How to prove that every open subset of metacompact space is also metacompact?

Question

Let $X$ be metacompact space.

Let $Y$ be an open subset of $X$.

How to show that $Y$ is also metacompact?

Answer 1

Suppose $Y$ is a locally compact Hausdorff space that is not metacompact.
Take $X$ to be its one-point compactification.
Then $Y \subset X$ is open but not metacompact.

Answer 2

For a concrete counterexample, the ordinal space $[0,\omega_1]$ is compact, so it is certainly metacompact, but its open subspace $[0,\omega_1)$ is not metacompact: the open cover $\mathscr{U}=\{[0,\alpha):0<\alpha<\omega_1\}$ has no point finite open refinement.

To see this, suppose that $\mathscr{R}$ is an open refinement of $\mathscr{U}$. For each limit ordinal $\alpha<\omega_1$ there are an ordinal $\varphi(\alpha)<\alpha$ and an $R(\alpha)\in\mathscr{R}$ such that $(\varphi(\alpha),\alpha]\subseteq R(\alpha)$. By a weak form of the pressing-down lemma there is a $\beta<\omega_1$ such that $A=\{\alpha<\omega_1:\varphi(\alpha)=\beta\}$ is uncountable. Then $\beta+1\in R(\alpha)$ for each $\alpha\in A$. Let $\mathscr{R}_0=\{R(\alpha):\alpha\in A\}$. Each member of $\mathscr{R}_0$ is countable, since it is contained in $[0,\gamma)$ for some $\gamma<\omega_1$, so if $\mathscr{R}_0$ were finite (or even countable), $\bigcup\mathscr{R}_0$ would be countable, and there would be some $\gamma<\omega_1$ such that $\bigcup\mathscr{R}_0\subseteq[0,\gamma)$. $A$ is uncountable, so there is an $\alpha\in A\setminus[0,\gamma)$. But then $\alpha\in R(\alpha)\in\mathscr{R}_0$, so $\alpha\in\bigcup\mathscr{R}_0\subseteq[0,\gamma)$, which is a contradiction. Thus, $\mathscr{R}_0$ cannot be finite (or even countable), and $[0,\omega_1)$ is not metacompact (or even metaLindelöf).