Countably Compact: Necessity & Sufficiency

[Bashyboy]

Homework Statement

Show that $X$ is countably compact if and only if every nested sequence $C_1 \supset C_2 \supset ...$ of closed nonempty sets of $X$ has a nonempty intersection.

Homework Equations

The Attempt at a Solution

Suppose that $X$ is not countably compact. Then there exists a countable open cover $\{U_n\}$ of $X$ that has no finite subcover. Consider the collection of closed sets $C_n = X - (U_1 \cup ... \cup U_n)$; If $x \in C_{n+1}$, then $x \in X-U_i$ for every $i=1,...,n+1$, so in particular $x \in X-U_i$ for $i=1,...,n$. Moreover, this collection consists entirely of nonempty sets, for if $C_n =X - (U_1 \cup ... \cup U_n)$ were empty, then we would have a finite subcover. Hence, $\{C_n\}$ is a nested sequence of closed nonempty sets. By way of contradiction, suppose that the intersection is nonempty. Then $x \in X - (U_1 \cup ... \cup U_n)$ for every $n$. Since $\{U_n\}$ is a cover, $x$ must be in $\bigcup U_i$, which implies there exists a $k$ such that $x \in U_k \subseteq U_1 \cup ... \cup U_k$. This contradicts the fact that $x \in X - (U_1 \cup ... \cup U_k)$. Hence, the intersection has to be empty.Now, suppose that there exists a nested sequence $\{C_i\}$ of nonempty closed sets whose intersection is empty. Then $U_i = X-C_i$ forms a collection of open sets, and since $\bigcup U_i = \bigcup (X-C_i) = X - \cap C_i = X - \emptyset = X$, we see moreover that it is an open cover. Now, if there were to exist a finite subcover, say $\{U_{k_1},...,U_{k_n} \}$, where $k_1 \le ... \le k_n$, then $X \subseteq \bigcup U_{k_i} = X - \bigcap C_{k_i} = X - \bigcap C_{k_i} = X - C_{k_n}$, implying that $C_{k_n} = \emptyset$, which is a contradiction. Hence, there cannot be a finite subcover.How does this sound?

 

[andrewkirk]

Yes that proof looks sound to me, and nicely intuitive too.