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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?

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  • $\begingroup$ "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. $\endgroup$ – Nate Eldredge Jun 9 '13 at 18:33
  • $\begingroup$ 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. $\endgroup$ – fierydemon Jun 9 '13 at 18:53
  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Jun 9 '13 at 18:56
  • $\begingroup$ @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. $\endgroup$ – fierydemon Jun 9 '13 at 18:58
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    $\begingroup$ "Every family of closed sets has the finite intersection property": This is not true. $\endgroup$ – Nate Eldredge Jun 9 '13 at 19:00
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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."

Let us assume we have a family of closed sets with the finite intersection property. This means that by taking inverses of a finite number of closed sets, we can't form an open covering of the space, as the intersection of all the closed sets will be non-empty. If a finite covering can't exist, by the above theorem, an infinite covering using the same sets can't exist either. Hence, the intersection of all closed sets will also be nonempty.

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    $\begingroup$ 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. $\endgroup$ – Brian M. Scott Jun 9 '13 at 20:32
  • $\begingroup$ @BrianM.Scott- Actually the textbook I'm following defines a countably compact space as one in which every infinite subset has a limit point. As for countable family of closed sets, I'm still working on why the "countable" requirement is there. I'm halfway there I think. Thanks $\endgroup$ – fierydemon Jun 10 '13 at 5:30
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    $\begingroup$ Ah. You should be aware, then that the textbook is using non-standard terminology: the property that it calls countable compactness is a strictly weaker property often called limit point compactness and sometimes the Bolzano-Weierstrass property. $\endgroup$ – Brian M. Scott Jun 10 '13 at 5:41

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