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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces