Urysohn Space is Completely Hausdorff Space
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Theorem
Let $\struct {S, \tau}$ be an Urysohn space.
Then $\struct {S, \tau}$ is also a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
Proof
Let $T = \struct {S, \tau}$ be an Urysohn space.
Then for any distinct points $x, y \in S$ (i.e. $x \ne y$), there exists an Urysohn function for $\set x$ and $\set y$.
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Thus:
- $\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$
which is precisely the definition of a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Additional Separation Properties
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.): Problems: $\S 2: \ 12$