbf{topology induced by the uniform structure $\fU$}.
\end{definition}
\begin{proof}
Using \ref{proposition:neighbourhoodcharacteristic}, it is sufficient to show that $\cn(x)$ is non-empty for all $x \in X$, and that it satisfies (F1), (F2), (V1), and (V2). Firstly, since $\fU \ne \emptyset$, $\cn(x) \ne \emptyset$ for all $x \in X$.
(F1): Let $U \in \fU$ and $V \supset U(x)$, then
\[
W = U \cup (\bracs{x} \times V) \supset U
\]
As $\fU$ satisfies (F1), $W \in \fU$. Thus
\[
V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
\]
(F2): Let $U, V \in \fU$, then $U(x) \cap V(x) = (U \cap V)(x)$. As $\fU$ satisfies (F2), $U \cap V \in \fU$ and $(U \cap V)(x) \in \cn(x)$.
(V1): Let $U \in \fU$. By (U1), $\Delta \subset U$, so $x \in U(x)$.
(V2): Let $U \in \fU$ and $x \in X$. By (U2), there exists $W \in \fU$ such that $W \circ W \subset U$. Let $y \in W$, then for any $z \in W(y)$, $(x, z) \in V$. Hence $W(y) \subset V(x)$ for all $y \in W$, so $V(x) \in \cn(y)$ for all $y \in W$.
\end{proof}
\begin{lemma}
\label{lemma:compositiongymnastics}
Let $(X, \fU)$ be a uniform space, $V \in \fU$ be a symmetric entourage, and $M \subset X \times X$, then the following are equivalent:
\begin{enumerate}
\item $(x, y) \in V \circ M \circ V$.
\item There exists $(p, q) \in M$ such that $(x, p) \in V$ and $(q, y) \in V$.
\item There exists $(p, q) \in M$ such that $x \in V(p)$ and $y \in V(q)$.
\item There exists $(p, q) \in M$ such that $(x, y) \in V(p) \times V(q)$.
\item There exists $(p, q) \in M$ such that $p \in V(x)$ and $q \in V(y)$.
\item There exists $(p, q) \in M$ such that $(p, q) \in V(x) \times V(y)$.
\end{enumerate}
\end{lemma}
\begin{proof}
$(1) \Rightarrow (2)$: By definition of composition, there exists $q \in X$ such that $(x, q) \in M$ and $(q, y) \in V$. Similarly, there exists $p \in X$ such that $(x, p) \in V$, $(p, q) \in M$, and $(q, y) \in V$.
$(2) \Rightarrow (3)$: By symmetry, $(p, x) \in V$. Thus $p \in V(x)$ and $q \in V(y)$.
$(3) \Leftrightarrow (4)$, $(5) \Leftrightarrow (6)$: By definition of product.
$(3) \Rightarrow (5)$: By symmetry, $a \in V(b)$ if and only if $b \in V(a)$.
$(5) \Rightarrow (1)$: By symmetry, $p \in V(x)$ implies that $(x, p) \in V$, and $(q, y) \in V$.
\end{proof}
\begin{proposition}[{{\cite[Proposition 2.1.2]{Bourbaki}}}]
\label{proposition:uniformneighbourhood}
Let $(X, \fU)$ be a uniform space, $V \in \fU$ be a symmetric entourage, and $M \subset X \times X$, then:
\begin{enumerate}
\item $V \circ M \circ V \in \cn(M)$.
\item Let $\fB$ be the set of all symmetric entourages, then $\ol{M} = \bigcap_{V \in \fB}V \circ M \circ V$.
\end{enumerate}
with respect to the product topology on $X \times X$.
\end{proposition}
\begin{proof}
(1): Let $(x, y) \in V \circ M \circ V$. By \ref{lemma:compositiongymnastics}, there exists $(p, q) \in M$ such that $(x, y) \in V(p) \times V(q) \in \cn(p, q)$. In particular, if $(x, y) \in M$, then this implies that $V \circ M \circ V \in \cn(x, y)$. Thus $V \circ M \circ V \in \cn(M)$.
(2): Let $(x, y) \in X \times X$. By \ref{lemma:compositiongymnastics}, the following are equivalent:
\begin{enumerate}
\item[(a)] $(x, y) \in V \circ M \circ V$ for all $V \in \fB$.
\item[(b)] For every $V \in \fB$, there exists $(p, q) \in M$ such that $(p, q) \in V(x) \times V(y)$.
\end{enumerate}
As $\bracsn{V(x) \times V(y)| V \in \fB}$ is a fundamental system of neighbourhoods at $(x, y)$, (b) is equivalent to $(x, y) \in \ol{M}$. Therefore
\[
\ol{M} = \bigcap_{V \in \fB}V \circ M \circ V
\]
\end{proof}
\begin{proposition}[{{\cite[Corollary 2.1.1]{Bourbaki}}}]
\label{proposition:uniformclosure}
Let $(X, \fU)$ be a uniform space and $A \subset X$, then:
\begin{enumerate}
\item For any symmetric entourage $V \in \fU$, $V(A) \in \cn(A)$.
\item Let $\fB \subset \fU$ be the family of all symmetric entourages, then $\ol{A} = \bigcap_{U \in \fB}U(A)$.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): For each $x \in A$, $x \in V(x) \subset V(A)$. By (F1) of $\cn(x)$, $V(A) \in \cn(x)$ for all $x \in A$. Thus $V(A) \in \cn(A)$.
(2): Let $x \in X$, then the following are equivalent:
\begin{enumerate}
\item[(a)] For all $V \in \fB$, $x \in V(A)$.
\item[(b)] For all $V \in \fB$, there exists $y \in A$ such that $x \in V(y)$.
\item[(c)] For all $V \in \fB$, $V(x) \cap A \ne \emptyset$.
\end{enumerate}
Since $\bracs{V(y): V \in \fB}$ is a fundamental system of neighbourhoods at $y$ (\ref{lemma:symmetricfundamentalentourage}), (c) is equivalent to $x \in \overline{A}$. Therefore $\ol{A} = \bigcap_{U \in \fB}U(A)$.
\end{proof}
\begin{proposition}[{{\cite[Corollary 2.1.2]{Bourbaki}}}]
\label{proposition:goodentourages}
Let $(X, \fU)$ be a uniform space, then the following famillies of sets form fundamental systems of entourages for $\fU$:
\begin{enumerate}
\item $\mathfrak{O} = \bracs{U^o| U \in \fU}$
\item $\mathfrak{K} = \bracsn{\overline{U}| U \in \fU}$.
\end{enumerate}
By \ref{lemma:symmetricfundamentalentourage}, there exists fundamental systems of entourages for $\fU$ consisting of symmetric and open/closed sets.
\end{proposition}
\begin{proof}
Let $U \in \fU$, then there exists a symmetric entourage $V \in \fU$ such that $V \circ V \circ V \subset U$ by (U2) and \ref{lemma:symmetricfundamentalentourage}. By (1) of \ref{proposition:uniformneighbourhood}, $V \circ V \circ V \in \cn(V)$. Since
\[
V \subset (V \circ V \circ V)^o \subset V \circ V \circ V \subset U
\]
the interior $(V \circ V \circ V)^o \in \fU$, and $U$ contains the interior of an entourage. Thus (1) is a fundamental system of entourages.
On the other hand, by (2) of \ref{proposition:uniformneighbourhood},
\[
V \subset \overline{V} \subset V \circ V \circ V \subset U
\]
So $\overline{V} \in \fU$ and is contained in $U$. Therefore (2) is also a fundamental system of entourages.
\end{proof}
\begin{lemma}
\label{lemma:openentourageneighbourhoods}
Let $(X, \fU)$ be a uniform space, $U \in \fU$ be a symmetric, open entourage, and $M \subset X$, then $U(M)$ is open.
\end{lemma}
\begin{proof}
Let $U \in \fV$ and $y \in U(M)$. Since $U \subset X \times X$ is open, there exists $x \in M$, $V \in \cn(x)$, and $V' \in \cn(y)$ such that $(x, y) \subset V \times V' \subset U$. In which case, $U(M) \supset V' \in \cn(y)$. Hence $U(M)$ is open.
\end{proof}
\begin{proposition}
\label{proposition:uniform-neighbourhoods}
Let $X$ be a uniform space and $x \in X$, then the closed neighbourhoods of $x$ form a fundamental system of neighbourhoods at $x$.
\end{proposition}
\begin{proof}
By \ref{proposition:goodentourages} and \ref{lemma:openentourageneighbourhoods}, the closed neighbourhoods form a fundamental system of neighbourhoods.
\end{proof}
\begin{proposition}
\label{proposition:uniform-regular}
Let $(X, \fU)$ be a uniform space, then the following are equivalent:
\begin{enumerate}
\item $X$ is T0.
\item $X$ is T1.
\item $X$ is Hausdorff.
\item $X$ is regular.
\item $\Delta = \bigcap_{U \in \fU}U$.
\end{enumerate}
\end{proposition}
\begin{proof}
$(1) \Rightarrow (5)$: Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U(x) \in \cn(x)$ such that $y \not\in U$. In which case, $(x, y) \not\in U$ and $\Delta \supset \bigcap_{U \in \fU}U$.
$(5) \Rightarrow (2)$: By \ref{proposition:goodentourages}, $\ol \Delta \subset \bigcap_{U \in \fU}\ol U = \Delta$, so $\ol \Delta$ is closed. By (6) of \ref{definition:hausdorff}, $X$ is Hausdorff.
$(1) \Rightarrow (4)$: $X$ is T1 and satisfies (2) of \ref{definition:regular} by \ref{proposition:uniform-neighbourhoods}, so $X$ is regular.
$(4) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1)$: (T3) $\Rightarrow $ (T2) $\Rightarrow$ (T1) $\Rightarrow$ (T0).
\end{proof}
\begin{proposition}
\label{proposition:subspace-entourage}
Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $\bracsn{\overline{U}: U \in \fU_A}$ forms a fundamental system of entourages for $X$.
\end{proposition}
\begin{proof}
Let $U \in \fU$ be an open entourage, then by (3) of \ref{definition:dense}, $\overline{U \cap (A \times A)} = \overline{U}$ for all $U \in \fU$, so $\overline{U \cap (A \times A)}$ is an entourage. By \ref{proposition:goodentourages}, every closed entourage of $X$ contains an element of $\bracsn{\overline{U}: U \in \fU_A}$.
\end{proof}

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\chapter{Uniform Spaces}
\label{chap:uniform-spaces}
\input{./src/topology/uniform/definition.tex}
\input{./src/topology/uniform/uc.tex}
\input{./src/topology/uniform/complete.tex}
\input{./src/topology/uniform/completion.tex}

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\section{Uniform Continuity}
\label{section:uc}
\begin{definition}[Uniform Continuity]
\label{definition:uniformcontinuity}
Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f: X \to Y$, then the following are equivalent:
\begin{enumerate}
\item For every $V \in \mathfrak{V}$, there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$.
\item For every $V \in \mathfrak{V}$, $(f \times f)^{-1}(V) \in \fU$.
\end{enumerate}
If the above holds, then $f$ is a \textbf{uniformly continuous} function.
The collection $UC(X; Y)$ denotes the set of all uniformly continuous functions from $X$ to $Y$.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: Since $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$, $V' \subset f^{-1}(V)$. By (F1) of $\fU$, $f^{-1}(V) \in \fU$.
$(2) \Rightarrow (1)$: Take $V' = (f \times f)^{-1}(V)$, then $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$.
\end{proof}
\begin{proposition}
\label{proposition:uniform-continuous}
Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f \in UC(X; Y)$, then $f \in C(X; Y)$.
\end{proposition}
\begin{proof}
Let $x \in X$ and $V(f(x)) \in \cn(f(x))$, then since $f^{-1}(V(f(x))) =[f^{-1}(V)](x)$, by (2) of \ref{definition:uniformcontinuity}, $f^{-1}(V(f(x))) \in \cn(x)$. As this holds for all $x \in X$, $f$ is continuous.
\end{proof}
\begin{definition}[Initial Uniformity]
\label{definition:initial-uniformity}
Let $X$ be a set, $\bracsn{(Y_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, and $\seqi{f}$ be a family of maps such that $f_i: X \to Y_i$ for each $i \in I$, then there exists a uniformity $\fU$ on $X$ such that:
\begin{enumerate}
\item For each $i \in I$, $f_i \in UC(X; Y_i)$.
\item[(U)] If $\mathfrak{V}$ is a uniformity on $X$ satisfying $(1)$, then $\mathfrak{V} \supset \fU$.
\end{enumerate}
Moreover,
\begin{enumerate}
\item[(3)] The family
\[
\fB = \bracs{\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \fU_j}
\]
is a fundamental system of entourages for $\fU$.
\end{enumerate}
known as the \textbf{initial uniformity} on $X$ generated by $\seqi{f}$.
\end{definition}
\begin{proof}
(3): Since the diagonal is mapped to the diagonal and $\fB$ is closed under intersections, it is sufficient to verify (UB3) for $\fB$. Let $J \subset I$ be finite and $\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j) \in \fB$, then there exists $\bracs{V_j}_{j \in J}$ such that $V_j \circ V_j \subset U_j$ for each $j \in J$. In which case, for any $(x, y), (y, z) \in f_j^{-1}(V_j)$, $(f(x), f(y)), (f(y), f(z)) \in V_j$ and $(f(x), f(z)) \in U_j$. Thus $(f_j \times f_j)^{-1}(V_j) \circ (f_j \times f_j)^{-1}(V_j) \subset (f_j \times f_j)^{-1}(U_j)$, and
\[
\paren{\bigcap_{j \in J}(f_j \times f_j)^{-1}(V_j)} \circ \paren{\bigcap_{j \in J}(f_j \times f_j)^{-1}(V_j)} \subset \bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j)
\]
By \ref{proposition:fundamental-entourage-criterion}, there exists a uniformity $\fU$ such that $\fB$ is a fundamental system of entourages for $\fU$.
(1): $\fU \supset (f_i \times f_i)^{-1}(\fU_i)$ for all $i \in I$.
(U): For any $i \in I$, $\mathfrak{V} \supset (f_i \times f_i)^{-1}(\fU_i)$. By (F2), $\mathfrak{V} \supset \fB$, so $\mathfrak{V} \supset \fU$.
\end{proof}
\begin{proposition}
\label{proposition:initial-uniform-topology}
Let $X$ be a set, $\bracsn{(Y_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, and $\seqi{f}$ where $f_i: X \to Y_i$ for each $i \in I$, then the initial topology $\topo$ on $X$ coincides with the topology $\topo_U$ induced by the initial uniformity.
\end{proposition}
\begin{proof}
By \ref{proposition:uniform-continuous} and (1) of \ref{definition:initial-uniformity}, $f_i: X \to Y_i$ is continuous with respect to $\topo_U$ for all $i \in I$. By (U) of \ref{definition:initial-topology}, $\topo_U \supset \topo$.
On the other hand, let $x \in X$ and $U \in \cn_{\topo_U}(x)$, then there exists an entourage $V$ such that $U = V(x)$. By (3) of \ref{definition:initial-uniformity}, there exists $J \subset I$ finite and $\seqj{V}$ such that $\bigcap_{j \in J}(f_j \times f_j)^{-1}(V_j) \subset V$ and each $V_j \in \fU_j$. In which case, $f_j^{-1}(V_j(f_j(x))) \in \cn(x)$ for all $j \in J$. Using (F2),
\[
W = \bigcap_{j \in J}f_j^{-1}(V_j(f_j(x))) \in \cn(x)
\]
where for any $y \in W$, $(f_j(x), f_j(y)) \in V_j$ for all $j \in J$. Thus $f(y) \in V(x) = U$, and $\topo_U \subset \topo$.
\end{proof}
\begin{definition}[Product Uniformity]
\label{definition:product-uniform}
Let $\bracs{(X_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, then the \textbf{product uniformity} $\fU$ on $X = \prod_{i \in I}X_i$ is the initial uniformity generated by the projections $\seqi{\pi}$, and:
\begin{enumerate}
\item The family
\[
\fB = \bracs{\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \fU_j}
\]
forms a fundamental system of entourages for $\fU$.
\item[(U)] For any uniform space $(Y, \mathfrak{V})$ and $\seqi{f}$ where $f_i \in UC(Y; X_i)$ for all $i \in I$, there exists a unique $f \in UC(Y; X)$ such that the following diagram commutes
\[
\xymatrix{
Y \ar@{->}[d]_{f} \ar@{->}[rd]^{f_i} & \\
X \ar@{->}[r]_{\pi_i} & X_i
}
\]
for all $i \in I$.
\end{enumerate}
\end{definition}
\begin{proof}
(1): By (3) of \ref{definition:initial-uniformity}.
(U): Let $f = \prod_{i \in I}f_i$, then $f: Y \to X$ is the unique function such that the diagram commutes for all $i \in I$.
For each $J \subset I$ finite and $\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j) \in \fU$,
\[
(f \times f)^{-1}\paren{\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_i)} = \bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j) \in \mathfrak{V}
\]
by (F2) of $\mathfrak{V}$.
\end{proof}
\begin{proposition}
\label{proposition:product-uniform-topology}
Let $\bracs{(X_i, \fU_i)}_{i \in I}$ be a family of uniform spaces, then the topology $\topo_U$ induced by the product uniformity coincides the product topology $\topo_P$.
\end{proposition}
\begin{proof}
By \ref{proposition:initial-uniform-topology}.
\end{proof}
\begin{proposition}
\label{proposition:initial-product-uniformity}
Let $X$, $\seqi{Y}$ be uniform spaces, $\seqi{f}$ where $f_i \in UC(X; Y_i)$ for each $i \in I$, then
\begin{enumerate}
\item There exists a unique $f \in UC(X; \prod_{i \in I}Y_i)$ such that the following diagram commutes
\[
\xymatrix{
& \prod_{i \in I} Y_i \ar@{->}[rd]^{\pi_i} & \\
X \ar@{->}[rr]_{f_i} \ar@{->}[ru]^{f} & & Y_i
}
\]
for all $i \in I$.
\item If $X$ is T0 and equipped with the initial uniformity induced by $\seqi{f}$, then $f$ is an isomorphism onto its image.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): By (U) of \ref{definition:product-uniform}.
(2): By (2) of \ref{proposition:initial-product-topology}, $f$ is injective.
Let $U$ be an entourage in $X$. Assume without loss of generality that there exists $J \subset I$ finite and $\bracs{U_j}_{j \in J}$ such that $U = \bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j)$. In which case, $\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j)$ is open in $\prod_{i \in I}Y_i$ and
\[
(f \times f)(U) = (f \times f)(X) \cap \bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j)
\]
\end{proof}