In the paper


<http://topo.math.auburn.edu/tp/reprints/v05/tp05011.pdf>


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



Lindelöf Property for Open Covers for Compact Sets
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.


