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.
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"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.
Best Answer
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 $C$.