# A possible contradiction to "A $T_{1}$ space is countably compact iff every countable family of closed sets has a nonempty intersection"?

My textbook says "A $T_{1}$-space is countably compact iff every countable family of closed sets having the finite intersection property has a non-empty intersection" (Principles of General Topology, Pervin).

Let the $T_{1}$-space be $X$. Then what I understand from the condtition is that if it is countably compact, then finite unions of its open sets exist, and the union of all open sets does not cover the whole of $X$ (except $X$ of course). We will assume that no finite covering of $X$ can exist, as "every countable family of closed sets has the finite intersection property".

Let $X$ be an infinite set; then, it will have a limit point "$l$" by virtue of it being a countably compact space, and every open set containing $l$ will contain an infinite number of other points from $X$ by virtue of it being a $T_{1}$-space.

Say we take an open set $A$ containing $l$, which contains infinite other points from $X$. $X$ might contain infinite other points not contained within $A$. This latter subset will again have a limit point, which will again be contained in an open set containing infinite points. Going this way we can find an infinite number of sets covering $X$ (not finite, as every family of closed sets has the finite intersection property).

So here we have a countably compact $T_{1}$-space with every family of closed sets having the finite intersection property, without the whole family having a nonempty intersection. Isn't this a contradiction?

• "Finite unions of its open sets exist": What do you mean by exist here? "The union of all open sets does not cover the whole of X": What makes you think that? It's almost never true. Commented Jun 9, 2013 at 18:33
• Let $A,B,C\dots N$ be closed sets in $X$. If all famililies of closed sets have the finite intersection property, then $A\cap B$, $A\cap B\cap C$, $B\cap C, \dots$ all of these are nonempty. Also, $A', B'\dots$, all these are compements of $A,B\dots$, and are hence open sets. If $A\cap B$ is nonempty, then $A'\cup B'=(A\cap B)'$ also exists. Hence, if finite intersections exist, then complements of those intersections also exist, which are finite unions of open sets.
– user67803
Commented Jun 9, 2013 at 18:53
• Countable compactness doesn't assert that every family of closed sets has the finite intersection property; only that, for those families that do happen to have this property, their intersection is nonempty. There can be families that don't have the finite intersection property at all, and countable compactness makes no claims about them. Commented Jun 9, 2013 at 18:56
• @NateEldredge- Let there be a finite cover of the space- $\cup_{i}G_{i}$. If $G_{i}'$ is the complement of $G_{i}$, then $(\cup_{i}G_{i})'=\cap_{i}G_{i}'$ should be empty. However, as every family of closed sets has the finite intersection property, and this is a finite cover, $\cap_{i}G_{i}'$ can't be empty. This is a contradiction. Hence we can't have a finite cover.
– user67803
Commented Jun 9, 2013 at 18:58
• "Every family of closed sets has the finite intersection property": This is not true. Commented Jun 9, 2013 at 19:00

This theorem is a corollary of the theorem "if a $T_{1}$-space is countably compact, then every countable open cover has to have a finite subcover."
• Two things. First, if a $T_1$-space is countable compact, then every countable open cover has a finite subcover isn’t really a theorem, and $T_1$ isn’t necessary here: the definition of countable compactness is that every countable open cover has a finite subcover. Secondly, you want to assume that you have a countable family of closed sets, since it’s only countable open covers that are guaranteed to have finite subcovers. Commented Jun 9, 2013 at 20:32