Metacompactness of the Euclidean space

Question

Does anyone know how to prove that every Euclidean space is countably metacompact? In particular, my interest is in $R^2$. Thanks,

Answer 1

Every metric space is paracompact, hence metacompact, hence countably metacompact. Thus, $\Bbb R^n$ is countably metacompact for every $n$.

Added: A direct proof is easy if you know the following classic characterization of countable metacompactness due to Ishikawa: a space $X$ is countably metacompact iff whenever $\mathscr{F}=\{F_k:k\in\Bbb N\}$ is a family of closed sets in $X$ such that $\bigcap\mathscr{F}=\varnothing$ and $F_k\supseteq F_{k+1}$ for each $k\in\Bbb N$, then there is a family $\mathscr{U}=\{U_k:k\in\Bbb N\}$ such that $\bigcap\mathscr{U}=\varnothing$ and $U_k\supseteq F_k$ for each $k\in\Bbb N$. If $\langle X,d\rangle$ is a metric space, and $\mathscr{F}=\{F_k:k\in\Bbb N\}$ is a family of closed sets in $X$ such that $\bigcap\mathscr{F}=\varnothing$ and $F_k\supseteq F_{k+1}$ for each $k\in\Bbb N$, let

$$U_k=\bigcup_{x\in F_k}B(x,2^{-k})\;,$$

where $B(x,r)$ is the open $d$-ball of radius $r$ centred at $x$; clearly $U_k\supseteq F_k$. Let $\mathscr{U}=\{U_k:k\in\Bbb N\}$, and suppose that $x\in\bigcap\mathscr{U}$; then for each $k\in\Bbb N$ there is an $x_k\in F_k$ such that $B(x,x_k)<2^{-k}$. Clearly $\langle x_k:k\in\Bbb N\rangle\to x$. For each $n\in\Bbb N$ we have $\{x_k:k\ge n\}\subseteq F_n$, and $F_n$ is closed, so $x\in F_n$ for each $n\in\Bbb N$, and hence $x\in\bigcap_{n\in\Bbb N}F_n=\varnothing$, which is absurd. Thus, $\bigcap\mathscr{U}=\varnothing$, and $X$ is countably metacompact. $\dashv$