Urysohn Space is Completely Hausdorff Space

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Theorem

Let $\left({S, \tau}\right)$ be an Urysohn space.


Then $\left({S, \tau}\right)$ is also a $T_{2 \frac 1 2}$ (completely Hausdorff) space.


Proof

Let $T = \left({S, \tau}\right)$ be an Urysohn space.

Then for any distinct points $x, y \in S$ (i.e. $x \ne y$), there exists an Urysohn function for $\left\{{x}\right\}$ and $\left\{{y}\right\}$.



Thus:

$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \varnothing$

which is precisely the definition of a $T_{2 \frac 1 2}$ (completely Hausdorff) space.

$\blacksquare$


Sources