Discrete Space satisfies all Separation Properties

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Theorem

Let $T = \left({S, \mathcal P \left({S}\right)}\right)$ be the discrete topological space on $S$.


Then $T$ fulfils all separation axioms:

$T$ is a $T_0$ (Kolmogorov) space
$T$ is a $T_1$ (Fréchet) space
$T$ is a $T_2$ (Hausdorff) space
$T$ is a semiregular space
$T$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space
$T$ is a $T_3$ space
$T$ is a regular space
$T$ is an Urysohn space
$T$ is a $T_{3 \frac 1 2}$ space
$T$ is a Tychonoff (completely regular) space
$T$ is a $T_4$ space
$T$ is a normal space
$T$ is a $T_5$ space
$T$ is a completely normal space
$T$ is a perfectly $T_4$ space
$T$ is a perfectly normal space


Proof

We have that a Standard Discrete Metric induces Discrete Topology.

Then we use Metric Space fulfils all Separation Axioms.

$\blacksquare$


Sources