Metric Space fulfils all Separation Axioms

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Theorem

Let $M = \struct {A, d}$ 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.

$\blacksquare$


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


Sources