separation axioms

Quote:

Originally Posted by flower3

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

Definition. A space X is a $T_{2\frac{1}{2}}$ -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 $x \in U$ and $y \in V$ .

Lemma 1. X is regular Hausdorff ( $T_3$ ) 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 $\bar{W} \subset U$ .

Assume X is regular Hausdorff ( $T_3$ ). 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 $\bar{W} \subset U$ and a neighborhood Y of y such that $\bar{Y} \subset V$ by lemma 1. Since U and V are disjoint, we see that $\bar{W}$ and $\bar{Y}$ 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 $x \in W$ and $y \in Y$ .Thus, X is $T_{2\frac{1}{2}}$ -space.

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