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.

Nuno · Accepted Answer · 2010-12-06T23:10:01.890

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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.

edited Dec 06 '10 at 23:10

answered Dec 06 '10 at 02:17

Nuno

3,787

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@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... – Arturo Magidin Dec 06 '10 at 02:20
@Arturo: I believe that we don't need the T1 hypothesis to prove 2. But, I'll check it carefully to see if I am missing something. – Nuno Dec 06 '10 at 02:36
Thank you for answering this question! Could you please clarify why the set A in the first part does not have a limit point? – Mark Dec 06 '10 at 02:39
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@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. – George Lowther Dec 06 '10 at 02:43
@Mark: Ah! I think I thought of a wrong argument to guarantee that $A$ doesn't have a limit point. We probably need to use the $T_1$ hypothesis there. I'll check and repost in a few minutes. – Nuno Dec 06 '10 at 02:44
@Nuno: Consider a non $T_0$ space whose points are the integers and whose nontrivial open sets are the intervals of the form $(-\infty, m]$ for each integer $m$. Then the collection of all these intervals has no finite subcover, but every cover of the space with more than one open set has a proper subcover, because the open sets are linearly ordered by $\subseteq$. Also, any set containing more than one point has a limit point; the larger number is in the closure of the smaller one. – Carl Mummert Dec 06 '10 at 02:51
@George Lowther: But the OP says he is having trouble with question 2; I assumed, as apparently Nuno did, that he is not asking for help with 1, but this assumption could be mistaken... – Arturo Magidin Dec 06 '10 at 02:52
@All: I believe that there is a flaw in my proof, since I'm not sure that A, in the first part, doesn't have a limit point. I'll try to correct this, but I'd appreciate if someone find an alternative to that. – Nuno Dec 06 '10 at 03:00
Yes, I understand question 1 and the proof can be found by visiting the link given in the question. I included question 1 as motivation/intuition for question 2. – Mark Dec 06 '10 at 03:00
@Nuno: That is unfortunate. The second half of the proof seems to be correct and demonstrates that any space (we don't care whether it is T1) in which infinite open covers have proper subcovers is countably compact. – Mark Dec 06 '10 at 03:10
@Mark: No, that doesn't hold in non-T1 spaces. See Carl's example. The sequence $x_n=n$ has no limit point. Nuno's proof used the T1 assumption at the point where he assumed the existence of an open set $U_x$ with $U_x\cap A={x}$. – George Lowther Dec 06 '10 at 03:20
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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. – George Lowther Dec 06 '10 at 03:23
So...the second part of Nuno's proof looks good to me. The first part is also good, except that the T1 hypothesis is required to show that A has no limit points. – George Lowther Dec 06 '10 at 03:31
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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. – Carl Mummert Dec 06 '10 at 03:33
@George Lowther: I'm just trying to think a way to prove that A has no limit points. How can we do that? The argument I thought when wrote the answer is wrong. – Nuno Dec 06 '10 at 03:38
@Carl Mummert: Thanks. I think it's right. Why don't you post it as an answer? – Nuno Dec 06 '10 at 03:43
@Mark: It may not be needed in the first part. For instance, in the first problem one of the implications should hold without having to rely on the $T_1$ hypothesis, while the other implication does require it. The same thing should happen with the second problem. That is: one of the two implications holds in more generality than in the $T_1$ case, but in the $T_1$ case you also get the other implication. – Arturo Magidin Dec 06 '10 at 03:44
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@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. – George Lowther Dec 06 '10 at 03:45
@Carl: Your subcover is not actually a subcover because you took a large chunk of the original cover and union-ed them together. A similar argument could show that every cover as a subcover of one set. Just union all the sets in the cover. – Mark Dec 06 '10 at 03:48
@Nuno: please feel free to simply edit your post to replace the first part with the argument I sketched. The second part of your argument is correct, I believe, so that resolves the original question 2 from the question. – Carl Mummert Dec 06 '10 at 03:49
@Mark: T1 is used implicitly in the first part, in the claim that A does not have a limit point. Every point x lies in one of the open sets $U\in\mathcal{U}$, and $U\cap A={p_U}$. If $x=p_U$ then it is not a limit point of $A$. If $x\not=p_U$, then $U\setminus{p_U}$ is open (using T1) and contains x, so x is not a limit point. – George Lowther Dec 06 '10 at 03:49
@Mark: split them back apart at the end, of course. That's why you only get a proper subcover in the end, not a finite subcover. – Carl Mummert Dec 06 '10 at 03:50
@George: You said "Nuno's proof used the T1 assumption at the point where he assumed the existence of an open set Ux with Ux∩A={x}." I believe, however, that this follows from the assumption that A has no limit points. – Mark Dec 06 '10 at 03:53
@Carl: Oh! That is so slick! – Mark Dec 06 '10 at 03:55
@Mark: I did say that. But then I corrected myself (after realizing that the definition given of countable compactness is slightly different from what I was assuming). – George Lowther Dec 06 '10 at 03:57
@All: Thanks everyone, I'll edit the question and add the useful information you added. I'll also add a proof for (1) and will explicitly point when we need T1. – Nuno Dec 06 '10 at 03:57
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@Nuno/Mark: 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 T0 but not T1. It is also countably compact, but the infinite cover $\mathcal{U}={[n,n+1)\colon n\in\mathbb{Z}}$ has no proper subcovers. So, T1 is needed. – George Lowther Dec 06 '10 at 04:05
@George: Yes, sorry for the confusion over the definition of countable compactness. Good example also. – Mark Dec 06 '10 at 04:13
@George: I am a little confused about the second part of this comment - "If x=pU then it is not a limit point of A. If x/=pU, then U∖{pU} is open (using T1) and contains x, so x is not a limit point." I agree that U∖{pU} would be an open set around x, but I do not see why it does not intersect A at some point. – Mark Dec 06 '10 at 04:18
@Mark: each open set $U$ in the cover contains exactly one point in $A$, namely $p_U$; this is how $A$ is constructed. So once we remove that point, the remaining open set must be disjoint from $A$. – Carl Mummert Dec 06 '10 at 04:25
@George Lowther: The proof I originally thought of 'A has no limit points' is similar to yours. But I forgot to consider the case when x isn't in A. That's the reason I believed we didn't need the $T_1$ hypothesis. – Nuno Dec 06 '10 at 04:25
@Carl and George: Sorry to prolong this; Here is how I understand the argument: Choose a point in the space. If this point is in A, it is not a limit point. If this point is not in A, we need to show that it is also not a limit point. We should do this by exhibiting an open set containing the point that such that the set is disjoint from A. Are you saying that A - {pU} demonstrates this set? – Mark Dec 06 '10 at 04:35
@Mark: Any point $x \not \in A$ is still in some open set $U$ from the cover, because it's a cover. Now there is exactly one point of $A$ in $U$, namely $p_U$, and so $U - {p_U}$ is an open neighborhood of $x$ disjoint from $A$. – Carl Mummert Dec 06 '10 at 04:38
@Mark: Yes, I was saying that the open set $U\setminus p_U$ is disjoint from A and contains x, so x is not a limit point of A. – George Lowther Dec 06 '10 at 04:40
@All: Ah! I believe that this question has been thoroughly proven. Thank you for your explanations. – Mark Dec 06 '10 at 04:43
@George/Carl/Mark: I've edited the answer. Let me know what can be done to improve it. I'll probably add the example George gave that we really need T_1 for the first implication. I'm just checking the details. I'll also try to add a proof for (1) tomorrow, since I need to sleep now. Thanks again for your help. – Nuno Dec 06 '10 at 04:50
@Nuno: A proof of (1) is provided with in the link. It might be helpful to also include Carl's elegant solution of reducing the cover to a countable cover. – Mark Dec 06 '10 at 05:30
@Mark: Ok. I'll add it. Is there something else you think I should add? – Nuno Dec 06 '10 at 05:35
@All: I think I have added everything now. Let me know if there is something imprecise/unclear/wrong. – Nuno Dec 06 '10 at 06:14
Looks good to me. – Carl Mummert Dec 06 '10 at 12:27
@Nuno After so many years this thread perhaps presents no interest any longer, however still I would like to make some clarifications. The discussion concerns the following 3 properties a topological space might have: 1) countable compactness, defined by requiring any countable open cover to admit a finite subcover (with no separation axioms included whatsoever); 2) the property that any infinite subset has nonempty derivative (where by derivative I mean the set of all accumulation points); 3) the claim that any infinite open cover admits a proper subcover. (to be cont.) – ΑΘΩ Apr 14 '21 at 09:23
@Nuno (cont.) Your goal in the above was to prove that for any $\mathrm{T}_1$ space countable compactness is equivalent to property 3) as listed above. The fact that a countably compact space always enjoys property 3) is indeed proved as you have in your answer above and does not require the $\mathrm{T}_1$ axiom. In order to prove the converse implication it you correctly show that any space with property 3) necessarily also has property 2), which once again does not require any $\mathrm{T}_1$ assumption. (to be cont.) – ΑΘΩ Apr 14 '21 at 09:29
@Nuno (cont.) Where the $\mathrm{T}_1$ separation axiom does come into play is in justifying the claim that property 2) entails property 1)! Indeed, any countably compact space will enjoy property 2) however to claim the reverse implication one requires the additional assumption of $\mathrm{T}_1$, in other words a $\mathrm{T}_1$-space in which any infinite subset has accumulation points is countably compact. – ΑΘΩ Apr 14 '21 at 09:31

A challenging question about T1 spaces and countable compactness

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