Recall that a space $X$ is metaLindelof if every open cover of $X$ has a point-countable open refinement. A space $X$ is metacompact if every open cover of $X$ has a point-finite open refinement.

I have the following two questions, because the questons are similar, I put together here:

Is there a $\sigma$-metaLindelof space which is a not metaLindelof?

Is there a $\sigma$-metacompact space which is not metacompact?

Thanks for your help.


1 Answer 1


In fact, there is a space which is the union of two paracompact spaces which is not metaLindelof. Let $X$ be a $\psi$-space, that is, a locally compact, pseudocompact space having a countable dense set of isolated points and such that the set of non-isolated points is uncountable, closed, and discrete. If $A$ is the set of isolated points and $B = X \setminus A$, then each of $A$ and $B$, being metrizable, is paracompact. For each $x \in B$ let $U_x$ be a neighborhood of $x$ such that $U_x \cap B = \{x\}$. Let ${\mathcal U}$ be the cover $\{U_x : x \in B\} \cup \{ \{n\}: n \in X \setminus \cup\{U_x : x \in B\}\}$. Since $B$ is uncountable and discrete and $A$ is countable, for every refinement $\mathcal V$ of $\mathcal U$ there is an $n \in A$ which is contained in uncountably many elements of $\mathcal V$, so $X$ is not metaLindelof.


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