# A question on linearly Lindelöf space

A space $X$ is linearly lindelöf if for every open cover of $X$, linearly ordered by the subset relation, has a countable subcover.

Question is this: How could we see that $X$ is linearly lindelöf iff for any uncountable regular cardinal $\kappa$, any decreasing $\kappa$-sequence of closed non-empty sets has a nonempty intersection?

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@Jyrki: If you are editing, why not capitalize Lindelof's name, and while you bumped this question... while not edit my answer as well? :-) –  Asaf Karagila Mar 8 '13 at 1:52

Assume that $C_\alpha$ is a decreasing sequence of closed sets. Take the complements $D_\alpha=X\setminus C_\alpha$. If the intersection is empty then $D_\alpha$ form a linear open cover.
If $X$ is linearly Lindelof (and $\kappa$ is regular and uncountable) then the sequence must stabilize at a bounded ordinal, and therefore the closed sets must become empty at a certain point, so if all the closed sets are non-empty it is impossible that their intersection is non-empty; on the other hand if $X$ is not linearly Lindelof we can find a linear open cover without a countable subcover and by taking complements this is a decreasing sequence of closed sets, none of which is empty and the intersection is empty.
The part is this:If $X$ is linearly Lindelof (and $\kappa$ is regular and uncountable) then the sequence must stabilize at a bounded ordinal, and therefore the closed sets must become empty at a certain point, so if all the closed sets are non-empty it is impossible that their intersection is non-empty. –  Paul Jan 18 '13 at 2:06
Paul, if $\kappa>\omega$ is regular then every countable subset is bounded. This means that if $D_\alpha$ actually cover $X$ then there is some $\beta<\kappa$ such that $D_\beta=X$ and so $C_\beta=\varnothing$, but the $C_\beta$ were decreasing so the subsequent closed sets have to be empty. Therefore all the sets are non-empty, and $X$ is linearly Lindelof then it is impossible that the intersection is empty, as this would witness an open cover without a countable subcover. –  Asaf Karagila Jan 18 '13 at 2:09