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.

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?


1 Answer 1


Proof looks OK. The basic ideas: any countable open cover can be made increasing, without changing the existence of a finite (which becomes singleton) subcover. De rest is de Morgan : take complements, unions become intersections and vice versa, and a cover (union is $X$) in the complement means empty intersection and vice versa.

A streamlined version to illustrate this point:

The following are equivalent:

  1. $X$ has a countable open cover $\mathcal{U}$ without a finite subcover.
  2. $X$ has a sequence of open sets $(U_n)_{n \in \mathbb{N}}$ such that $\forall n: U_n \neq X \land U_{n} \subseteq U_{n+1}$, and also $\bigcup_n U_n = X$.
  3. $X$ has a countable sequence of non-empty closed sets $(C_n)_{n \in \mathbb{N}}$ such that $\forall n: C_{n+1} \subseteq C_n$, and also $\bigcap_n C_n =\emptyset$.

Proof: suppose 1. holds. To see 2: index $\mathcal{U}$ as $\{O_n: n \in \mathbb{N}\}$ to witness its countability. Then define $U_n = \cup_{i=1}^n O_n$. The fact that $\mathcal{U}$ has no finite subcover shows that $U_n \neq X$ for all $n$, and $U_n \subseteq U_{n+1}$ is true by construction. So 2. holds.

Suppose 2. holds. Then define for all $n$: $C_n = X \setminus U_n$, which are closed as the $U_n$ is open and non-empty as no $U_n$ equals $X$.

$$U_{n} \subseteq U_{n+1} \implies X\setminus U_n \supseteq X\setminus U_{n+1}$$ which says that $C_{n+1} \subseteq C_n$.

By de Morgan's laws: $$\bigcap_n C_n = \bigcap_n (X\setminus U_n) = X \setminus \bigcup_n U_n = X\setminus X = \emptyset$$ as required.

Suppose 3. holds. Then 2 holds, in an entirely symmetric way: define $U_n = X\setminus C_n$ which is open as $C_n$ is closed, and $U_n \neq X$ as $C_n \neq \emptyset$. Again $$C_{n+1} \subseteq C_n \implies U_{n+1} = X\setminus C_{n+1} \supseteq X\setminus C_n= U_n$$ which says $U_n \subseteq U_{n+1}$, as required. De Morgan again: $$\bigcup_n U_n = \bigcup_n (X\setminus C_n) =X \setminus \bigcap_n C_n = X\setminus \emptyset =X$$ as required. Finally: assume 2. holds. Then define $\mathcal{U}=\{U_n: n \in \mathbb{N}\}$which is an open cover of $X$. Suppose we have a finite subcover $U_{n_1},\ldots,U_{n_k}$ of $\mathcal{U}$. Suppose WLOG that $n_1 < n_2 \ldots < n_k$. Then $\bigcup_{i=1}^k U_{n_i} = U_{n_k} \neq X$, where the first holds by increasingness of the $U_n$ and the last is a property of all $U_n$. So these sets do not form a subcover, so there can be no finite subcover for $\mathcal{U}$. So 1. holds. and all are equivalent. Now note that your original statement was not 1 iff not 3. Which now follows.

As an aside: It's standard in compactness as well, but there we use closed sets with the finite intersection property instead (or their extension, filters of closed sets). We could do decreasing "sequences" as well,but then one gets into ordinals and cardinals and such, and we have to consider cofinalities.

  • $\begingroup$ To the O.P. Re: de Morgan's laws: If $\cup\{U_n:n\in N\}=X$ then by the very basic properties of sets we have $\cap \{C_n:n\in N\}=\phi$ so you can shorten the proof of the first part, although your work is logical as it is. $\endgroup$ Jul 16, 2017 at 3:43

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