# Real Number Space satisfies all Separation Axioms

## Theorem

Let $\left({\R, \tau_d}\right)$ be the real number line considered as a topological space under the usual (Euclidean) topology.

Then $\left({\R, \tau_d}\right)$ fulfils all separation axioms:

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

## Proof

$\blacksquare$