Is Alexandroff Duplicate A(X) of X paracompact?

Question

Prove or disprove:
If $X$ is a paracompact space, then Alexandroff Duplicate $A(X)$ of $X$ is paracompact. Thanks for any help.

Answer 1

Using the notation I set up in my previous answer, suppose that $X$ is a paracompact space. To show that the Alexandroff double $A(X) = X \times \{ 0,1 \}$ is paracompact, consider the following

Hints. Suppose that $\{ W_i = ( V_i \times \{ 0 , 1 \} \setminus ( A_i \times \{ 1 \} ) : i \in I \}$ is a family of basic open subsets of $A(X)$ where each $U_i$ is open in $X$, and $A_i \subseteq X$ is finite.

1. If $\{ V_i : i \in I \}$ is a locally finite family of open subsets of $X$, then $\{ W_i : i \in I \}$ is a locally finite family of open subsets of $A(X)$.
2. If $\bigcup_{i \in I} V_i = X$, then $A(X) \setminus \bigcup_{i \in I}W_i$ is a closed discrete subset of $A(X)$.