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Prove that if $K$ and $L$ are compact subsets of a Hausdorff space $X$, then $K \cap L$ is a compact subset of $X$.

I understand that since $K$ and $L$ are compact subsets, they each have finite coverings. Do I have to show that the intersection of $2$ finite coverings is a compact subset?

Davide Giraudo
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general1597
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1 Answers1

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Theorem:Any compact subset of a Hausdorff space is closed.

You can find a proof here.

Let $\mathcal{A}$ be an open cover of $K \cap L$. Since $K$ and $L$ are closed sets ,$K \cap L$ is also closed. So ${(K \cap L)}^C$ is open. Then $\mathcal{B}=\mathcal{A}\cup {(K \cap L)}^C$ is an open cover of $K$(and $L$). Since $K$ is compact,$\mathcal{B}$ has a finite subcover,say,$\{A_1,A_2,\ldots,A_n,{(K \cap L)}^C\}$. So $\{A_1,A_2,\ldots,A_n\}$ is a finite subcover of $\mathcal{A}$ that covers $K \cap L$. Hence, $K \cap L$ is compact.

cqfd
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  • Why don't you have 2 cases to account for whether or not $(K\cap L)^C$ is included in the finite subcover? –  Dec 05 '18 at 06:19
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    @JackBauer Even if $(K\cap L)^C $ is not included in the finite subcover, still the union of that subcover and $(K\cap L)^C $ will cover $K$, right? So basically I was trying to incorporate the two cases into one by taking the above mentioned subcover. If there is any mistake , please let me know. – cqfd Dec 05 '18 at 09:04
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    I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant. –  Dec 12 '18 at 20:11