# Example of a paracompact space that is not metrizable

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.

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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
$[0,\omega_1] = \omega_1 + 1$ 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