Characterising the discrete topology with compact subsets

Question

If a set is endowed with the discrete topology then a subset is compact iff it is finite.

Is the converse true? That is, given a Hausdorff topological space such that every compact subset is finite, then is the topology necessary discrete?

Answer 1

The converse is not true. Consider the set of continuous functions $[0,1]\to \mathbb{R}$ in the box topology. $X$ is Hausdorff, and a sequence $x_1,x_2,\ldots$ of points of $X$ converges if and only if it is eventually constant by a diagonal argument using the fact that if two continuous functions differ at only finitely many points then they are equal. Therefore any infinite subset of $X$ cannot be compact because there exist sequences of points in the set with no convergent subsequences. $X$ is not discrete because one point sets are not open.

Answer 2

There is also another example, kind of standard, Arens space.

As yet another example consider the Stone-Cech compactification of the integers $\beta \Bbb N$, and take $x$ in the remainder and consider the space $X=\{x\}\cup \Bbb N$. It is known that every compact subset of $\beta \Bbb N$ mist be either finite, or have cardinality $2^\frak c$, so compact subsets of $X$ must be finite.

These two examples are not compactly generated, where a space $X$ is compactly generated, also called a $k$-space if it satisfies the following condition: A subspace $A$ is closed in $X$ if and only if $A\cap K$ is closed in $K$ for all compact subspaces $K\subseteq X$.

It is clear that if $X$ a is compactly generated Hausdorff and all compact subsets are finite the $X$ has the discrete topology. The class of compactly generated spaces includes all metrizable spaces, as well as all 1st-countable spaces, and all locally compact spaces.