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As the title says, I am trying to find an example of a topological space which is, perfectly normal, Lindelöf, but not metacompact. Can anyone help me with an example for such a space, or if there isn't, show why?

Thank you!

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up vote 1 down vote accepted

Actually, every regular Lindelöf space is paracompact (which clearly implies metacompact).

Here is a proof:

Suppose $X$ is a regular Lindelöf space. Given an open cover $\mathcal{U}$ of $X$, for each $x \in X$ we can choose open neighbourhoods $U_x , V_x$ of $x$ such that $\overline{V}_x \subseteq U_x$, and $U_x \in \mathcal{U}$. Then by Lindelöfness there are $\{ x_i : i \in \mathbb{N} \}$ such that $\bigcup_{i \in \mathbb{N}} V_{x_i} = X$. For $i \in \mathbb{N}$ set $$W_i = U_{x_i} \setminus {\textstyle \bigcup_{j<i}} \overline{V_{x_j}}.$$ Clearly each $W_i \subseteq U_{x_i}$ is open. To see that it covers $X$, given $x \in X$ let $i \in \mathbb{N}$ be least such that $x \in U_{x_i}$, and note that $x \in W_i$. To see that it is locally finite, given $x \in X$ let $i \in \mathbb{N}$ be such that $x \in V_{x_i}$, and note that $V_{x_i} \cap W_j = \varnothing$ for all $j > i$.

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ok got it. Thank you!! – topsi Jan 29 '14 at 16:54

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