Thread: separation axioms

prove that :
Every space is

A good way to start: write out the definitions of and

Originally Posted by flower3

prove that :
Every space is

Definition. A space X is a -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 and .

Lemma 1. X is regular Hausdorff ( ) 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 .

Assume X is regular Hausdorff ( ). 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 and a neighborhood Y of y such that by lemma 1. Since U and V are disjoint, we see that and 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 and .Thus, X is -space.

The converse ("Every -space is ") is not necessarily true. The example where a space X is -space but not can be found in the K-topology on , where a closed set (Note that K is closed in the K-topology on ) and {0} cannot be separated by disjoint open sets.