A challenging question about T1 spaces and countable compactness

Question

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.

Answer 1

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.