# Homework Help: Topology: $\mathscr{T}_{2.5}\Rightarrow\mathscr{T}_2$

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1. May 6, 2015

### Kevin_H

The book I am using for my Introduction to Topology course is Principles of Topology by Fred H. Croom.

We are going over separation axioms in class when we were asked to prove that every Urysohn Space is a Hausdorff.

What I understand:
• A space $X$ is Urysohn space provided whenever for any two distinct points of $X$ there are neighborhoods of $U$ of $x$ and $V$ of $y$ such that $\overline{U}\cap \overline{V}=\emptyset.$
• A space $X$ is Hausdorff if for any two distinct points of $X$ can be separated by open neighborhoods $(x\in U,y\in V,U\cap V=\emptyset)$.
My rough attempt at proving this:
• Given $(X,\mathscr{T})$ to be a Urysohn space, let $x,y\in X$ be distinct points. Then there are neighborhoods of $U$ of $x$ and $V$ of $y$ such that $\overline{U}\cap \overline{V}=\emptyset$. We seek to prove there exists $x\in B$ open $\subset X$ and $y\in A$ open $\subset X$ such that $A\cap B=\emptyset$. Consider $X\setminus\overline{U}$. This is an open neighborhood of $y$, thus there exists $y\in A\subset\overline{A}\subset X\setminus\overline{U}$. Consider $X\setminus \overline{A}$. This is an open neighborhood of $x$, thus there exist $x\in B\subset X\setminus\overline{A}.$ Thus $A\cap B=\emptyset$. QED.
Is my proof okay? Am I missing anything? Any suggestions? If you need me to clarify anything, please let me know. Thank you for taking the time to read this post. I greatly appreciate any assistance you may provide.

2. May 6, 2015

### Dick

The proof is ok. I might do a little more direct approach. If $\overline U$ is a neighborhood of $x$, doesn't that mean that it contains an open set that contains $x$? Why not just use that?

3. May 6, 2015

### Kevin_H

Thank You very much for the feedback. I was originally thinking of approaching it that way, but I wasn't sure if I was allowed to assume the closure had an open subset containing $x$. Now that I know, I can assume both $\overline{U}$ and $\overline{V}$ have open neighborhoods containing their respective point, $x\in B$ and $y\in A$. Since the closures are disjoint, then their open subsets should also be disjoint; satisfying the conditions for Hausdorff.