# Carothers Ch. 8 q. 32: equivalence of $\bigcap_{i=1}^\infty N_i \ne \emptyset$ for all closed/nested $N_i\subset M$ and covering compactness.

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.

• 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). May 7 '20 at 5:44
• I apologize for omitting that. It looks like Carothers is referring to a metric space $M$ here.
– akm
May 7 '20 at 6:06

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).