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?

  • $\begingroup$ 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 $O_2$? $\endgroup$ Commented Aug 17, 2023 at 21:55
  • $\begingroup$ Yes, the tricky part is on the boundary of the open sets. $\endgroup$
    – PatrickR
    Commented Aug 17, 2023 at 21:58

1 Answer 1


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 $U \cup V = X$.

  • $\begingroup$ Very nice example! $\endgroup$
    – PatrickR
    Commented Aug 18, 2023 at 22:17
  • $\begingroup$ For reference, the space is described in Engelking 2.3.36, where it is shown it is not normal. $\endgroup$
    – PatrickR
    Commented Aug 19, 2023 at 3:23

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