Totally Separated Space is Completely Hausdorff and Urysohn
Theorem
Let $T = \struct {S, \tau}$ be a topological space which is totally separated.
Then $T$ is completely Hausdorff and Urysohn.
Proof
Let $T = \struct {S, \tau}$ 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 $f: S \to \closedint 0 1$ such that:
- $\Img {f {\restriction_U} } = \set 0$
- $\Img {f {\restriction_V} } = \set 1$
where:
- $f {\restriction_U}$ and $f {\restriction_V}$ denote the restrictions of $f$ to $U$ and $V$ respectively
- $\Img g$ 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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness