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?



Does the class of Hausdorff spaces have a shared "Coordinate space"?
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 $f: \kappa+1 \rightarrow C$.


