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src/topology/main/hausdorff.tex

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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}