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Carothers, 8.32:

Prove that

A metric space $M$ is compact $\Longleftrightarrow$ every countable open cover admits a finite subcover by showing that the following two statements are equivalent:

  1. Every decreasing sequence of nonempty closed sets in $M$ has a nonempty intersection.
  2. Every countable open cover of $M$ admits a finite subcover; that is, if $(G_n)$ is a sequence of open sets in $M$ satisfying $\bigcup_{n=1}^\infty G_n \supset M$ then $\bigcup_{n=1}^N G_n \supset M$ for some (finite) $N$,

I was hoping someone could check my proof for the first direction and could give me hints to prove the other direction.

Proof: $(1) \rightarrow (2)$ Suppose $(G_n)$ is an infinite open cover of $M$. Because it is an open cover, $M \cap \bigcup_{n=1}^\infty G_n = M$ and so $M \cap \bigcap_{n=1}^\infty (G_n)^c = \emptyset$. Consider the sequence $(Z)_j$ where $$Z_j = M \cap (\bigcup_{n=1}^{j} G_n)^c = M \cap \bigcap_{n=1}^{j} G_n^c$$ where $Z_j\to M\cap (\bigcup_{n=1}^\infty G_n)^c = \emptyset$. Because $Z_j$ is the complement of a union of open sets intersected with another open one, it is closed. Yet, because it has an empty intersection, the conditions for $(1)$ do not hold, meaning that at least one set in $(Z)_j$ is empty, or that that $Z_{k+1} \not\subset Z_k$ for some integer $k$, or some combination thereof.

By construction, $Z_{k+1} \subseteq Z_k$. If $Z_{k+1} = Z_k$ then $G_{k+1} = \emptyset$. If $Z_n = Z_k$ for all $n>k$, we are left with a finite cover $\mathcal{G} = \{G_1, \ldots, G_k\}$ and so intersection of the compliments of the sets do not form a sequence, as they are finite. If the number of empty sets is finite, we are left with a subsequence of closed sets $(Z_j)_k$ where $Z_n \supset Z_{n+1}$ for all $n$. So the union of the remaining open sets continues to form an infinite open cover, since we only removed empty sets from the original sequence.

Having removed the possibility that $Z_{n+1} \not\subset Z_n$, it must be the case that at least one set from the subsequence must be empty. If $Z_k = \emptyset$ then because $Z_k \supset \bigcup_{j=k+1}^\infty Z_j$, the union $\bigcup_{j=k+1}^\infty Z_j = \emptyset$. Then $M \cap (\bigcup_{n=k+1}^j G_n)^c = \emptyset$ and so a finite cover can be obtained by restricting $(G)_j$ to all sets before the first empty set.

Edited to clarify that $M$ refers to a metric space.

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  • $\begingroup$ What is $M$ here? The property "$M$ is compact iff every countable cover contains a finite subcover" is false for general topological spaces. For example, $\omega_1$ (the first uncountable ordinal) is not compact with the order topology, but it is countably compact (i.e. every countable cover has a finite subcover). $\endgroup$
    – Reveillark
    May 7 '20 at 5:44
  • $\begingroup$ I apologize for omitting that. It looks like Carothers is referring to a metric space $M$ here. $\endgroup$
    – akm
    May 7 '20 at 6:06
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Hint for (2) => (1): Try to reformulate the statement (1) using the complements in $M$ of each of the closed sets in the given sequence, then prove the reformulation using (2).

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  • $\begingroup$ Thanks. If anyone is familiar with MathStackExchange rules, would it make sense for me to edit the original question with the proof in both directions for reference to others, or should I post it as a comment or a separate answer? $\endgroup$
    – akm
    May 8 '20 at 1:39

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