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Usually one says that a collection $\mathscr{C}$ of subsets of a topological space $X$ is locally finite if each point $x \in X$ has a neighborhood $U_x \subset X$ such that the set $\{ Z \in \mathscr{C} : U_x \cap Z \ne \emptyset \}$ of subsets of the collection intersecting $U_x$ is finite.

Is this equivalent to the condition that for every quasi-compact open subset $U \subset X$, the set $\{ Z \in \mathscr{C} : U \cap Z \ne \emptyset \}$ is finite?

I think this should be true, but I haven't seen such a characterization anywhere.

(By quasi-compact, I mean compact but not necessarily Hausdorff.)

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up vote 1 down vote accepted

It is not equivalent in general. For a connected non-compact Hausdorff space $X$ (in particular, the real line) the only (quasi-)compact open set is $\emptyset$, and for every family $\mathscr{C}$ of subsets of $X$ we have $\{ Z \in \mathscr{C} : \emptyset \cap Z \neq \emptyset \} = \emptyset$ is finite, and thus satisfies the condition you have given.

On the other hand (as I am sure you have noticed) local-finiteness does imply the condition you have given.

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Thanks. I wonder if there are "reasonable" assumptions on the space to make the reverse implication true... – Adeel Feb 14 '13 at 11:51
@Adeel: The obvious assumption would be that the family of quasi-compact open sets forms a base for the topology. But except for trivialities (e.g., cofinite spaces) I am having trouble thinking of examples of such spaces, so it is probably not correct to call this a reasonable assumption The cheating way is to say "every quasi-compact subset has an open neighbourhood $U$ which meets only finitely many elements of $\mathscr{C}$". But as finite sets are quasi-compact, this is more of an obfuscation than a characterisation. – arjafi Feb 14 '13 at 12:01
Ah, of course; and this is precisely what I needed, because schemes have such topologies. Thanks! – Adeel Feb 14 '13 at 15:25

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