# Content-Type: text/shitpost

Date: 2020-07-09T16:38:11
Newsgroup: comp.protocols.tcp-ip.compact-subspaces
Message-ID: <a50438d45cf11695@shitpost.plover.com>
Content-Type: text/shitpost

Finite sets are always compact. Suppose we have an infinite compact space. Is it possible that the only compact proper subspaces that it possesses are the finite ones? The answer turns out to be no. Every infinite compact space has an infinite compact proper subspace. So by repetition, every infinite compact space has an infinite descending chain of compact subspaces.

I thought hey, here we have a partial order (of infinite compact subspaces) in which every descending chain has a lower bound, so we can apply Zorn's lemma… except no, not every chain has a lower bound, what was I thinking? If we include the finite subspaces then every chain does have a lower bound and we can apply Zorn's lemma, but the result is the empty subspace so that was not useful.

Interesting: The example of !![0,1]!! shows that you can't always obtain a smaller compact subspace by deleting a finite set from the original space.

The compact subspaces of an infinite compact set form a rather interesting lattice structure. At the bottom are the finite subsets, arranged like a very ordinary Boolean lattice, but with no maximum element. (Isn't there a name for a complete associative lattice that may or may not have a maximum element?) Floating above this are the infinite compact subspaces, in which there are no minimal elements, and the original space at the top.

Consider just a relatively simple example. Let !!X!! be the one-point compactification of the natural numbers. That is, $$X = \Bbb N \cup {\infty}$$ where a subset !!G!! of !!X!! is open if and only if !!G!! is a finite set that does not contain !!∞!! or !!G!! is a cofinite set that does contain !!∞!!. An infinite compact subspace of !!X!! must contain !!∞!!.