Totally Separated Space is Completely Hausdorff and Urysohn

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

Then $T$ is completely Hausdorff and Urysohn.

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

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

Consider a mapping $f: S \to \left[{0 \,.\,.\, 1}\right]$ such that:
 * $\operatorname {Im} \left({f {\restriction_U} }\right) = \left\{ {0}\right\}$
 * $\operatorname {Im} \left({f {\restriction_V} }\right) = \left\{ {0}\right\}$

where:
 * $f {\restriction_U}$ and $f {\restriction_V}$ denote the restrictions of $f$ to $U$ and $V$ respectively
 * $\operatorname {Im} \left({g}\right)$ denotes the image set of a mapping $g$.

From Constant Mapping is Continuous, both $f {\restriction_U}$ and $f {\restriction_V}$ are continuous on $U$ and $V$ respectively.

From Continuous Mapping of Separation, it follows that $f$ is continuous on $S$ 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.