Urysohn Space is Completely Hausdorff Space
Theorem
Let $\left({X, \vartheta}\right)$ be an Urysohn space.
Then $\left({X, \vartheta}\right)$ is also an $T_{2 \frac 1 2}$ (completely Hausdorff) space.
Proof
Let $T = \left({X, \vartheta}\right)$ be an Urysohn space.
Then for any distinct points $x, y \in X$ (i.e. $x \ne y$), there exists an Urysohn function for $\left\{{x}\right\}$ and $\left\{{y}\right\}$.
Then we do some stuff.
You can help Proof Wiki by expanding it.
To discuss it in more detail, feel free to use the talk page.
When this page/section has been completed, {{Stub}} should be removed from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page (see the proofread template for usage).
Thus:
- $\forall x, y \in X: x \ne y: \exists U, V \in \vartheta: x \in U, y \in V: U^- \cap V^- = \varnothing$
which is precisely the definition of an $T_{2 \frac 1 2}$ (completely Hausdorff) space.
$\blacksquare$
Sources
- Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 2$: Additional Separation Properties
- Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970): Problems: $\S 2: \ 12$
