Hausdorff Bookkeeping

In this exercise, we are going to verify various properties of Hausdorff spaces.

Problem 1. Suppose $K$ is a Hausdorff space. Prove that any subspace $L \subseteq K$ equipped with the subspace topology is also Hausdorff.

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Solution. Recall that $U \subseteq L$ is open if and only if there exists an open set $V \subseteq K$ such that $U = L \cap V$ . Let $x_1,x_2 \in L$ be distinct points. Since $K$ is Hausdorff, there exist disjoint neighbourhoods $V_i$ of $x_i$ in $K$ . Then $U_i := L \cap V_i$ are disjoint neighbourhoods of $x_i$ in $L$ , as required.

Problem 2. Let $I$ be any index set and $\{K_\alpha\}_{\alpha \in I}$ be a collection of Hausdorff spaces. Prove that $\prod_{\alpha \in I} K_\alpha$ is Hausdorff in the product topology.

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Solution. Fix distinct points $\mathbf x, \mathbf y \in \prod_{\alpha \in I} K_\alpha$ . Then there exists $\alpha \in I$ such that $\mathbf x_\alpha \neq \mathbf y_\alpha$ . Find disjoint neighbourhoods $U_\alpha, V_\alpha$ of $\mathbf x_\alpha, \mathbf y_\alpha$ respectively. Define the disjoint neighbourhoods $U = \prod_{\beta} W_\beta, V = \prod_{\beta} X_\beta$ of $\mathbf x, \mathbf y$ by

$W_\beta := \begin{cases} U_\beta & \alpha = \beta, \\ K_\beta & \alpha \neq \beta, \end{cases}\quad X_\beta := \begin{cases} V_\beta & \alpha = \beta, \\ K_\beta & \alpha \neq \beta. \end{cases}$

as required.

Problem 3. Prove that $K$ is a Hausdorff space if and only if its diagonal

$\Delta_K := \{t \times t : t \in K\} \subseteq K \times K =: K^2$

is closed.

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Solution. For the direction $(\Rightarrow)$ , suppose $K$ is a Hausdorff space. Fix $s \times t \in K^2 \backslash \Delta_K$ . Since $s \times t \notin \Delta_K$ , $s \neq t$ . Hence, there exist disjoint neighbourhoods $U, V$ of $s, t$ respectively. Then $U \times V$ is a neighbourhood of $s \times t$ .

We claim that $U \times V \subseteq K^2 \backslash \Delta_K$ . Suppose otherwise. Then there exists $s' \times t' \in (U \times V) \cap \Delta_K$ . Hence, $s' = t'$ . Thus, $\emptyset = U \cap V = \{s'\}$ , a contradiction. Therefore, $U \times V \subseteq K^2 \backslash \Delta_K$ . Since $s \times t$ has been chosen arbitrarily, $K^2 \backslash \Delta_K$ is open, so that $\Delta_K$ is closed, as required.

For the direction $(\Leftarrow)$ , suppose $\Delta_K$ is closed. Then $K^2 \backslash \Delta_K$ is open. Fix distinct $s, t \in K$ . Then $s \times t \in K^2 \backslash \Delta_K$ . Thus, there exist open sets $U,V \subseteq K$ such that $s \times t \in U \times V \subseteq K^2 \backslash \Delta_K$ . We observe that if $U \cap V$ is not empty, then $U \times V \not\subseteq K^2 \backslash \Delta_K$ , a contradiction. Therefore, $U \cap V = \emptyset$ , as required.