Compact objects in the category of topological spaces

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

Definition. Let $\mathscr C$ be a cocomplete category. Then an object $X \in \ob \mathscr C$ is compact if $\Hom(X,-)$ commutes with filtered colimits.

Exercise. An $R$ -module $M$ 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 $X \in \Top$ be a compact object. Then $X$ is finite.

Proof. Let $Y$ be the set $X$ with the indiscrete topology, i.e. $\mathcal T_Y = \{\varnothing, Y\}$ . It is the union of all its finite subsets, and this gives it the colimit topology because a subset $U \subseteq Y$ is open if and only if its intersection with each finite subset is. Indeed, if $U$ were neither $\varnothing$ nor $Y$ , then there exist $y_1, y_2 \in Y$ with $y_1 \in U$ and $y_2 \not\in U$ . But then $U \cap \{y_1,y_2\}$ is not open, because $\{y_1,y_2\}$ inherits the indiscrete topology from $Y$ .

Therefore, if $X$ is a compact object, then the identity map $X \to Y$ factors through one of these finite subsets, hence $X$ is finite. $\qedsymbol$

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 $X \in \Top$ be a compact object. Then $X$ is a compact topological space.

Proof. It is finite by the lemma above. Every finite topological space is compact. $\qedsymbol$

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

Lovely little lemmas

Or maybe ‘amusing observations’

Compact objects in the category of topological spaces

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