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Every Tychonoff space is an image of a moscow space under a continuous open mapping.

A space $X$ is called Moscow if the closure of every open set $U\subset X$ is the union of a family of $G_\delta$-subsets of $X$.

Reference: A. Arhangel'skii and M. Tkachenko, Topological Groups and Related Structures, p358, Ex 6.3.b .

I've tried to answer the question but I did not succeed. For example, For Tychonoff space $X$, if we can construct a space $Y$ such that $Y\times X$ is extremally disconnected (Or Moscow) then $\pi:X\times Y\to X$ is continuous open mapping but It is impossible .

Thanks for any help and advice.

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I’ve not solved the problem, but I can say that it can’t be solved by constructing a suitable space $Y$ and looking at $\pi:X\times Y\to X$: if a product is Moscow, so is each factor. – Brian M. Scott Apr 7 '13 at 21:17
This is Exercise 6.3.b from the book “Topological groups and related structures” by Arhangel'skii and Tkachenko. Thus it should not be very hard. :-) – Alex Ravsky Apr 12 '13 at 19:36
@AlexRavsky , Yes This question is clearly obvious, but only for teachers :-). – TXC Apr 13 '13 at 3:46
I'll try to test it at my teacher, who is a pupil of Arhangel'skii. :-) – Alex Ravsky Apr 16 '13 at 15:42
up vote 5 down vote accepted

This is a story.

I did not read the book by Arhangel'skii and Tkachenko, I only glanced at some pages of it related with my investigations. Since I found that even some of “open problems” related to my investigations are easily solvable or already solved, I decided that the exercises should not be very hard. :-) But I am not a specialist in Moscow spaces. :-( Even more, maybe I firstly read about them from the same book. :-) So, maybe there is a trick in the solution of this exercise.

Searching the answer to the you question, I looked in Bible of general topologists – “General topology” [Eng] by Ryszard Engelking, and found unexpectedly few helpful claims. In fact, there was only a remark in the Exercise 4.2.D in my (Russian) edition, claiming that Isbell in [Isb] proved that each topological space is an open image of a hereditarily paracompact and strongly zero-dimensional space. But we need somewhat stronger result, because, by Theorem 6.2.25 from [Eng], we have that each nonempty extremally disconnected Tychonoff space in strongly zero-dimensional, but the opposite inclusion is not true (the space of rational numbers is strongly zero-dimensional but not extremally disconnected).

So I tried to look at Isbell’s paper [Isb]. And I successfully found it here, at Springer. But this day was not mine, :-) because article’s free preview ends exactly at the searched theorem, :-) and idea of paying for such the things is completely alien to my mentality. :-)

So I had to find other ways. Unexpectedly, I found few results in the Internet too.

According to [Haz, p.524, Tychonoff spaces are perfect images of extremally disconnected spaces. But this should not work, because, accordingly to [Eng], an open image of an of extremally disconnected space is extremally disconnected too.

From the other side, it seems that the original problem can be reduced to compact spaces. Indeed, let $X$ be a Tychonoff space. There exists a compact space $bX$ such that $X$ is a dense subset of $bX$. If $f:Y\to bX$ is a continuous open map from a Moscow space $Y$ onto $bX$, then $f^{-1}(X)$ is dense in $Y$. Since a dense suspace of a Moscow space is Moscow [AT, Prop. 6.1.2], we see that $f^{-1}(X)$ is a Moscow space. But, by the above remark, $f^{-1}(X)$ cannot be extremally disconnected. In particular, Martin’s construction of the map from $\beta X$ showing that Every compact space is a continuous image of a compact Moscow space. fails for the present proof. Moreover, if $f$ is an open continuous map of a Moscow space $Y$ onto a space $X$ such that $f^{-1}(x)$ is compact for each $x\in X,$ then $X$ is also a Moscow space [AT, Th. 6.3.1]. So, maybe for the construction of the desired space $Y$ we should “spread vertically” the space $X$ in the product $X\times Z$ for some space $Z$.

At last I got it. I found the theorem which should imply the positive answer.

A (Hausdorff) space is called strictly $\sigma$-discrete, if it is a union of countably many of its closed discrete subspaces. It is easy to show that each point of a strictly $\sigma$-discrete space is a $G_\delta$-set. Therefore each strictly $\sigma$-discrete space is Moscow. According to [Tka], Junilla (who also was mentioned in the Engelking’s remark) showed in [Jun] that every topological space is an open image of a strictly $\sigma$-discrete Tychonoff space. $\square$

Update. I have just found an other reference in [Tka’]: “Every topological space $X$ can be represented as an open continuous image of a completely regular submetrizable space $Y$ (in other words, $Y$ admits a continuous one-to-one mapping onto a metrizable space) – the corresponding construction is given on p. 331 of [Eng2]”.


[AT] A.V. Arhangel'skii, M. Tkachenko, Topological groups and related structures, Atlantis Press, Paris; World Sci. Publ., NJ, 2008.

[Eng] R. Engelking. General Topology. -- M.: Mir, 1986 (in Russian).

[Eng2] R. Engelking. General Topology. -- Heldermann Verlag, Berlin, 1989.

[Haz] M. Hazewinkel (ed.), Encyclopaedia of Mathematics (9).

[Isb] J.R. Isbell. A note on complete closure algebras. -- Math. Systems. Theory 3 (1969), 310-312.

[Jun] H.J.K. Junnila, Stratifiable preimages of topological spaces, Colloq. Math. Soc. J. Bolyai 23. Topology, Budapest, 1978, 689--703. MR **81m:*54019

[Tka’] M. G. Tkachenko. Topological groups for topologists: part I, Bol. Soc. Mat. Mexicana (3), 5, 1999, 237-279.

[Tka] V.V. Tkachuk. When do connected spaces have nice connected preimages? -- Proc. Amer. Math. Soc, 125:11 (November 1998), 3437--3446.

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A Great happy ending story :-), Thank you for trying to solve this question. Do you know how to get this ( Stratiable preimages of topological spaces ) paper? – M.Sina Apr 17 '13 at 6:50
Oops, this was “stratifiable”, no “stratiable” (the joined letter “fi” was dropped while the pasting from the pdf-file). :-) Unfortunately, I have no access to this paper. Maybe my colleague, who has a huge mathematical electronic library, has the article. I’ll ask him about it. But the hope is weak, because this journal is not in the list of his library’s journals, and, moreover, I must wait some weeks before my colleague return to our city and his library. :-( – Alex Ravsky Apr 17 '13 at 7:34
@M.Sina I asked my colleague about Junnila's article. Unfortunately, he doesn't have it. – Alex Ravsky May 27 '13 at 3:51
@AlexRavsky: Thank you very much for following up on this subject. – M.Sina May 27 '13 at 14:06
@TXC I have no files of these articles. If I remember it right, I have obtained them from my teacher. Unfortunately, now he is on vacations. When he will return, I'll ask him about the files. If he will not have them then I'll scan them and upload. – Alex Ravsky Jul 16 '13 at 3:08

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