**Problem Statement:** Use the open cover definition of compact to show that is a compact subset of .

**Proof:** Let be an open cover of , for some indexing set . If is a finite cover then we’re done, so suppose that is not a finite open cover of .

Since there is at least one such that . Since converges to there is an such that for every . Furthermore, since is an open cover of there is at least one such that . Pick one of these and call it . We may do this for and find .

Let which is contained in . We claim that is a finite open cover of . First note that each of the are open and so their union is open. Let . If then , otherwise, . Either way, . Thus, is a finite open cover of .

Since such a can be constructed from any open cover of it follows that every open cover of in has a finite subcover in , thus, is compact in .

**Reflection:** The most important part of this proof rests in the fact that is the limit of the sequence . This allows us to cover infinitely many of the points in with a single open set and leave only finitely many left to cover. After that, the rest is a snap 🙂

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Ooooooooh snap!

Wow, someone knows how to spend their sunday nights!

We’re the ones reading math blogs on Sunday, Wadesworth!

Squeaky clean.