Is the union of two open paracompact subspaces of a space $X$ paracompact?
A space is called paracompact if every open cover of the space has a locally finite open refinement.
Proof attempt: Suppose $X=O_1\cup O_2$ with each $O_i$ open and paracompact. Given an open cover $\mathcal U$ of $X$, intersect every element of $\mathcal U$ with $O_1$ to get an open cover of $O_1$. Then take a refinement of that cover that is locally finite in $O_1$. And similarly for $O_2$. The union of the two refinements is an open cover of $X$ that refines $\mathcal U$. But I am having difficulty showing the result is locally finite.
Presumably the union is not paracompact in general.
What would be a (Hausdorff if possible) counterexample?
Best Answer
No, the so-called Thomas plank provides a counter-example:
Let $A(\omega_1) = \omega_1 \cup \{\infty\}$ and $A(\omega) = \omega \cup \{\infty\}$ be the one-point compactifications of the discrete spaces $\omega_1$ and $\omega$, respectively.
$X := (A(\omega_1) \times A(\omega)) \setminus \{(\infty, \infty)\}$ is completely regular, T2. It is well-known that $X$ is not even normal (the closed, disjoint sets $\omega_1 \times \{\infty\}$ and $\{\infty\} \times \omega$ cannot be separated). In particular, it is not paracompact.
$U := A(\omega_1) \times \omega$ and $V := \omega_1 \times A(\omega)$ are products of a compact and a discrete space. Hence, they are paracompact (in fact $U, V$ are hereditarily ultraparacompact).
Of course, $U, V$ are open in $X$ and $U \cup V = X$.