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I'd really appreciate some input on these two proofs regarding unions and intersections of compact subsets (under additional necessary conditions). Namely, are my proofs valid? Could they be improved?

"The union of a finite number of compact subsets of a space $X$ is compact."

Let $C_1,...,C_n$ be compact subsets of $X$. Suppose to the contrary that $\displaystyle\bigcup_{i=1}^n C_i$ is not compact. Then it must be that $C_i$ is not compact for some $i$, contradicting our assumption that $C_1,...,C_n$ are compact. $$\,$$

"If $X$ is Hausdorff, then the intersection of any family of compact subspaces is compact."

Let $X$ be a Hausdorff space and let $\{C_\alpha\}_{\alpha \in I}$ be a family of compact subspaces of $X$. Let $C:=\displaystyle\bigcap_{\alpha \in I} C_\alpha$. Since $X$ is Hausdorff, every compact subset is closed. It follows that $C$ is closed as well. Since $C$ is the set of points in $X$ that are in $C_\alpha$ for every $\alpha$, $C \subseteq C_\alpha$ for all $\alpha$. Then since $C$ is a closed subset of $C_\alpha$, a compact subspace, we have that $C:=\displaystyle\bigcap_{\alpha \in I}C_\alpha$ is compact.

I knew my first proof felt very 'thin' but I wasn't sure how to fill in the details. Here's my second try given the help I've received. How is it?

Let $\mathcal{O}:=\{O_\alpha\}_{\alpha \in I}$ be an open cover for $C:=\displaystyle\bigcup_{i=1}^n C_i$. Since $C \subseteq \displaystyle\bigcup_{\alpha \in I} O_\alpha$ we also have that $C_i \subseteq \displaystyle\bigcup_{\alpha \in I} O_\alpha$ for all $i$. Since $C_i$ is compact there is a finite subcover $\{O_j\}_{j=1}^k$ for $C_i$. Since $C_m$ is compact for all $m$, the unions of these finite subcovers yields a finite subcover of $C$ derived from $\mathcal{O}$. Therefore, $C$ is compact.

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Second one seems fine. First one should be a bit more detailed - you don't explain too well why $C_i$ mustn't be compact. Try starting with a cover for the union, think about how this naturally covers each piece, and apply your knowledge of what happens when you take the finite union of finite sets. –  user17794 Oct 1 '12 at 0:32

2 Answers 2

up vote 2 down vote accepted

First proof

You need to explain why a subset $C_i$ cannot be compact if the union is not compact.

Another way to approach it: Every open cover of the union is an open cover of each compact subset $C_i$. What can you say about the union of the finite subcovers of all compact subsets?

Second proof

Looks good!

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I prefer a direct proof of your first proposition.

Let $C_1,...,C_n$ be compact subsets of $X$. If $V$ is an open cover of $C :=\bigcup_{i = 1}^n C_i$, by compactness each $C_i$ has a finite subcover $V_i$, so $\bigcup_{i=1}^n V_i$ is a finite subcover of $C$.

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