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Let $X$ be a Tychonoff space. How can we construct a space $Y$ such that $Y\times X$ is extremally disconnected (Or Moscow) ?

TXC
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    I don't think this is possible in general. If $X$ is not extremally disconnected, then there is an open $U \subseteq X$ such that $\overline{U}$ is not open. Then for any space $Y$ the set $Y \times U \subseteq Y \times X$ is open, and $\overline{Y \times U} = Y \times \overline{U}$ is not open in the product. Thus $Y \times X$ is not extremally disconnected. – user642796 Mar 28 '13 at 14:08
  • @ArthurFischer , Perhaps this problem could be avoided by adding conditions. I tried to answer the question when $Y\times X$ is a Moscow space, But I did not succeed. – TXC Mar 28 '13 at 16:50
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    Since retracts of Moscow spaces are themselves Moscow spaces and since $X$ is a retract of $X \times Y$ it follows that there can't be a space $Y$ such that $X \times Y$ is a Moscow space unless $X$ is already a Moscow space. // What is the motivation of this question? – Martin Apr 03 '13 at 13:34
  • @Martin : First, I am trying to solve this Exercise 6.3.b. – TXC Apr 04 '13 at 05:10
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    Also Every retract of extremally disconnected space is a extremally disconected space. – M.Sina Apr 04 '13 at 07:16

1 Answers1

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As stated in the comments, It is impossible. Since Every retract of (Moscow) extremally disconnected space is a (Moscow) extremally disconected space.

TXC
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