(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.