Let $ (X, \tau) $ be a paracompact space $ T_2 $ and $ A \subset X $. If $A$ is closed, then the quotient space $X / A$ is paracompact and $T_2$.

Question

Let $ (X, \tau) $ be a paracompact space $ T_2 $ and $ A \subset X $. If $A$ is closed, then the quotient space $X / A$ is paracompact and $T_2$.

proof: Let $ \pi: X \rightarrow X / A$ with $x \rightarrow [x]$ and let $ \mathcal{U} = \{U_j; j \in J \} $ an open coverage of $ X / A $, let $ U_0 \in \mathcal{U} $ such that $ [A] \in U_0 $ and define:

$\mathcal{V}= \{U_0\} \cup \{ \mathcal{U}/\{[A]\} \ : \ U \in \mathcal{U}\setminus \{U_0\} \} $

$ \mathcal{V} $ is an open coverage of $ X / A $ that refines $ \mathcal{U} $ and $ U_0 $ is the only member of $ \mathcal{V} $ that contains $ [A] . $

Note that $ \{\pi^{-1} [V] \: \ V \in \mathcal{V} \} $ is an open coverage of $ X, $ in addition to $ A \subseteq \pi^{- 1} [ U_0] $ y $ \pi^{- 1} [V] \cap A = \emptyset $ for each $ V \in \mathcal{V} \setminus \{U_0 \}. $

By hypothesis $ X $ is paracompact, so $ \{\pi^{- 1} [V] \: \ V \in \mathcal{V} \} $ has a locally finite refinement $ \mathcal {W} $. ...

How can I continue with the demo? how should I define $ \mathcal{W} $ that allows me insanely finite refinement of $ \mathcal{U}. $

Answer 1

This is to complete my comment. The following statement appears as Theorem 1 in E. Michael's paper Another Note on Paracompact Spaces, Proc. Amer. Math. Soc., 8 (1957), 822-828.

A regular Hausdorff space is paracompact if and only if each open covering of it has a closure-preserving closed refinement.

We apply this to answer your question as follows. We assume $X$ is paracompact $T_2$ and $A\subset X$ a closed subspace. Then the quotient map $\pi:X\rightarrow X/A$ is closed. Since $X$ is paracompact $T_2$ it is normal Hausdorff, and thus so is $X/A$ (see, for instance, Willard's General Topology Th. 15.4). Thus to show that $X/A$ is paracompact it will suffice to show that an arbitrary open covering $\mathcal{U}$ of it has a closure-preserving closed refinement.

To this end consider the covering $\pi^{-1}\mathcal{U}$ of $X$. Let $\mathcal{C}$ be a closure-preserving closed refinement of $\pi^{-1}\mathcal{U}$. Then $\pi(\mathcal{C})=\{\pi(C)\mid C\in\mathcal{C}\}$ is a closed covering of $X/A$ which clearly refines the original covering $\mathcal{U}$. It remains to see that $\pi(\mathcal{C})$ is closure-preserving. For this it will suffice to take an arbitrary subfamily $\mathcal{C}'\subset\mathcal{C}$ and notice that $\bigcup\pi(\mathcal{C}')=\pi(\bigcup\mathcal{C}')$ is closed. $\square$

That answers your question. In case you feel a bit cheated with me using a result which you may not have seen before, I will supply a second answer. It begins as all good things do, by quoting Engelking's book.

A $T_1$ space $Y$ is paracompact $T_2$ if and only if every open covering of it has a subordinated partition of unity.

This is Theorem 5.1.9 in R. Engelking's General Topology. By subordinated here I mean that any covering $\mathcal{U}$ of $Y$ is refined by the cozero sets of some (not necessarily locally-finite) partition of unity $\{\xi_i:Y\rightarrow I\}_{i\in\mathcal{I}}$ on $Y$.

We apply it as follows. We start with an arbitrary open covering $\mathcal{U}$ of $X/A$. We pull this back to an open covering $\pi^{-1}\mathcal{U}$ of $X$ and find a partition of unity $\{\xi_i:X\rightarrow I\}_{i\in\mathcal{I}}$ subordinated to it.

Next find $U\in\mathcal{U}$ such that $[A]\in U$. Then $\{\pi^{-1}U,X\setminus A\}$ covers $X$ so has a subordinated partition of unity. This is exactly a single function $\phi:X\rightarrow I$ with $\phi(A)=0$ and $\phi(X\setminus U)=1$ (so that $\phi+(1-\phi)=1$).

Write $\xi'_i=\phi\cdot\xi_i$ and $\xi_{i'}=1-\phi$ and put $\mathcal{J}=\mathcal{I}\cup\{i'\}$. We check that $\{\xi'_j\}_{j\in\mathcal{J}}$ is a partition of unity on $X$, in that it sums to $1$. Since each $\xi'_j$ is constant on $A$, for each $j\in\mathcal{J}$ there is an induced map $\overline\xi_j:X/A\rightarrow I$ satisfying $\overline \xi_j\pi=\xi'_j$.

Then $\{\overline\xi_j\}_\mathcal{J}$ is a partition of unity on $X/A$. The sum $\sum_\mathcal{J}\overline\xi_j$ is the continuous function induced by $\sum_\mathcal{J}\xi_j'=1$, so in particular is constant at $1$. For $i\in\mathcal{I}\subseteq\mathcal{J}$ we have $\pi^{-1}(\overline\xi_i^{\,-1}(0,1])={\xi_i'}^{-1}(0,1]\subset\xi_i^{-1}(0,1]$, whilst $\pi^{-1}(\overline\xi_{i'}^{\,-1}(0,1])={\xi_{i'}'}^{-1}(0,1]\subseteq \pi^{-1}U$. In either case we can conclude that $\overline\xi_j^{\,-1}(0,1]$ is contained in some member of $\mathcal{U}$.

Thus we have shown that $\mathcal{U}$ has a subordinated partition of unity, which was what we needed to do to show that $X/A$ is paracompact. $\square$