# Countably Compact vs Compact vs Finite Intersection Property

There is this exercise: Show that countable compactness is equivalent to the following condition. If ${C_n}$ is a countable collection of closed sets in S satisfying the finite intersection hypothesis, then $\bigcap_{i=1}^\infty C_i$ is nonempty.

Definitions:

• A Space S is countably compact if every infinite subset of S has a limit point in S.
• A space S has the finite intersection property provided that if ${C_\alpha}$ is any collection of closed sets such that any finite number of them has a nonempty intersection, then the total intersection $\bigcap_\alpha C_\alpha$ is non-empty
• A family of closed sets, in any space, such that any finite number of them has a nonempty intersection, will be said to satisfy the finite intersection hypothesis.

Now there is also a related theorem in the book: Compactness is equivalent to the finite intersection property.

Sounds to me countable compactness and compactness are pretty much the same.

I am not asking for a solution to the exercise. My question is this:

What is the difference between Compactness and Countable Compactness in terms of closed Collections? Both things sound to me like this: Given a collection of closed sets, when a finite number of them has a nonempty intersection, all of them have a nonempty intersection.

BTW: the definitions, the theorems and the exercise are from Topology by Hocking/Young.

The difference is that if $X$ is compact, every collection of closed sets with the finite intersection property has a non-empty intersection; if $x$ is only countably compact, this is guaranteed only for countable collections of closed sets with the finite intersection property. In a countably compact space that is not compact, there will be some uncountable collection of closed sets that has the finite intersection property but also has empty intersection.

An example is the space $\omega_1$ of countable ordinals with the order topology. For each $\xi<\omega_1$ let $F_\xi=\{\alpha<\omega_1:\xi\le\alpha\}=[\xi,\omega_1)$, and let $\mathscr{F}=\{F_\xi:\xi<\omega_1\}$. $\mathscr{F}$ is a nested family: if $\xi<\eta<\omega_1$, then $F_\xi\supsetneqq F_\eta$. Thus, it certainly has the finite intersection property: if $\{F_{\xi_0},F_{\xi_1},\dots,F_{\xi_n}\}$ is any finite subcollection of $\mathscr{F}$, and $\xi_0<\xi_1<\ldots<\xi_n$, then $F_{\xi_0}\cap F_{\xi_1}\cap\ldots\cap F_{\xi_n}=F_{\xi_n}\ne\varnothing$. But $\bigcap\mathscr{F}=\varnothing$, because for each $\xi<\omega_1$ we have $\xi\notin F_{\xi+1}$. This space is a standard example of a countably compact space that it not compact.

Added: Note that neither of them says:

Given a collection of closed sets, when a finite number of them has a nonempty intersection, all of them have a nonempty intersection.

The finite intersection property is not that some finite number of the sets has non-empty intersection: it says that every finite subfamily has non-empty intersection. Consider, for instance, the sets $\{0,1\},\{1,2\}$, and $\{0,2\}$: every two of them have non-empty intersection, but the intersection of all three is empty. This little collection of sets does not have the finite intersection property.

Here is perhaps a better way to think of these results. In a compact space, if you have a collection $\mathscr{C}$ of closed sets whose intersection $\bigcap\mathscr{C}$ is empty, then some finite subcollection of $\mathscr{C}$ already has empty intersection: there is some positive integer $n$, and there are some $C_1,\dots,C_n\in\mathscr{C}$ such that $C_1\cap\ldots\cap C_n=\varnothing$. In a countably compact space something similar but weaker is true: if you have a countable collection $\mathscr{C}$ of closed sets whose intersection $\bigcap\mathscr{C}$ is empty, then some finite subcollection of $\mathscr{C}$ already has empty intersection. In a countably compact space you can’t in general say anything about uncountable collections of closed sets with empty intersection.

• Thanks for this great explanation, Brian. So it's really all about the attribute "countable"... Ironically I encountered the standard example some pages before in an exercise; there it was in the context of subbases though. – Philip Jun 20 '12 at 19:10