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http://www.proofwiki.org/wiki/Closed_Subspace_of_Compact_Space_is_Compact

ProofWiki provides the following proof that a closed subspace of a compact space is compact:

Let $T$ be a compact space. Let $C$ be a closed subspace of $T$. Let $\mathcal U$ be an open cover of $C$. Since $C$ is closed, it follows by definition of closed that $T \setminus C$ is open in $T$. So if we add $T \setminus C$ to $\mathcal U$, we see that $\mathcal U \cup \left({T \setminus C}\right)$ is also an open cover of $T$. As $T$ is compact, there is a finite subcover of $\mathcal U$, say $\mathcal V = \left\{{U_1, U_2, \ldots, U_r}\right\}$. This covers $C$ by the fact that it covers $T$. If $T \setminus C$ is an element of $\mathcal V$, then it can be removed from $\mathcal V$ and the rest of $\mathcal V$ still covers $C$. Thus we have a finite subcover of $\mathcal U$ which covers $C$, and hence $C$ is compact.

My question is about this line: "As $T$ is compact, there is a finite subcover of $\mathcal U$, say $\mathcal V = \left\{{U_1, U_2, \ldots, U_r}\right\}$." But they just said that $\mathcal U \cup \left({T \setminus C}\right)$ is an open cover of $T$, so shouldn't they be saying that $\mathcal V = \left\{{U_1, U_2, \ldots, U_r}\right\}$ is a subcover of $\mathcal U \cup \left({T \setminus C}\right)$? That is, not just of $\mathcal U$?

Thanks for clearing this up.

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Yes, you are right. And I don't know why "we see that $\mathcal U\cup(T\setminus C)$ is also an open cover of $T$", why the also ? The wording should really be changed. –  Stefan Hamcke Nov 6 '13 at 22:36
    
But if $\mathcal V$ is a subcover of $\mathcal U \cup (T \backslash C)$ and not $\mathcal U$, the rest of the proof doesn't make sense, right? –  user66360 Nov 6 '13 at 22:51
    
The proof makes sense. $\cal V$ covers $C$. –  Stefan Hamcke Nov 6 '13 at 22:53
    
Ok. So I guess you take $\mathcal V$, which is a subcover of $\mathcal U \cup (T \backslash C)$, remove from it any element in in $(T \backslash C)$, and what we're left with is a subset of $\mathcal U$ that still covers $C$? –  user66360 Nov 6 '13 at 22:59
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We start with a cover of $C$, that is a family of open subsets of $T$ such that $C$ is in their union. This cover alone might already cover $T$, in this case $T-C$ may be absent from $\cal V$, otherwise $T-C$ must be an element of $\cal V$. Then since each point of $T$ is in some element of $\cal V$, and $T-C$ is disjoint from $C$, each point in $C$ must lie in some element from $\mathcal V-\{T-C\}$. –  Stefan Hamcke Nov 6 '13 at 23:12

1 Answer 1

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We start with an open cover $\cal U$ of $C$, that is a family of open subsets of $T$ such that $C$ is in their union. This cover alone might already cover $T$, in this case $T−C$ may be absent from $\mathcal V$, the finite subcover of $\mathcal U\cup\{T-C\}$, otherwise $T−C$ must be an element of $\cal V$. Then since each point of $T$ is in some element of $\cal V$, and $T−C$ is disjoint from $C$, each point in $C$ must lie in some element from $\mathcal V−\{T−C\}$, which is therefore a subfamily of $\cal U$ covering $C$.

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