# separation axioms

• May 28th 2009, 11:56 AM
flower3
separation axioms
prove that :
Every$\displaystyle \displaystyle \tau \scriptstyle3$ space is $\displaystyle \displaystyle \tau \scriptstyle 2 \frac{1}{2}$
• May 28th 2009, 01:37 PM
HallsofIvy
A good way to start: write out the definitions of $\displaystyle T_3$ and $\displaystyle T_{2 1/2}$
• May 30th 2009, 02:05 AM
aliceinwonderland
Quote:

Originally Posted by flower3
prove that :
Every$\displaystyle \displaystyle \tau \scriptstyle3$ space is $\displaystyle \displaystyle \tau \scriptstyle 2 \frac{1}{2}$

Definition. A space X is a $\displaystyle T_{2\frac{1}{2}}$-space provided that for each pair x,y of distinct points of X, there exists open sets U and V with disjoint closures such that $\displaystyle x \in U$ and $\displaystyle y \in V$.

Lemma 1. X is regular Hausdorff ($\displaystyle T_3$) if and only if given a point x of X and a neighborhoood U of x, there is a neighborhood W of x such that $\displaystyle \bar{W} \subset U$.

Assume X is regular Hausdorff ($\displaystyle T_3$). Since X is also Hausdorff, for each pair x,y of distinct points of X, there exists disjoint open sets U and V containing x and y, respectively. By hypothesis, it follows that there is a neighborhood W of x such that $\displaystyle \bar{W} \subset U$ and a neighborhood Y of y such that $\displaystyle \bar{Y} \subset V$ by lemma 1. Since U and V are disjoint, we see that $\displaystyle \bar{W}$ and $\displaystyle \bar{Y}$ are disjoint. Now, for each pair x,y of distinct points of X, there exists open sets W and Y with disjoint closures such that $\displaystyle x \in W$ and $\displaystyle y \in Y$.Thus, X is $\displaystyle T_{2\frac{1}{2}}$-space.

The converse ("Every $\displaystyle T_{2\frac{1}{2}}$-space is $\displaystyle T_3$") is not necessarily true. The example where a space X is $\displaystyle T_{2\frac{1}{2}}$-space but not $\displaystyle T_3$ can be found in the K-topology on $\displaystyle \mathbb{Re}$, where a closed set $\displaystyle K=\{1/n | n \in \mathbb{Z}^+\}$ (Note that K is closed in the K-topology on $\displaystyle \mathbb{Re}$ ) and {0} cannot be separated by disjoint open sets.