Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
1  
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. –  Thomas E. Jan 11 '12 at 10:54
3  
$[0,\omega_1] = \omega_1 + 1$ 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. –  Thomas E. Jan 11 '12 at 11:06

1 Answer 1

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|improve this answer
    
Thanks Henno. The Sorgenfrey line seems like a good counter-example for many cases, more than I had suspected. –  Thomas E. Jan 11 '12 at 11:08

Your Answer

 
discard

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.