# Union of two open paracompact subspaces is paracompact?

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?

• If you live in $O_1\cap O_2$, you can use the intersection of neighborhoods witnessing locally finite in each. If you live in $O_1\setminus cl(O_2)$, you can use the neighborhood from $O_1$ subtracting $cl(O_2)$, Likewise for $O_2\setminus cl(O_1)$. So it'd seem the trickiness is when you're on the boundary of $O_1$ or $O_2$? Commented Aug 17, 2023 at 21:55
• Yes, the tricky part is on the boundary of the open sets. Commented Aug 17, 2023 at 21:58

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 $$U \cup V = X$$.