We compare two competing notions of compactness for topological spaces. Besides the usual notion, there is the following:

**Definition.** Let be a cocomplete category. Then an object is *compact* if commutes with filtered colimits.

**Exercise.** An -module is compact if and only if it is finitely generated.

We want to study compact objects in the category of topological spaces. One would hope that this corresponds to compact topological spaces. However, this is very far off:

**Lemma.** *Let be a compact object. Then is finite.*

*Proof.* Let be the set with the indiscrete topology, i.e. . It is the union of all its finite subsets, and this gives it the colimit topology because a subset is open if and only if its intersection with each finite subset is. Indeed, if were neither nor , then there exist with and . But then is not open, because inherits the indiscrete topology from .

Therefore, if is a compact object, then the identity map factors through one of these finite subsets, hence is finite.

However, the converse is not true. In fact the indiscrete space on a two element set is not a compact object, as is explained here.

**Corollary.** *Let be a compact object. Then is a compact topological space.*

*Proof.* It is finite by the lemma above. Every finite topological space is compact.

Originally, this post relied on the universal open covering of my previous post to show that a compact object in is compact; however the above proof shows something much stronger.