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What is a counterexample to show that the class of Moscow spaces is not closed hereditary?

(A space $X$ is called Moscow if the closure of every open $U \subseteq X$ is the union of a family of G$‎_{‎\delta‎‎‎}$‎-subsets of $X$.)

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This is essentially copied from A. V. Arhangel'skii, Moscow spaces and topological groups, Top. Proc., 25; pp.383-416:

Let $D ( \tau )$ be an uncountable discrete space, and $\alpha D ( \tau )$ the one point compactification of $D ( \tau )$. Then $D ( \tau )$ is a Moscow space, and $D ( \tau )$ is G$_\delta$-dense in $\alpha D ( \tau )$, while $\alpha D ( \tau )$ is not a Moscow space. Indeed, let $U$ be any infinite countable subset of $D ( \tau )$. Then $U$ is open in $\alpha D ( \tau )$, and $\overline{U} = U \cup \{ \alpha \}$, where $\alpha$ is the only non-isolated point in $\alpha D ( \tau )$. Every G$_\delta$-subset of $\alpha D ( \tau )$ containing the point $\alpha$ is easily seen to be uncountable; therefore, $\overline{U}$ is not the union of any family of G$_\delta$-subsets of $\alpha D ( \tau )$. Since $\alpha D ( \tau )$ is a closed subspace of a Tychonoff cube, we conclude that the class of Moscow spaces is not closed hereditary.

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$1.$ Why $\alpha D ( \tau )$ is a closed subspace of tychonoff cube? $2.$Tychonoff cube is a moscow space? –  M.Sina Nov 20 '12 at 19:33
    
@MohammaDSina: The Tychonoff cube $[0,1]^\kappa$ is universal for Tychonoff spaces of weight $\kappa$, and so $\alpha D ( \tau )$ is embeddable in an appropriate $[0,1]^\kappa$. Since $\alpha D ( \tau )$ is compact, the image must be closed. –  Arthur Fischer Nov 20 '12 at 19:40
    
@MohammaDSina: (Sorry, missed your second question.) If you look at the paper I referenced/linked to, you will notice that arbitrary products of metric spaces are Moscow (this is Corollary 1.2 on p.386). –  Arthur Fischer Nov 20 '12 at 19:46
    
Thank you and Best wishes;) –  M.Sina Nov 20 '12 at 19:55

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