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 axioms 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 $X$?
