Metrizable implies perfectly normal

Statement
Any metrizable space, i.e., any space realized as the topological space for a metric space, is a perfectly normal space -- it is a normal space and every closed subset of it is a G-delta subset (it is a countable intersection of open subsets).

Facts used

 * 1) uses::Metrizable implies normal

Proof
Given: A metric space $$(X,d)$$. with the topology arising from the metric.

To prove: $$X$$ is a perfectly normal space: $$X$$ is a normal space and for every closed subset $$A$$ of $$X$$, there is a countable collection of open subsets $$U_n$$ of $$X$$ such that $$A$$ equals the intersection of the $$U_n$$s.

Proof: By fact (1), $$X$$ is a normal space, so we show the second part of the definition. For the closed subset $$A$$, define $$U_n$$ as the set of all points $$p \in X$$ such that there exists a point $$a \in A$$ such that $$d(p,a) < (1/n)$$. Then:


 * 1) Each $$U_n$$ is open: $$U_n$$ is the union of the open balls of radius $$1/n$$ about all the points of $$A$$. Hence, it is a union of open subsets, hence open.
 * 2) The intersection of the $$U_n$$s contains $$A$$.
 * 3) If $$p$$ is not in $$A$$, there is some $$U_n$$ such that $$p \notin U_n$$: Since $$A$$ is closed, there exists $$\epsilon$$ such that the ball of radius $$\epsilon$$ about $$p$$ does not intersect $$A$$. In other words, there is no point of $$A$$ whose distance from $$p$$ is less than $$\epsilon$$. Let $$n$$ be a positive integer greater than $$1/\epsilon$$. Then, $$U_n$$ does not contain $$p$$.

Together, (1), (2) and (3) complete the proof.