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In Munkres, he proves that every closed subspace Y of a compact space X is compact. In the proof, he adjoins the open set X - Y to an open cover of Y, and then he goes on to prove Y is compact.

Why did he adjoin X - Y to this open cover? Could he have adjoined any open set that already includes X - Y, such as X? What is wrong with adjoining X? Are there counterexamples that demonstrate why X isn't a good choice?

Britt K
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    Then ${X}$ is a finite subcover of the new covering ... one that tells you nothing. – Angina Seng Apr 21 '19 at 18:18
  • @kimchilover but could he have thrown in the set X instead of X-Y? – Britt K Apr 21 '19 at 18:28
  • @LordSharktheUnknown yes but the original cover of Y, now adjoined with X, could have a finite subcover other than {X}, right? – Britt K Apr 21 '19 at 18:29
  • This is exactly the point in the proof where $Y$ being closed is used. – Berci Apr 21 '19 at 18:33
  • @Berci I get that the closedness of Y is being used there (then X-Y is open), but why couldn't the open set X (instead of the open set X-Y) be adjoined to the original open cover of Y? – Britt K Apr 21 '19 at 18:35
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    @kimchilover I understand that we need to get an open cover of X... that is why X-Y was adjoined to the original open cover of Y. You write "If the original open cover of Y did not contain Y...". But the original open cover of Y does include Y (Y is a subset of the union of this open cover). Also, we can use, as our original open cover of Y, a collection of open sets in X (not open in Y). But again I don't understand why we MUST use X-Y (and not simply X) to construct an open cover of X from the original open cover of Y. – Britt K Apr 21 '19 at 18:54

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As you say in the comments, the extended open cover could have a finite subcover other than $\{X\},$ but we have no way to know that without using the very result that we're trying to prove!

The goal, instead, is to turn the open cover $\mathcal U$ of $Y$ into an open cover $\mathcal U'=\mathcal U\cup\{U\}$ of $X$ by adjoining some open subset $U$ of $X,$ then to use compactness to reduce obtain a finite subcover of $\mathcal U',$ say $\mathcal S,$ such that $\mathcal S\cap\mathcal U$ is a cover of $Y.$

If we let $U$ cover $Y$ even in part, then we might lose that possibility, due to throwing out sets that weren't necessary to cover $X,$ but which we needed to keep to cover $Y$! For example, consider the real intervals $X=[-2,2]$ and $Y=[-1,1].$ Cover $Y$ with $\mathcal U=\bigl\{(-2,1),(0,2)\bigr\}.$ Then $\mathcal U\cup\bigl\{[-2,-1)\cup(0,2]\bigr\}$ is an open cover of $X,$ and has the subcover $\bigl\{(-2,1),[-2,-1)\cup(0,2]\bigr\},$ but $\bigl\{(-2,1)\bigr\}$ doesn't cover $Y.$

Thus, we must take $U\subseteq X-Y.$ On the other hand, in order for $U'$ to be guaranteed to be a cover of $X,$ we also need $U\supseteq X-Y.$ After all, we could have $Y=\bigcup\mathcal U.$ Thus, the proof used $X-Y$ to extend the cover.

Cameron Buie
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