Can compact sets completey determine a topology?

Suppose that $\tau_1$ and $\tau_2$ are two topologies on a set $X$ with the property that $K\subset X$ is compact with respect to $\tau_1$ if and only if $K$ is compact with respect to $\tau_2$. Then is this enough information to determine whether $\tau_1 = \tau_2$?

If not (which is suspect to be the case) is there a nice counterexample, and what is the minimum amount of extra information required for $\tau_1 = \tau_2$?

I feel like there would be an issue regarding 'points at infinity.'

Solution 1:

It is not enough information. A simple (and perhaps frightening) counterexample is as follows: if $X$ is an uncountable set, then in both the discrete and co-countable topologies the compact sets are exactly the finite sets. It's frightening because these two topologies share virtually no properties: one is metrizable (hence perfectly normal), the other isn't even Hausdorff; one is Lindelöf, and the other has maximal Lindelöf number.

I'm not certain what additional information would be required to conclude that two topologies are the same. (I'll think about this some more.)

Addition 1

Hausdorffness is not sufficient. The discrete space on $\mathbb{N}$ and the Arens–Fort space are both Hausdorff, and the compact subsets are exactly the finite subsets, but these spaces are not homeomorphic. (Since they are both countable, we can transfer the topology of one onto the underlying set of the other, and the compact sets will be the same.)

Actually, this example shows that perfect normality, Lindelöfness, and paracompactness are not sufficient.

Solution 2:

Consider a finite set $X$,then every subset $A$ of $X$ is compact in the trivial topology $\{\emptyset,X\}$, but every subset $A$ of $X$ is also compact in the discrete topology on X (the topology where all subsets of X are considered open). If $|X|>1$, then these topologies are obviously not equal.

I will return to your second question if it happens that I have anything useful to say about it.