prove that :

Every space is

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- May 28th 2009, 12:56 PMflower3separation axioms
prove that :

Every space is - May 28th 2009, 02:37 PMHallsofIvy
A good way to start: write out the definitions of and

- May 30th 2009, 03:05 AMaliceinwonderland
**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.