Discrete Space satisfies all Separation Properties

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 Discrete Metric induces Discrete Topology.

Then we use Metric Space fulfils all Separation Axioms.