Metrizable implies perfectly normal
This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., metrizable space) must also satisfy the second topological space property (i.e., perfectly normal space)
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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).
Given: A metric space . with the topology arising from the metric.
To prove: is a perfectly normal space: is a normal space and for every closed subset of , there is a countable collection of open subsets of such that equals the intersection of the s.
Proof: By fact (1), is a normal space, so we show the second part of the definition. For the closed subset , define as the set of all points such that there exists a point such that . Then:
- Each is open: is the union of the open balls of radius about all the points of . Hence, it is a union of open subsets, hence open.
- The intersection of the s contains .
- If is not in , there is some such that : Since is closed, there exists such that the ball of radius about does not intersect . In other words, there is no point of whose distance from is less than . Let be a positive integer greater than . Then, does not contain .
Together, (1), (2) and (3) complete the proof.