Compact subsets and Hausdorffness of Topology

2012-06-01T20:03:34Z

We Know that all closed subsets of a compact topological space $(X,\tau)$ are compact. But if we add The Hausdorff 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)

Know for asking the converse of the above fact we could or not omit the compactness of the space$(X,\tau)$ as follows:

(STATEMENT) If all compact subsets of a topological space $(X,\tau)$ are closed then $(X,\tau)$ is Hausdorff.
If the above statement is not valid, Is there a separation axiom weaker than Hausdorffness on the space $X$ that compact subsets are closed?

For the first statement If we add the condition of compactness of $(X,\tau)$, it changes as follows:

Is The space $(X,\tau)$ Hausdorff,If closed subsets and compact subsets are equivalent in $X$?

Answer by David White for Compact subsets and Hausdorffness of Topology

2013-01-12T18:56:38Z

The Statement in the question is false. A counter-example can be found in this MO answer. As for a separation axiom weaker than Hausdorff which make compact subsets closed, one such notion is that of a Weak Hausdorff space.

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).

Compact subsets and Hausdorffness of Topology - MathOverflow

Compact subsets and Hausdorffness of Topology

Answer by David White for Compact subsets and Hausdorffness of Topology