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Let X be a topological space. Show that X is $ T_2 $ if, and only if,

$ \alpha $ = {(x, x) $ \in $ X $\times$ X : x $ \in $ X}

is closed at X $\times$ X.

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By Hausdorff definition we have x, y $ \in $ X, with x $ \neq $ y, there are U, V $ \subset $ X open such that x $ \in $ U, y $ \in $ V e U $ \cap $ V = $ \emptyset $.

Assuming that close in X $ \times $ X = A and that to be close $ A^c $ needs to be open. But if X is Hausdorff that means that exist x $\in$ X, U $\subset$ X open, where x $\in$ U. But that doesn't mean X $\times$ X is open?

Gabriel
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1 Answers1

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There is an open set about $(x,y)$ not hitting $\alpha$ iff there is an open set of the form $U\times V$ about $(x,y)$ not hitting $\alpha$, where $U,V\subset X$ are open.
$U\times V$ lies outside $\alpha$ iff $U\cap V=\varnothing$.

John Hopfensperger
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