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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}