A space is Hausdorff iff the diagonal is closed

Let $𝑋$ be a topological space, $𝑋 \times 𝑋$ its product, and $Δ = {(𝑥, 𝑥) : 𝑥 \in 𝑋} \subseteq 𝑋 \times 𝑋$ its diagonal. Then $𝑋$ is Hausdorff iff $Δ \subseteq 𝑋 \times 𝑋$ is closed.

Proof

Let $𝐴 = 𝑋 \times 𝑋 ∖ Δ$ ,

First let $𝑋$ be Hausdorff. and $(𝑥, 𝑦) \in 𝐴$ , so $𝑥 \neq 𝑦$ . Since $𝑋$ is Hausdorff, there exist open neighbourhoods $𝑈$ and $𝑉$ of $𝑥$ and $𝑦$ respectively such that $𝑈 \cap 𝑉 = \emptyset$ , and thus $𝑈 \times 𝑉$ is an open neighbourhood of $(𝑥, 𝑦)$ contained in $𝐴$ . Hence every point in $𝐴$ has an open neighbourhood contained in $𝐴$ , therefore $𝐴$ is open, whence $Δ$ is closed.

Now let $Δ$ be closed, i.e. $𝐴$ be open. Let $𝑥, 𝑦 \in 𝑋$ with $𝑥 \neq 𝑦$ . Then there exists an open neighbourhood $𝑈 \subseteq 𝐴$ of $(𝑥, 𝑦)$ . Canonical projections are open, hence $𝜋 1 (𝑈)$ and $𝜋 2 (𝑈)$ are open neighbourhoods of $𝑥$ and $𝑦$ respectively, which do not intersect. Hence $𝑋$ is Hausdorff.

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