Metric Space Fulfils All Separation Axioms
From ProofWiki
Theorem
Let $M = \left({A, d}\right)$ be a metric space.
Then $M$, considered as a topological space, fulfils all separation axioms:
- $M$ is a $T_0$ (Kolmogorov) space
- $M$ is a $T_1$ (Fréchet) space
- $M$ is a $T_2$ (Hausdorff) space
- $M$ is a semiregular space
- $M$ is a $T_3$ space
- $M$ is a regular space
- $M$ is an Urysohn space
- $M$ is a $T_{3 \frac 1 2}$ space
- $M$ is a Tychonoff (completely regular) space
- $M$ is a $T_4$ space
- $M$ is a normal space
- $M$ is a $T_5$ space
- $M$ is a completely normal space
- $M$ is a perfectly $T_4$ space
- $M$ is a perfectly normal space
Proof
We have that:
- A metric space is a $T_2$ (Hausdorff) space.
- A metric space is a perfectly $T_4$ space.
- A metric space is a $T_5$ space.
From Sequence of Implications of Separation Axioms we then have:
- $T_2$ (Hausdorff) Space is $T_1$ (Fréchet) Space.
- $T_1$ (Fréchet) Space is $T_0$ (Kolmogorov) Space.
By definition, a perfectly normal space is:
So $M$ is a perfectly normal space.
By definition, a completely normal space is:
So $M$ is a completely normal space.
Then from Sequence of Implications of Separation Axioms we can complete the chain:
- Completely Normal Space is Normal Space.
- Normal Space is Tychonoff (Completely Regular) Space.
- Completely Regular (Tychonoff) implies $T_{3 \frac 1 2}$ by definition.
- Completely Regular (Tychonoff) Space is Regular Space.
- Completely Regular (Tychonoff) Space is Urysohn Space.
- Regular implies $T_3$ by definition.
- Regular Space is Completely Hausdorff Space.
- Regular Space is Semiregular Space.
$\blacksquare$
There are other chains of implications which can be used, but the above are sufficient to prove the hypothesis.
Sources
- Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$
