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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.

  1. How can we show that every compact space is a continuous image of a compact moscow space?
  2. 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 $X$ .

Asaf Karagila
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TXC
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    I did hear about Polish spaces, but never heard about a Moscow space. Interesting :) – SBF Mar 20 '13 at 10:59
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    In the standard topology on $\Bbb{R}$ (and thus, in any subspace with the induced topology), a single point is a $G_\delta$ set. Therefore, any set is a union of a family of $G_\delta$ subsets (since any set is a union of all of its singleton subsets); in particular, any subset is Moscow. However, not every compact space is metric so this isn't enough to get your second conclusion. – Yoni Rozenshein Mar 20 '13 at 11:14
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    Somewhat more general than @YoniRozenshein's comment: in a first-countable Hausdorff (even T$1$) space all singletons are G$\delta$, and thus all subsets are unions of G$_\delta$ sets. – user642796 Mar 20 '13 at 11:52

1 Answers1

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Here's one way to answer your questions:

  1. 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.

  2. 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.

Martin
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