
> 
> 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.



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


