TITLE: Universal sets in metric spaces
QUESTION [11 upvotes]: (I am cross-posting this from math.SE as it seems to be slightly over the top for that site.)
I saw in the class the theorem:
Suppose $X$ is a separable metric space, and $Y$ is a polish space (metric, separable and complete) then there exists a $G\subseteq X\times Y$ which is open and has the property:
For all $U\subseteq X$ open, there exists $y\in Y$ such that $U = \{x\mid\langle x,y\rangle\in G\}$.
$G$ with this property is called universal.
The proof is relatively simple, however the $y$ we have from it is far from unique, in fact it seems that it is almost immediate that there are countably many $y$'s with this property.
My question is whether or not this $G$ can be modified such that for every $U\subseteq X$ open there is a unique $y\in Y$ such that $U = \{x\mid\langle x,y\rangle\in G\}$?
Perhaps we need to require more, or possibly even less, from $X$ and $Y$?
Some thoughts:
Firstly $X$ cannot be finite, otherwise there are less than continuum many open subsets, and since $G$ is open we have that the projection on $Y$ is open, since $Y$ is Polish we have that this projection is of cardinality continuum, which in turn implies there are continuum many $y$'s with the same cut.
Secondly, as the usual proof goes through a Lusin scheme over $Y$, and using it to define $G$, I thought at first that using the axiom of choice we can select a set of points on which the mapping to open sets of $X$ is 1-1, and somehow remove some of the sets from the scheme. This proved to be a bad idea, as we remove sets that can be used for other open sets.
Thirdly, I thought about enumerating the open sets according to a rational enumeration so $A_i\subseteq A_j$ if and only if $q_i\le q_j$, and then instead of just placing the open sets of $X$ arbitrarily by the Lusin scheme, we use the rationals somehow.

REPLY [4 votes]: While idly browsing around I stumbled over the follwing paper and remembered this question:
A.W. Miller, Uniquely Universal Sets, Topology and its Applications 159 (2012), pp. 3033–3041. It's available in various formats here.
Let me quote the abstract (to avoid confusion: Miller's terminology reverses the rôles of $X$ and $Y$ in your question):

We say that $X \times Y$ satisfies the Uniquely Universal property (UU)
  iff there exists an open set $U \subseteq X \times Y$ such that for
  every open set $W \subseteq Y$ there is a unique cross section of
  $U$ with $U_x=W$.  Michael Hrušák raised the question of when does
  $X \times Y$ satisfy UU and noted that if $Y$ is compact, then $X$
  must have an isolated point.  We consider the problem when
  the parameter space $X$ is either the Cantor space $2^\omega$ or the Baire
  space $\omega^\omega$.
  We prove the following:

If $Y$ is a locally compact zero dimensional
  Polish space which is not compact, then $2^\omega\times Y$ has UU.
If $Y$ is Polish, then
  $\omega^\omega \times Y$ has UU iff $Y$ is not compact.
If $Y$ is a $\sigma$-compact subset of a Polish space which is
  not compact, then $\omega^\omega \times Y$ has UU.


His results are mostly positive: “a certain space or family of spaces has UU” and various permanence properties. One nice “negative” result:

Proposition 30: There exists a partition $X\cup Y=2^\omega$ into Bernstein sets
  $X$ and $Y$ such that for every Polish space $Z$ neither
  $Z\times X$ nor $Z\times Y$ has UU.

He also raises a few questions, e.g.:

Question 4: Does $(2^\omega\oplus 1) \times [0,1]$ have UU?
Question 6: Does either $\mathbb{R} \times \omega$
or $[0,1]\times \omega$ have UU?
Or more generally, is there any example of UU for a
connected parameter space?
Question 11: Is the converse of Corollary 10 false?
That is: Does there exist $Y$ such that $\omega^\omega \times Y$ has UU but
$2^\omega\times Y$ does not have UU?