> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. $(X, \topo)$ is a **Hausdorff space** if the following equivalent conditions hold: > 1. For every $x, y \in X$ with $x \ne y$, there exists [[Neighbourhood|neighbourhoods]] $U \in \cn(x)$ and $V \in \cn(y)$ such that $U \cap V = \emptyset$. > 2. For every $x \in X$, $\bigcap_{U \in \cn(x)}\overline{U} = \bracs{x}$. > 3. Every [[Net|net]] in $X$ converges to at most one point. > 4. Every [[Filter|filter]] in $X$ converges to at most one point. > 5. Every convergent net in $X$ has exactly one cluster point. > 6. Every convergent [[Filter|filter]] in $X$ has exactly one cluster point. > 7. The diagonal of the product $X^I$ is closed for any index set $I$. > 8. The diagonal of the product $X \times X$ is [[Closed Set|closed]]. > > *Proof*. $(1) \Rightarrow (2)$: Let $y \ne x$ and $U, V \subset X$ be open neighbourhoods as in $(1)$, then $\overline{U}^c = (U^c)^p \supset V$. Hence $y \not\in \overline{U} \supset \bigcap_{W \in \cn(x)}\overline{U}$. > > $(2) \Rightarrow (3), (4)$: Let $x \ne y$, then there exists $U \in \cn(x)$ such that $y \not\in \overline{U}$, and $\overline{U}^c \in \cn(y)$. > > Let $\net{x}$ with $x_\alpha \to x$ and $\alpha_0 \in A$ such that $x_\alpha \in \overline{U}$ for all $\alpha \gtrsim \alpha_0$, then $x_\alpha \not\in \overline{U}^c$ for all $\alpha \ge \alpha_0$. Thus $x_\alpha \not\to y$. > > Let $\fF \supset \cn(x)$ be a filter, then $\overline{U} \in \cn(x)$, and $\overline{U}^c \not\in \fF$ and $\fF \not\supset \cn(x)$. > > $(3) \Rightarrow (5)$, $(4) \Rightarrow (6)$: Convergent nets/filters have their limit points as their cluster points. > > $(5) \Rightarrow (7)$: Let $\net{x}$ be a net in the diagonal such that $x_\alpha \to y$ in $X^I$, then $\pi_i(x_\alpha) \to \pi_i(y)$ for all $i \in I$. For each $i, j \in I$, $\pi_i(x_\alpha) = \pi_j(x_\alpha)$, so by uniqueness of cluster points, $\pi_i(y) = \pi_j(y)$. > > $(6) \Rightarrow (7)$: Let $\fF$ be a filter in the diagonal such that $\fF \to y$ in $X^I$. For every $i \in I$, $\pi_i(\fF) \to \pi_i(y)$ as a filter base. For each $i, j \in I$, $\pi_i(\fF) = \pi_j(\fF)$, so by uniqueness of cluster points, $\pi_i(y) = \pi_j(y)$. > > $(7) \Rightarrow (8)$: Take $I = \bracs{1, 2}$. > > $(8) \Rightarrow (1)$: Let $x, y \in X$ with $x \ne y$, then $(x, y)$ is not in the diagonal of $X \times X$. Thus there exists $U \in \cn(x)$ and $V \in \cn(y)$ such that $U \times Y$ does not intersect the diagonal, and $U \times V = \emptyset$. > [!theorem] > > Let $\bracs{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $(X, \topo)$ be their [[Product Topology|product topology]]. If $(X_i, \topo_i)$ is Hausdorff for each $i \in I$, then $(X, \topo)$ is also Hausdorff. > > *Proof*. Let $x, y \in X$, $x \ne y$. Then there exists $i \in I$ such that $\pi_i(x) \ne \pi_i(y)$. Since $(X_i, \topo_i)$ is Hausdorff, there exists disjoint open sets $U_i, V_i$ such that $\pi_i(x) \in U_i, \pi_i(y) \in V_i$. As the projection maps are [[Continuity|continuous]] $\pi_i^{-1}(U_i)$ and $\pi_i^{-1}(V_i)$ are disjoint open sets such that $x \in \pi_i^{-1}(U_i)$ and $y \in \pi_i^{-1}(V_i)$. Therefore $(X,\topo)$ is also Hausdorff. > [!theorem] > > Let $(X, \topo)$ be a Hausdorff space, $F$ [[Compactness|compact]] and $x \not\in F$. Then there exists disjoint open sets $U, V$ such that $x \in U$, $F \subset V$. > > *Proof*. For all $y \in F$, and take disjoint $U_y$ and $V_y$ such that $x \in U_y$ and $y \in V_y$. Then $\bracs{V_y}_{y \in F}$ is an open cover of $F$, and has a finite subcover $\bracs{V_i}_{1}^{n}$. Let $U = \bigcap_{i = 1}^{n}U_i$ and $V = \bigcup_{i = 1}^{n}V_i$, then $U, V$ are disjoint with $x \in U$ and $F \subset V$. > [!theorem] > > Let $(X, \topo)$ be a Hausdorff space and $F$ be a compact set, then $F$ is [[Closed Set|closed]]. > > *Proof*. Let $x \not\in F$. For each $y \in F$, there exists $U_y \in \cn^o(x)$ and $V_y \in \cn^o(y)$ such that $U_y \cap V_y \ne \emptyset$. By taking a subcover corresponding to $\seqf{y_j}$, $\bigcap_{j = 1}^n V_{y_j} \in \cn^o(y)$ does not intersect $F$. > [!theorem] > > Let $(X, \topo)$ be a [[Compactness|compact]] Hausdorff space, then $X$ is [[Normal Space|normal]]. > > *Proof*. Let $E, F$ be closed sets, then they are also compact. For each $x \in E$, choose $U_x, V_x$ disjoint such that $x \in U_x$, $F \subset V_x$. Then $\bracs{U_x}_{x \in E}$ is an open cover of $E$, and has a finite subcover $\bracs{U_i}_{1}^{n}$. Let $U = \bigcup_{i = 1}^{n}U_i$ and $V = \bigcap_{i = 1}^{n}V_i$, then $U, V$ are disjoint with $E \subset U$, $F \subset V$. > [!theorem] > > Let $X$ be a topological space. $X$ is **Hausdorff** if and only if every [[Net|net]] in $X$ converges to at most one point. > > *Proof*. Suppose that $X$ is Hausdorff and let $\net{x}$ be a net such that $x_\alpha \to x$. Let $y \ne x$, then there exists $U \in \cn(x)^o$ and $V \in \cn(y)^o$ such that $U \cap V = \emptyset$. Since $x_\alpha \to x$, there exists $\beta \in A$ such that $x_\alpha \in U, x_\alpha \not \in V$ for all $\alpha \gtrsim \beta$. Therefore $\net{x}$ is not eventually in $V$, and $x_\alpha \not\to y$. > > Suppose that $X$ is not Hausdorff. Let $x \ne y$ such that there exists no disjoint open neighbourhoods that separate them. Consider the [[Directed Set|directed set]] $\cn(x)^o \times \cn(y)^o$. For each $\alpha = (U, V)$ where $U \in \cn(x)^o$ and $V \in \cn(y)^0$, choose $x_\alpha \in U \cap V$, which is possible since $x$ and $y$ have no disjoint neighbourhoods. Then for any $U \in \cn(x)^o$, $x_{(U',V)} \in U$ for all $(U', V) \gtrsim (U, X)$. For any $V \in \cn(y)^o$, $x_{(U, V')} \in V$ for all $(U, V') \gtrsim (X, V)$. Therefore $x_\alpha \to x$ and $x_\alpha \to y$, and the limit is not unique.