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Can anyone give me an intuition for why projection along a compact space is a closed map (and in fact this characterizes compact spaces)? In other words, $$\pi: K \times X \to X \,\,\text{ is closed for all topological spaces }X \iff K \,\text{ compact.} $$

I understand the proof, I understand the counter-examples (especially $(k_\alpha, x_\alpha)$ where $x_\alpha$ is a convergent net and $k_\alpha$ is a net with no convergent subnet) and why $K$ being compact prevents them. I guess I'm looking for some sort of further intuition. One obstacle to an intuitive explanation might be that being a closed map is a relatively unintuitive property.

Eric Auld
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I don't really know what kind of "intuition" you're looking for here, but here's one take on it. You can see this as a generalization of the fact that a finite union of closed sets is closed. Indeed, when $K$ is finite and discrete, it reduces to that fact. More generally, one can think of a closed subset of $K\times X$ as a "continuous family of closed subsets of $X$" parametrized by $K$, and the image under the projection is then the union of all the sets in the family. So this result says that compact spaces are exactly the "index sets" for a continuous family of closed sets that guarantees that its union will still be closed.

(Actually, it's probably better to think of a closed subset of $K\times X$ not as a "continuous" family of closed sets but as an "upper semicontinuous" family of closed sets.)

Eric Wofsey
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