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A space $X$ is called perfectly $\kappa$-normal if the closure of any open set (that is, every canonical closed set) is a zero-set.

How can i prove this proposition directly?

$Proposition$: A Cartesian product of metric spaces is perfectly $\kappa$-normal.

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    Do you mean "is the zero set of some continuous function"?2012-11-21
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    @Rhys: Yes. Such spaces have also been called Oz-spaces (or $O_z$-spaces).2012-11-21
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    I’ve never seen Ščepin’s proof of the result, and I don’t see a proof at the moment. However, I did find [this paper](http://projecteuclid.org/euclid.jmsj/1230395789), which contains a proof of a more general result; see Theorem $2$.2012-11-21
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    This is equivalent to proving that every metric space is perfectly $\kappa$-normal, right? Having seen Brian's comment, I guess it isn't this easy, but my approach would have been, given some open set $U \subset (X,d_X)$, to define $f_U : X \to \mathbb{R}^+ ~,~ f(x) := \text{inf}\{d_X(x,y) ~|~ y \in U\}$. Then show that $f$ is continuous...2012-11-21
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    @Rhys: No, it isn’t: products of uncountably many non-trivial metric spaces aren’t metrizable, and the theorem is for arbitrary products. It **is** fairly trivial that metric spaces are $\kappa$-normal.2012-11-21
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    Thanks Brian; I see. This would be a much better question if it included some of these details.2012-11-21
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    @BrianM.Scott: Ščepin introduced the notions of $\kappa$-metrizability and proved that $\kappa$-metrizability is productive [ On $\kappa$-metrizable space, MAth. USSR-Izv. 14 (1980), 407-440 ]. Since every metrizable space is $\kappa$-metrizable and every $\kappa$-metrizable space is perfectly $\kappa$-normal then A cartesian product of metric spaces is perfectly $\kappa$-normal.2012-11-26

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