Compact subsets and Hausdorffness of Topology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:00:09Z http://mathoverflow.net/feeds/question/98601 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98601/compact-subsets-and-hausdorffness-of-topology Compact subsets and Hausdorffness of Topology AliReza Olfati 2012-06-01T20:03:34Z 2013-01-12T18:56:38Z <p>We Know that all closed subsets of a compact topological space $(X,\tau)$ are compact. But if we add The <strong>Hausdorff</strong> condition on the topology $\tau$ we could see the equivalence of these subsets.(i.e. in compact Hausdourff spaces closed subsets are the same as compact subsets)</p> <p>Know for asking the converse of the above fact we could or not omit the compactness of the space$(X,\tau)$ as follows:</p> <ul> <li><p>(<strong>STATEMENT</strong>) If all compact subsets of a topological space $(X,\tau)$ are closed then $(X,\tau)$ is Hausdorff.</p></li> <li><p>If the above statement is not valid, Is there a separation axiom weaker than Hausdorffness on the space $X$ that compact subsets are closed?</p></li> </ul> <p>For the first statement If we add the condition of compactness of $(X,\tau)$, it changes as follows:</p> <ul> <li>Is The space $(X,\tau)$ Hausdorff,If closed subsets and compact subsets are equivalent in $X$? </li> </ul> http://mathoverflow.net/questions/98601/compact-subsets-and-hausdorffness-of-topology/118750#118750 Answer by David White for Compact subsets and Hausdorffness of Topology David White 2013-01-12T18:56:38Z 2013-01-12T18:56:38Z <p>The Statement in the question is false. A counter-example can be found in <a href="http://mathoverflow.net/questions/88420/example-of-a-weak-hausdorff-space-that-is-not-hausdorff" rel="nofollow">this MO answer</a>. As for a separation axiom weaker than Hausdorff which make compact subsets closed, one such notion is that of a Weak Hausdorff space. </p> <p>As Gjergji points out, spaces where compact subsets are closed are called KC-spaces. Hausdorff implies KC, but not conversely (this answers the OP's third question).</p>