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As per a recent question of mine, $\omega_1 \times \beta \mathbb{N}$ is not normal. I'm wondering whether there's some sort of "natural" condition that describes when a space has a normal product with $\beta \mathbb{N}$, analagous to Dowker's characterisation that $X$ is countably paracompact iff $X \times [0, 1]$ is normal.

The context here is that I'm looking for something I can weaken to something sensible, as I have a property implied by $X \times \beta \mathbb{N}$ being normal which I am "surprised" isn't an equivalence ($\omega_1$ has this property) and would like to see if I can show its equivalenct to some slightly weaker topological condition.

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Fixed title to remove LaTeX –  Harry Gindi Apr 8 '10 at 10:30
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I don't follow. How is that a fix? LaTeX works fine in the titles. –  David R. MacIver Apr 8 '10 at 14:39
    
@David: in the past it hasn't for me, but that could be a browser issue –  Yemon Choi Apr 8 '10 at 16:26
    
Strange. I see e.g. the title of the linked post about $\omega_1$ working fine as LaTeX. –  David R. MacIver Apr 8 '10 at 16:31
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1 Answer

up vote 6 down vote accepted

The space $X\times\beta\mathbb{N}$ is normal if and only if $X$ is normal and $\mathfrak{c}$-paracompact. This follows from results of Morita (Paracompactness and product spaces, MR132525), where he generalizes Dowker's characterization of countable paracompactness.

First note that $X\times\beta\mathbb{N}$ is normal if and only if $X\times K$ is normal for every separable compact Hausdorff space $K$. This is because every separable compact Hausdorff space is a perfect image of $\beta\mathbb{N}$.

Morita's Theorem 2.2 shows that if $X$ is normal and $\mathfrak{c}$-paracompact, then $X \times K$ is normal for every compact Hausdorff space $K$ of weight at most $\mathfrak{c}$. Hence, $X\times K$ is normal for every separable compact Hausdorff space $K$ since these all have weight at most $\mathfrak{c}$.

Morita's Theorem 2.4 shows that a space $X$ is normal and $\mathfrak{c}$-paracompact if (and only if) $X\times[0,1]^{\mathfrak{c}}$ is normal. Since the space $[0,1]^{\mathfrak{c}}$ is a separable compact Hausdorff space, this closes the implication loop.

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(Note: $\mathfrak{c} = 2^{\aleph_0}$.) –  François G. Dorais Apr 11 '10 at 19:53
    
Lovely. Thanks. –  David R. MacIver Apr 11 '10 at 20:33
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