numerably paracompact spaces

Question

A numerably paracompact spaces if and only if, for every decreasing sequence of closed non-empty $\{F_n : n \in \mathbb{N}\}$ such that $\bigcap F_n= \emptyset$, there is a succession of open $\{G_n : n \in \mathbb{N}\}$ such that $F_n\subset G_n$ for every $n\in\mathbb{N}$ and $\bigcap \overline{G_n}=\emptyset$

Answer 1

Theorem 5.2.1. [En] For every Hausdorff space $X$ the following conditions are equivalent:

(i) The space $X$ is countably paracompact.

(ii) For every countable open cover $\{U_i\}_{i=1}^\infty$ of the space $X$ there exists a locally finite open cover $\{V_i\}_{i=1}^\infty$ of $X$ such that $V_i\subset U_i$ for $i = 1,2,\dots$

(iii) For every increasing sequence $W_1\subset W_2\subset\dots$ of open subsets of $X$ satisfying $\bigcup_{i=1}^\infty W_i=X$ there exists a sequence $F_1, F_2,\dots$ of closed subsets of $X$ such that $F_i\subset W_i$ for $i =1,2,\dots$ and $\bigcup_{i=1}^\infty \operatorname{Int} F_i=X$.

(iv) For every decreasing sequence $F_1\supset F_2\supset \dots $ of closed subsets of $X$ satisfying $\bigcap_{i=1}^\infty F_i=\varnothing$ there exists a sequence $W_1, W_2,\dots$ of open subsets of $X$ such that $F_i\subset W_i$ for $i = 1,2,\dots$ and $\bigcap_{i=1}^\infty \overline{W_i}=\varnothing$.

Proof. To show that $(i)\Rightarrow (ii)$ it suffices to take a locally finite open refinement $\mathcal V$ of the cover $\{U_i\}_{i=1}^\infty$, for every $V\in\mathcal V$ choose a natural number $i(V)$ such that $V\subset U_{i(V)}$, and let $V_i = \bigcup \{V : i(V) = i\}$.

We shall show that $(ii)\Rightarrow (iii)$. As $\{W_i\}_{i=1}^\infty$ is a countable open cover of $X$, there exists an open locally finite cover $\{V_i\}_{i=1}^\infty$ of $X$ such that $V_i\subset W_i$ for $i = 1,2,\dots$. The sets $F_i =X\setminus\bigcup_{j>i} V_j\subset \bigcup_{j\le i} V_j$ are closed and – since $\bigcup_{j\le i} V_j\subset \bigcup_{j\le i} W_j=W_i$ – we have $F_i\subset W_i$ for $i = 1,2,\dots$. The family $\{U_i\}_{i=1}^\infty$ being locally finite, every point $x\in X$ has a neighbourhood which is contained in some $F_i$, i.e., $\bigcup_{i=1}^\infty \operatorname{Int} F_i=X$.

From De Morgan's laws it follows easily that conditions $(iii)$ and $(iv)$ are equivalent; hence, to conclude the proof it suffices to prove that $(iii)\Rightarrow (i)$. Let $\{U_i\}_{i=1}^\infty$ be a countable open cover of the space $X$. Consider the increasing sequence $W_1\subset W_2\subset\dots$ of open subsets of $X$, where $W_i =\bigcup_{j\le i} U_j$; as $\bigcup_{i=1}^\infty W_i= X$, there exists a sequence $F_1,F_2,\dots$ of closed subsets of $X$ such that $F_i\subset W_i$ and $\bigcup_{i=1}^\infty \operatorname{Int} F_i=X$. The set $V_i= U_i\setminus \bigcup_{j<i} F_j\subset U_i$ is open for $i = 1,2,\dots$; since $\bigcup_{j<i} F_j\subset \bigcup_{j<i} W_j\subset \bigcup_{j<i} U_j$ we have $U_i\setminus \bigcup_{j<i} U_j\subset V_i$ which implies that the family $\{V_i\}_{i=1}^\infty$ is a cover of $X$. Every point $x\in X$ has a neighbourhood of the form $\operatorname{Int} F_j$; this neighbourhood is disjoint from all sets $V_i$ for $i > j$, so that the cover $\{V_i\}_{i=1}^\infty$ is locally finite. $\square$

References

[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.