I was working through a textbook on topology and I came across a problem I couldn't solve.

1) It is known that if a space is T1, it is countably compact if and only if every countable open cover has a finite subcover. (See below for the definition of countably compact, which might be different than the conventional definition.)

2) It is also true that if a space is T1, it is countably compact if and only if every infinite open cover has a proper subcover. (why?)

Intuitively, both properties seem to talk about how open covers can be removed of unnecessary elements and still work as a cover, under conditions where points are sufficiently close together. However, I cannot figure out a proof for the second statement. Because the cover may contain uncountably many sets, it is very hard to deal with.

This question appears in "Elements of Point Set Topology" by John D. Baum as exercise 3.33. The question and related hint can be viewed here.

Terminology used in this text:

A T1 space is a topological space such that, if x is an element of the space, the set {x} is closed.

A countably compact space is a space such that every infinite subset of the space has a limit point in the space. I am under the impression that other texts refer to this property as limit point compactness.

An infinite open cover is a collection of infinitely many open sets which cover the space.


We shall show that if $X$ is a $T_1$ space, then it is countably compact if and only if every infinite open cover has a proper subcover. The key idea is to proceed by contrapositive.

($\Rightarrow$ Needs $T_1$) Suppose that $ \ \mathcal{U} \ $ is an infinite open cover of $X$ with no proper subcover. Then for each $U \in \mathcal{U}$, there is a point $p_U \in U$ that doesn't belong to any other member of $ \ \mathcal{U} \ $. The set $A = \{p_U : U \in \mathcal{U} \}$ is infinite and doesn't have a limit point. Indeed, if $x \in X$, then there is $U \in \mathcal{U}$ containing $x$ and $ U \cap A = \{ p_U \}$. If $x = p_U$, then it isn't a limit point of $A$. If $x \ne p_U$, then, using $T_1$, $U- \{ p_U \}$ is open and $(U- \{ p_U \}) \cap A =\emptyset$. So $x$ isn't a limit point of $A$.

George Lowther gave an example showing that the $T_1$ hypothesis is fundamental. Since there are many comments, I'll reproduce it here:

"Consider the example of the real numbers where the open sets are unions of intervals $[n,a)$ for integer $n$ and real $a>n$. This is $T_0$ but not $T_1$. It is also countably compact, but the infinite cover $\mathcal{U}=\{[n,n+1)\colon n\in\mathbb{Z}\}$ has no proper subcovers. So, $T_1$ is needed."

A direct approach to prove this implication was suggested by Carl Mummert. Let $ \ \mathcal{U} \ $ be an infinite open cover of $X$ and $ \ \mathcal{U}_0 $ be a countably infinite subset of $ \ \mathcal{U} $. Now consider $ \ \mathcal{U}_1 = \mathcal{U} - \mathcal{U}_0$ and let $V \ $ be the union of all sets in $ \ \mathcal{U}_1$. Then $ \ \mathcal{U}_0 \cup \{ V \ \}$ is a countable open cover of $X$ and follows from $(1)$ that it has a finite subcover $U_1, \ldots, U_n$. Now the set consisting of all $U_i$, with possible exception of $V$, adjoined with the sets $ U \in \mathcal{U}_1$ is a proper subcover of $ \ \mathcal{U}$.

($\Leftarrow$ Doesn't need $T_1$) Now, suppose that $X$ isn't countably compact. Then there is an infinite set $A$ with no limit points. Thus $A$ is closed. By the definition of limit point, for each $x \in A$, there is an open set $U_x$ containing $x$ such that $ U_x \cap A = \{ x \} $. Consider $ \ \mathcal{U} \ $ the set of all $U_x$ plus, if necessary, $X-A$. Then $ \ \mathcal{U} \ $ is an infinite cover of $X$ with no proper subcovers.

P.S. Thanks Carl Mummert, George Lowther and Mark for your efforts to clarify and improve this answer.

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  • 3
    $\begingroup$ @Nuno: I was halfway through essentially the same argument. It might be worth pointing out where the hypothesis of $T_1$ will come into play... $\endgroup$ – Arturo Magidin Dec 6 '10 at 2:20
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    $\begingroup$ @Nuno: The question is asking for the equivalence of 3 different statements. So, that requires proving 3 (or 4) implications some, but not all, of which will require the T1 hypothesis. I think that proving (2) requires it in one direction but not the other. $\endgroup$ – George Lowther Dec 6 '10 at 2:43
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    $\begingroup$ Sorry, limit point of a set is not the same as limit point of a sequence, and the former was used in the definition of sequential compactness used in the question. A subtle difference, but it means that Carl's example is sequentially compact. $\endgroup$ – George Lowther Dec 6 '10 at 3:23
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    $\begingroup$ The claim that countable compactness implies "any infinite cover has a proper subcover" can be proved using 1. Split the infinite cover arbitrarily into a countable part and all the rest. Take the union of the rest to make it into a single open set. Now you have a countable cover, which has a finite subcover by assumption. This gives a proper subcover of the original cover. $\endgroup$ – Carl Mummert Dec 6 '10 at 3:33
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    $\begingroup$ @Mark: No, the T1 hypothesis is not used in the second part. If A has no limit points then $\bar A\setminus A$ is empty, so A is closed. Also, every $x\in A$ has an open neighbourhood $U_x$ containing no points of A other than x itself, so $X_x\cap A=\{x\}$ as you say. No use of T1 there. $\endgroup$ – George Lowther Dec 6 '10 at 3:45

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