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Given topological spaces $X$ and $C$ we call $C$ a coordinate space for $X$ to mean that every open set $U \subset X$ is of the form $f^{-1}(V)$ for some open $V \subset C$ and continuous $f \colon X \to C$.

More generally if $\mathscr X$ is a class of spaces we call $C$ a coordinate space for $\mathscr X$ to mean it is a coordinate space for every member of $\mathscr X$.

Two familiar examples of coordinate spaces:

  1. The Seirpinski space $\{0,1\}$ with exactly one closed point and one open point is a coordinate space for the class of all topological spaces.

  2. The closed unit interval is a coordinate space for the class of all completely-regular spaces. Moreover the closed unit interval is itself completely-regular.

But what about the class of all Hausdorff spaces? Obviously the Seirpinski space is still a coordinate space here. But what if we demand the coordinate space itself be Hausdorff -- as with example 2?

I would reckon such a space does not exist but have no idea how to prove that?

Does this notion already have a name and has the question been addressed before?

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    $\begingroup$ For starters, I wonder if there is a Hausdorff coordinate space for the class of all ordinals. $\endgroup$ – Nate Eldredge Jan 23 '17 at 16:51
  • $\begingroup$ There was a related classical result in the past by Stanisław Mrówka, celebrated in the P.S.Aleksandrov's surveys on general topology. $\endgroup$ – Włodzimierz Holsztyński Jan 23 '17 at 20:25
  • $\begingroup$ A pedantic remark: it is Wacław Sierpiński, not Seirpinski ("ie", not "ei"). You may apply Polish diacritics by "cut & paste" from any non-sloppy article, say, from Wikipedia. BTW, the Polish combination "ie" is pronounced on its own (as a vowel) just like "e" alone but it affects the preceding "S" turning it into a soft " S' ". $\endgroup$ – Włodzimierz Holsztyński Jan 23 '17 at 20:50
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First of all, let me point out that $[0,1]$ is not a coordinate space for the class of completely regular spaces.

The definition of completely regular spaces says that for any closed $C \subset X$ and any point $p \notin C$, there is a continuous $[0,1]$-valued function (say $f$) mapping $p$ to $1$ and mapping every point in $C$ to $0$. It does not say that $f^{-1}(0,1] = X-C$. In other words, $f$ does map $C$ to $0$, but it may also map some other points to $0$ as well.

For example, it can be shown that any continuous function $f: \omega_1 \rightarrow [0,1]$ is eventually constant. It follows that $[0,1]$ is not a coordinate space for $\omega_1+1$: the open set $\omega_1$ is never equal to $f^{-1}(U)$ for an open $U \subseteq [0,1]$ and continuous $f: \omega_1+1 \rightarrow [0,1]$, because $f$ maps a tail of $\omega_1$ to a single point $c$, and must (by continuity) map a tail of $\omega_1+1$ to $c$. So if $f^{-1}(U)$ contains $\omega_1$, it contains $\omega_1+1$.

A similar argument shows

There is no Hausdorff coordinate space for the class of Hausdorff spaces.

To prove this, suppose $C$ is such a space. Let $\kappa$ be a regular cardinal bigger than $|C|$. Any function $\kappa \rightarrow C$ must map an unbounded subset of $\kappa$ to a single point (there's no topology involved in this statement -- it's just because $\kappa$ is regular and bigger than $|C|$). Now consider the ordinal space $\kappa+1$. As above, the open set $\kappa$ is never equal to $f^{-1}(U)$ for an open $U \subseteq C$ and continuous $f: \kappa+1 \rightarrow C$.

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A classical notion was considered and solved in an elegant paper by Stanisław Mrówka: (1) instead of arbitrary open subspaces of $\ X\ $ one simply considers just X itself; and (2) instead of a single space $\ C,\ $ one considered the class of all powers $\ C^S.\ $ Then Stanisław Mrówka answered that this kind of embeddability problem is characterized by the property of separating points from closed subsets of $\ X\ $ by functions into $\ C^F,\ $ where this time the exponents $\ f\ $ are just finite sets--a very elegant conceptual theorem.

Acknowledgment--thank you, Will Brian, for pointing to my earlier error.

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    $\begingroup$ It seems to me that you are overstating your case in your first paragraph. For example, the class of discrete spaces contains arbitrarily large spaces, but it has the discrete two-point space as a coordinate space (in the sense of the OP). A little less trivially, the class of perfectly normal spaces has $[0,1]$ for a coordinate space. (Recall: "perfectly normal" means normal and all closed sets are $G_\delta$, and this allows you to construct, for any closed $C$, a $[0,1]$-valued function $f$ with $f(x) = 0$ if and only if $x \in C$.) $\endgroup$ – Will Brian Jan 23 '17 at 21:13
  • $\begingroup$ W.B., I did a quick fix; I may write more perhaps. $\endgroup$ – Włodzimierz Holsztyński Jan 23 '17 at 21:25
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    $\begingroup$ Yes, this looks correct now -- and thanks for pointing out this result, which I was not aware of, but I agree that it is elegant. $\endgroup$ – Will Brian Jan 23 '17 at 21:26
  • $\begingroup$ Will Brian, I have misread OP statement. I somehow saw "homeomorphism" which was absent from the original author's formulation. I'll edit my answer. Thank you. $\endgroup$ – Włodzimierz Holsztyński Jan 23 '17 at 22:43

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