paracompact heredity to quotient space. i want to show that if $ X $ is paracompact and if $A\subseteq X$ is closed set, then $X/A$ is paracompact.
the quotient space $X/A$ is the set of equivalence classes $[x]$ where $[x]=A$ if $x\in A$ and $[x]={x}$ if $x$ not lies in $A$.
Any help is very welcome.
 A: HINT: Let $q:X\to X/A:x\mapsto[x]$ be the quotient map, and let $\mathscr{U}$ be an open cover of $X/A$. Fix $U_0\in\mathscr{U}$ such that $[A]\in U_0$, and let
$$\mathscr{V}=\{U_0\}\cup\big\{U\setminus\{[A]\}:U\in\mathscr{U}\setminus\{U_0\}\big\}\;;$$
$\mathscr{V}$ is an open cover of $X/A$ that refines $\mathscr{U}$, and $U_0$ is the only member of $\mathscr{V}$ containing $[A]$.
Clearly $\{q^{-1}[V]:V\in\mathscr{V}\}$ is an open cover of $X$. Moreover, $q^{-1}[U_0]\supseteq A$, and $q^{-1}[V]\cap A=\varnothing$ for $V\in\mathscr{V}\setminus\{U_0\}$. $X$ is paracompact, so $\{q^{-1}[V]:V\in\mathscr{V}\}$ has a locally finite open refinement $\mathscr{R}$.

*

*Show that if $R\in\mathscr{R}$ and $R\cap A=\varnothing$, then $q[R]$ is open in $X/A$ and is a subset of some $V\in\mathscr{V}$.


*Let $W=\bigcup\{R\in\mathscr{R}:R\cap A\ne\varnothing\}$. Show that $q[W]$ is open in $X/A$, and $[A]\in q[W]\subseteq U_0$.


*Conclude that $\{q[W]\}\cup\big\{q[R]:R\in\mathscr{R}\text{ and }R\cap A=\varnothing\big\}$ is a locally finite open refinement of $\mathscr{U}$.
Edit: I was sloppy here: you need to do a bit more work to get the refinement in $X/A$ to be locally finite. You’ll need the fact that locally finite families are closure-preserving, meaning that if $\mathscr{F}$ is any locally finite collection of sets, then $$\operatorname{cl}\bigcup\mathscr{F}=\bigcup_{F\in\mathscr{F}}\operatorname{cl}F\;.$$ You’ll also want the following

Theorem. Let $X$ be a paracompact space, and let $\mathscr{U}=\{U_\alpha:\alpha\in A\}$ be an open cover of $X$. Then there is a locally finite open cover $\mathscr{V}=\{V_\alpha:\alpha\in A\}$ such that $\operatorname{cl}V_\alpha\subseteq U_\alpha$ for each $\alpha\in A$.
Sketch of Proof. $X$ is regular, so for each $x\in X$ there is an open $W_x$ such that $$x\in W_x\subseteq\operatorname{cl}W_x\subseteq U_\alpha$$ for some $\alpha\in A$. Let $\mathscr{R}$ be a locally finite open refinement of $\{W_x:x\in X\}$. For each $R\in\mathscr{R}$ there is an $\alpha(R)\in A$ such that $R\subseteq\operatorname{cl}R\subseteq U_{\alpha(R)}$. For each $\beta\in A$ let $$V_\beta=\bigcup\{R\in\mathscr{R}:\alpha(R)=\beta\}\;;$$ then $\operatorname{cl}V_\beta=\bigcup\{\operatorname{cl}R:\alpha(R)=\beta\}\subseteq U_\beta$, since $\mathscr{R}$ is locally finite. Let $\mathscr{V}=\{V_\beta:\beta\in A\}$. If $U$ is any open set, and $R\in\mathscr{R}$, then $U\cap R\ne\varnothing$ iff $U\cap V_{\alpha(R)}\ne\varnothing$, so if $U$ meets only finitely many members of $\mathscr{R}$, then $U$ meets only finitely many members of $\mathscr{V}$. Thus, $\mathscr{V}$ is locally finite. $\dashv$

In my original answer choose $\mathscr{R}$ so that $\{\operatorname{cl}R:R\in\mathscr{R}\}$ refines $\big\{q^{-1}[V]:V\in\mathscr{V}\big\}$. Then $$X\setminus\bigcup\{\operatorname{cl}R:R\in\mathscr{R}\text{ and }R\cap A=\varnothing\}$$ is an open set containing $A$ that is disjoint from all of the $R\in\mathscr{R}$ except $W$. Now the third bullet point should be pretty straightforward. (It helps to recognize that $X\setminus A$ and $(R/A)\setminus\big\{[A]\big\}$ are homeomorphic, so that it’s only what happens at the point $[A]\in X/A$ that might cause trouble.)
