In the paper


the author claimes (Theorem 2, without proof) that for a completely regular Hausdorff space $X$ the following are equivalent:

(1) The space of continuous functions $C(X)$ with the compact open topology is countably tight.

(2) Every open cover for compact subsets of $X$ has a countable subcover for compact sets.

I try to understand condition (2).

Here an "open cover for compact sets" is an open cover $\mathcal U$ such that every compact subset of $X$ is contained in some member of $\mathcal U$.

Obviously, the second condition looks a lot like the Lindelöf property:

(3) The space $X$ is Lindelöf, i.e. every open cover has a countable subcover.

I think that condition (3) looks stronger than condition (2), because every open cover for compact sets is an open cover and therefore has a countable subcover.

However in the paper it is stated (Proposition 5, also without proof) that (2) implies (3).

(My question is now: Can anybody explain to me what is going on here?)

EDIT: I agree, my question as I wrote it down is not a very well-posed question ;)

Another try: My question consists of two parts:

  • Is there a space $X$ which satisfies (3), i.e.$X$ is Lindelöf, but does not satisfy (2) ? (this would imply that $C(X)$ is not countably tight with the compact open topology)

  • Why does (2) imply (3) ? (I know the reference claims that but since there seems to be no proof in the paper, I have problems to believe it)

Thank you in advance! Tom

  • $\begingroup$ It's not clear to me where your problem is. Anyway, it may be useful to you to note that any open cover closed for finite unions is an open cover for compact sets. $\endgroup$ – Renan Maneli Mezabarba Dec 20 '16 at 16:05
  • 1
    $\begingroup$ @Renan Maneli Mezabarba: Thanks for your comment! I reread my question ("Can anybody tell me what is going on here?") and I have to admit that it was not really clear what I wanted to know... ;-) Sorry for that. I hope it is clearer now. $\endgroup$ – Tom Dec 21 '16 at 8:17

The following is adressed to the second part of your question.

For brevity, let's call a collection $\mathcal{U}$ of open sets of $X$ as $k$-cover if for any compact subset $K\subset X$ there exists $U\in\mathcal{U}$ such that $K\subset U$. Keep in mind that (clearly) any $k$-cover of $X$ is an open covering for $X$.

Now, suppose $X$ satisfies condition (2), that is, any $k$-cover for $X$ has a countable $k$-subcover. To show that $X$ is a Lindelöf space it is enough to prove that any open covering closed for finite unions has a countable subcover, but any such open covering is itself a $k$-cover, so it contains a countable $k$-subcover and we are done.

Right now I can't remember a counterexample to the converse.

| cite | improve this answer | |
  • 1
    $\begingroup$ Thanks for your answer. Maybe I am asking something really stupid, but one question: Why is it enough to prove that any open coverging closed for finite unions has a countable subcover? If someone gives me an uncountable cover, it may not be closed under finite unions... $\endgroup$ – Tom Dec 21 '16 at 11:39
  • $\begingroup$ Suppose $\mathcal{U}$ is an open covering for $X$ and let $\mathcal{V}=\{\bigcup G:G$ is a finite subset of $\mathcal{U}\}$. If $\{\bigcup G_n:n\in\omega\}$ is a countable subcover of $\mathcal{V}$, then $\bigcup_{n\in\omega}G_n$ is a countable subcover of $\mathcal{U}$, because each $G_n\subset\mathcal{U}$ is finite. $\endgroup$ – Renan Maneli Mezabarba Dec 21 '16 at 11:41
  • $\begingroup$ Okay, that was too easy for me to see. Thanks! $\endgroup$ – Tom Dec 21 '16 at 11:44

Example 14 in the paper you cite shows the Sorgenfrey line $\mathbb{R}_\ell$ (that is, $\mathbb{R}$ with the lower limit topology), which is a standard example of a Lindelöf space, does not satisfy (2).

| cite | improve this answer | |
  • $\begingroup$ This is what I was looking for! It seems I was unable to read that paper carefully enough (plus I was not aware of the fact that the Sorgenfrey line is Lindelöf). THank you very much! $\endgroup$ – Tom Dec 21 '16 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.