It seems likely that for any open cover, we can construct a locally finite refinement using the local compactness of the space. I can't figure out how to work the construction though, and I'm not yet convinced that there is no counterexample.

  • 2
    $\begingroup$ I can tell you that every second-countable locally compact Hausdorff space is paracompact, as should be shown here. $\endgroup$
    – MickG
    Dec 2, 2015 at 21:09
  • $\begingroup$ Yes, theorem 2.6, pp. 2-3. $\endgroup$
    – MickG
    Dec 2, 2015 at 21:10
  • $\begingroup$ But without second-countability I wouldn't know… $\endgroup$
    – MickG
    Dec 2, 2015 at 21:11

3 Answers 3


A locally compact Hausdorff space need not be paracompact. In fact, it need not be normal (and all paracompact Hausdorff spaces are normal). A basic example is the deleted Tychonoff plank. Consider the product $[0,\omega] \times [0,\omega_1]$ of these two ordinal spaces ($\omega$ denotes the least infinite ordinal, and $\omega_1$ the least uncountable ordinal), and let $X$ be the subspace obtained by removing the poing $\langle \omega , \omega_1 \rangle$.

  • As a subspace of a Hausdorff space, it is clearly Hausdorff.
  • As an open subspace of a compact space, it is locally compact.
  • It is not normal because the closed sets $E = [0,\omega) \times \{ \omega_1 \}$ and $F = \{ \omega \} \times [0,\omega_1)$ cannot be separated by disjoint open sets.

Another example of a locally compact Hausdorff space which is not paracompact is the open ordinal space $[0,\omega_1)$.

  • As an ordered space, it is Hausdorff.
  • As an open subspace of the compact space $[0,\omega_1]$, it is locally compact. (In fact, for each $\alpha < \omega_1$, $[0,\alpha]$ is a compact neighborhood of $\alpha$.)
  • Suppose that $\mathcal U$ is an open refinement of the open cover $\{ [0,\alpha) : \alpha < \omega_1 \}$ (essentially it is an open cover with no unbounded sets). Setting $\alpha_0 = 0$, for each $i$ pick $U_i \in \mathcal{U}$ so that $\alpha_i \in U_i$, and let $\alpha_{i+1}$ be the least ordinal not in $\bigcup_{j \leq i} U_j$. Then $( \alpha_i )_i$ is an increasing sequence of ordinals in $\omega_1$, and so has a limit (supremum) $\alpha$. But any open neighborhood of $\alpha$ contains infinitely many $\alpha_i$, and therefore meets infinitely many $U_i$, so $\mathcal U$ could not be locally finite.

$\pi$-Base is an online encyclopedia in the spirit of Steen and Seebach's Counterexamples in Topology. According to its database, the following spaces are locally compact Hausdorff but not paracompact. (You can view the search result to learn more.)

$[0,\Omega) \times I^I$

Deleted Tychonoff Plank

Open Ordinal Space $[0,\Omega)$

Rational Sequence Topology

The Long Line

The Open Long Line

Thomas' Plank

  • $\begingroup$ That is a fantastic resource. Thank you! $\endgroup$ Dec 2, 2015 at 22:07

Just notices this was already discussed in the first posted answer, let me complement it.

Consider the space of all countable ordinals $[0,\omega_1)=\omega_1$ (this is collection-wise normal, as a linear order, but not paracompact).

One possible way to prove that $\omega_1$ is not paracompact is to use that it is countably compact (= limit point compact) but not compact, and that countably compact paracompact spaces are compact

Alternatively, you could use the Pressing-down Lemma to prove that $\omega_1$ is not even metacompact. Specifically the open cover $\{[0,\alpha):\alpha<\omega_1\}$ has no point-finite open refinement.


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