Recall the following theorem: Given a set $X$ with two topologies $\cal U$ and $\cal V$, with $\cal U$ weaker than $\cal V$, if $\cal U$ is Hausdorff and $\cal V$ is compact then in fact $\cal U = \cal V$.

In general, is there a "right" topology, in this sense? Specifically, given a Hausdorff topology $\cal U$ can you weaken it to a compact topology which is still Hausdorff, and given a compact topology $\cal V$ can you strengthen it to a Hausdorff topology which is still compact?

For example, consider the cofinite topology on $\omega$. The cofinite topology is always compact. Can we extend this topology to a Hausdorff topology? Yes we can: $\omega$ could just as well have been any countably infinite set, say $\{0\}\cup\{1/n: n\in{\bf Z}^+\}$, and now one checks that the usual topology (induced by $\mathbb{R}$, say) is compact, Hausdorff, and stronger than the cofinite topology (because finite sets are closed).

EDIT: A more or less complete answer can be found on http://mathoverflow.net/questions/15841/how-do-the-compact-hausdorff-topologies-sit-in-the-lattice-of-all-topologies-on-a.