Every compact space is a continuous image of a compact Moscow space. From WikiPedia,

every compact metric space is a continuous image of the Cantor set.

and in Topological Groups and Related Structures Ex 6.3.a.

every compact space is a continuous image of a compact moscow space.



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*How can we show that every compact space is a continuous image of a compact moscow space? 

*Cantor set is a Moscow space?


Thanks.

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 $X$ .
 A: Here's one way to answer your questions:


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*A space is said to be extremely disconnected if the closure of every open set is open. Extremely disconnected spaces are Moscow spaces. The Čech-Stone compactification $\beta D$ of a discrete space $D$ is extremely disconnected and hence a compact Moscow space. Now let $X$ be compact and let $X_\delta$ be the set $X$ endowed with the discrete topology. Then the identity mapping $X_\delta \to X$ is continuous and surjective and by the universal property of the Čech-Stone compactification it extends to a continuous and surjective function $\beta X_\delta \to X$. This shows that every compact space $X$ is (canonically) the continuous image of a compact Moscow space.

*As an alternative to the arguments suggested in the comments, observe that in a metric space closed sets are $G_\delta$-sets, so the closure of an open set is a $G_\delta$-set and hence metric spaces are Moscow spaces. Since the Cantor set is a metric space, it is a Moscow space.
