Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A space $X$ is called $Moscow$, if for each open subset $U$ of $X$, the closure of $U$ in $X$ is the union of a family of $G‎_{δ}$‎‎‎ ‎-subsets of $X$ .

For example, Every first countable $T_1$ space is a moscow space.

Please help me to find more examples. Thanks

share|improve this question
3  
If $X$ is Hausdorff and either perfectly $\kappa$-normal, of countable pseudocharacter, or extremally disconnected, then $X$ is a Moscow space; all of these are clear from the definitions. There’s a theorem that an arbitrary product of first countable spaces is a Moscow space, but I’ve not seen a proof of it. –  Brian M. Scott Jan 28 '13 at 6:30
    
@BrianM.Scott First, thank you for the advice. I can show that every Extremally disconnected and every perfectly $\kappa$-normal space is a moscow space by definitions, But i can not comprehend the definition of "space of countable pseudocharacter". Is it true that a space $X$ of countable pseudocharacter means every point of the space $X$ is an intersection of countably many of its neighbourhoods? –  M.Sina Jan 29 '13 at 17:33
1  
Yes: or as I would put it, every singleton is a $G_\delta$-set. Thus, every set in the space is a union of $G_\delta$-sets. –  Brian M. Scott Jan 29 '13 at 18:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.