Urysohn Space is Completely Hausdorff Space

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 $\set y$.



This theorem requires a proof.
In particular: Then we do some stuff.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.



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: $\S 2: \ 12$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.