Metric Space fulfils all Separation Axioms

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 $T_{2 \frac 1 2}$ (completely Hausdorff) 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:
 * $T_5$ space is $T_4$ space.


 * $T_2$ (Hausdorff) Space is $T_1$ (Fréchet) Space.
 * $T_1$ (Fréchet) Space is $T_0$ (Kolmogorov) Space.

By definition, a perfectly normal space is:
 * a perfectly $T_4$ space
 * a $T_1$ (Fréchet) space

So $M$ is a perfectly normal space.

By definition, a completely normal space is:


 * a $T_5$ space
 * a $T_1$ (Fréchet) space

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.

There are other chains of implications which can be used, but the above are sufficient to prove the hypothesis.