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I'm having trouble understanding a point in the following standard argument to show that closed subsets of compact spaces are compact.

The proof usually goes as follows.

Let $X$ be compact and let $C$ be closed. Let $\{U_\alpha \}$ be some open cover of $C$. Now, $\{U_\alpha \}$ with $X \setminus C$ is an open cover of $X$. As such, we get a finite open cover of $X$ from this collection, from which we get finite subcover of $C$.

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The point I am having trouble understanding is that if $\{U_\alpha \}$ is an open cover of closed set $C$, it does not seem necessary that all $U_\alpha$ is open in $X$. The example I am thinking of is $[0,1]$ in $\mathbb{R}$. I believe there is an open cover(with respect to subspace topology of $[0,1]$) of $[0,1]$ with $[0,0.6)$ and $(0.4,1]$. However, both of these are not open sets in $\mathbb{R}$. So I am troubled by the point that $\{U_\alpha \}$ with $X \setminus C$ forms an open cover of $X$.

Thanks for your help!

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  • $\begingroup$ The cover $\mathcal{U}$ is open in $X$, not in the subspace topology. $\endgroup$ Commented Dec 31, 2020 at 19:59
  • $\begingroup$ @BrianM.Scott the thing that confuses me is then, shouldn't the cover be open in $C$, since we want to prove that $C$ is a compact subset, and it should be compact with respect to the subspace topology of $C$? $\endgroup$
    – Phil
    Commented Dec 31, 2020 at 20:02
  • $\begingroup$ It doesn’t matter, for the reason given in leoli1’s answer. (The author whom you’re quoting probably should have made that explicit.) $\endgroup$ Commented Dec 31, 2020 at 20:05

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You are right, if $U$ is open in $C$ then it is not necessarily open in $X$. However there will be an open subset $U'$ of $X$ such that $U=C\cap U'$. Then you can simply replace the $U_\alpha$ with the $U_\alpha'$.

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  • $\begingroup$ Thanks so much! $\endgroup$
    – Phil
    Commented Dec 31, 2020 at 20:12

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