4.8 Hausdorff Spaces
Definition 4.8.1 (Hausdorff). Let $X$ be a topological space, then the following are equivalent:
For any $x, y \in X$ with $x \ne y$, there exists $U \in \cn(x)$ and $V \in \cn(y)$ such that $U \cap V = \emptyset$.
For every $x \in X$, $\bracs{x}= \bigcap_{U \in \cn(x)}\overline{U}$.
Every convergent filter in $X$ has exactly one cluster point.
Every filter in $X$ converges to at most one point.
For any index set $I$, the diagonal $\Delta$ is closed in $X^{I}$.
The diagonal $\Delta$ is closed in $X \times X$.
If the above holds, then $X$ is a T2/Hausdorff space.
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 Definition 4.5.2, $y \not\in \overline{U}\subset \bigcap_{U \in \cn(x)}\overline{U}$.
$(2) \Rightarrow (3)$: Let $\fF \subset 2^{X}$ be a filter and $x \in X$ such that $\cn(x) \subset \fF$, then
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 Definition 4.5.2, 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 Proposition 4.7.2, $\pi_{i}(\fF)$ 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.$\square$
Proposition 4.8.2. 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 $F = G$.
Proof. Let $x \in X$. By (4) Definition 4.5.2, 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 Definition 4.8.1.$\square$
Proposition 4.8.3. Let $\seqi{X}$ be Hausdorff spaces, then $\prod_{i \in I}X_{i}$ is Hausdorff.
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 $\pi_{i}^{-1}(U) \cap \pi_{i}^{-1}(V) = \emptyset$.$\square$