Metric Space is Perfectly T4

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Theorem

Let $M = \struct {A, d}$ be a metric space.

Then $M$ is a perfectly $T_4$ space.


Proof

We have that a metric space is $T_4$.

We also have that every closed set in a metric space is a $G_\delta$ set.

Hence the result, by definition of a perfectly $T_4$ space.

$\blacksquare$


Sources