Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm looking for an example of a space which is paracompact but not metrizable. The definition of paracompactness that I'm working with is that $(X,\tau)$ is paracompact if it is Hausdorff ($T_{2}$) and for every open cover there exists a locally finite open refinement.

I'd also like to know how well paracompactness is preserved in products. Thanks in advance.

share|cite|improve this question
Compact Hausdorff spaces are certainly paracompact. See this thread for non-metrizable compact spaces. – t.b. Jan 11 '12 at 10:26
Thanks, I agree that Compact Hausdorff spaces are paracompact, and I found one good counter example from there. But to be honest, most of the examples in that thread do not convince me. E.g. $[0,\omega_{1}]$ is certainly not a Compact Hausdorff space with the order topology, but sequentially compact which is not equivalent with compactness in general. It is not even Lindelöf. Also, I fail to follow why $\{0,1\}^{\mathbb{R}}$ is not metrizable with the given argument there? But thanks alot, that link was very helpful. – T. Eskin Jan 11 '12 at 10:54
$[0,\omega_1] = \omega_1 + 1$ is certainly compact and Hausdorff (do you confuse it with $[0,\omega_1)$?). For the second example $\{0,1\}^{\mathbb{R}}$ note that it's not even first countable (but I don't understand the argument in one of the answers either). In fact, a product $X = \prod_{i \in I} X_i$ of metrizable spaces with more than two points each is metrizable if and only if $I$ is at most countable. – t.b. Jan 11 '12 at 10:57
Yeah, I did confuse it with $[0,\omega_{1})$. Thanks. – T. Eskin Jan 11 '12 at 11:06
up vote 4 down vote accepted

The Sorgenfrey line is a classical example (besides the compact examples mentioned in the thread from the comments), also because its square (the Sorgenfrey plane) is not even normal, let alone paracompact, which shows that products of even 2 relatively nice paracompact spaces can fail to be paracompact.

As a positive result, the product of a paracompact and a compact Hausdorff space is paracompact.

share|cite|improve this answer
Thanks Henno. The Sorgenfrey line seems like a good counter-example for many cases, more than I had suspected. – T. Eskin Jan 11 '12 at 11:08

$\pi$-Base is an online encyclopedia of topological spaces inspired by Steen and Seebach's Counterexamples in Topology. It lists the following paracompact, Hausdorff spaces that are not metrizable. You can learn more about any of them by visiting the search result.

Alexandroff Square

Appert Space

Arens-Fort Space

Closed Ordinal Space $[0,\Omega]$

Concentric Circles

Fortissimo Space

Helly Space


Lexicographic Ordering on the Unit Square

Radial Interval Topology

Right Half-Open Interval Topology

Single Ultrafilter Topology

Stone-Cech Compactification of the Integers

The Extended Long Line

Tychonoff Plank

Uncountable Fort Space

share|cite|improve this answer
You need to add Hausdorff, as the OP want Hausdorff to be part of the definition. – Henno Brandsma Nov 22 '14 at 21:13
And the Baire product metric seems metrisable by definition? – Henno Brandsma Nov 22 '14 at 21:27
I added Hausdorff to the search. Thanks. Regarding the Baire product space, the culprit might be that it was asserted (without proof) that this space is not perfectly normal. Since metrizable spaces are perfectly normal, the database deduced that the Baire product space is not metrizable. – Austin Mohr Nov 22 '14 at 21:41
You need some more proofs etc there – Henno Brandsma Nov 22 '14 at 21:42
But it's a nice site; still working on it? – Henno Brandsma Nov 22 '14 at 21:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.