Totally Separated Space is Completely Hausdorff and Urysohn

Theorem
Let $T = \left({X, \vartheta}\right)$ be a topological space which is totally separated.

Then $T$ is completely Hausdorff and Urysohn.

Proof
Let $T = \left({X, \vartheta}\right)$ be a totally separated space.

Then for every $x, y \in X: x \ne y$ there exists a partition $U \mid V$ of $T$ such that $x \in U, y \in V$.

Consider the mapping $f: X \to \left[{0..1}\right]$ defined as:
 * $f \restriction_U = 0$
 * $f \restriction_V = 1$

where $f \restriction_U$ and $f \restriction_V$ are the restrictions of $f$ to $U$ and $V$ respectively.

From Continuous Mapping of Partition and Constant Mapping is Continuous, it follows that $f$ is continuous on $U$ and $V$, and also on $X$ itself.

By definition, $f$ is then an Urysohn function on $x$ and $y$.

Hence $T$ is Urysohn.

Having proved that $T$ is an Urysohn space, it follows from Urysohn Space is Completely Hausdorff Space that $T$ is also completely Hausdorff.