(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.



Universal sets in metric spaces
The possibly unsatisfying answer to your question is "sometimes." I will instead discuss the obviously equivalent question about universal closed subsets (it will let me use more standard notation later). Moreover, I will focus on the special case that $X$ and $Y$ are both Polish, since that has been examined more in the literature.


First, let me point out an oversight in your analysis of the case that $X$ is finite. Certainly $X$ must have the discrete topology, so every subset of $X$ is closed. However, $Y = \mathcal{P}(X)$ is a perfectly fine Polish space when endowed with its own discrete topology. Then the set $\{(x,A) \in X \times \mathcal{P}(X) : x \in A\}$ is "uniquely" universal closed.


This may seem pedantic, but it actually generalizes to large $X$. Suppose now that $X$ is a compact Polish space, and endow its space of compact (equiv., closed) subsets $\mathcal{K}(X)$ with the *Vietoris topology*, generated by sets of the form


$\{K : K \subseteq U\}$ and $\{K : K \cap U = \emptyset\}$,


where $U\subseteq X$ is open. For Polish $X$, this is a Polish topology on $\mathcal{K}(X)$. Note that in the special case where $X$ is finite (thus compact), this coincides with the discrete topology on $\mathcal{P}(X)$. Motivated by this analogy, we proceed as before and choose our uniquely universal closed set to equal $G = \{(x,K) \in X \times \mathcal{K}(X) : x \in K\}$. The only thing left to check is that this set is indeed closed. You can see this directly by assuming $(x\_0,K\_0) \notin G$, fixing a little open neighborhood $U$ around $x\_0$ disjoint from $K\_0$, and then checking that $U \times \{K : K \cap U = \emptyset\}$ is an open neighborhood of $(x\_0, K\_0)$ disjoint from $G$.


The obvious place to look for more information about this is Kechris' descriptive set theory text. Unfortunately I don't have a copy on hand at the moment (which makes me feel like a child without a security blanket), so I can't give more specific references.


Moving on. For noncompact Polish spaces $X$ you can endow the space $\mathrm{CL}(X)$ of closed subsets of $X$ with a topology called the *Wijsman topology*. Well, really there are several such topologies, since the definition relies on a choice of compatible complete metric $d$ on $X$. This topology is the weakest topology making the functions $f\_x : A \mapsto d(x, A)$ continuous for each $x \in X$. It is a result of Gerald Beer's that this topology is Polish for $(X,d)$ as above. (This might well be in Kechris' book, but as I mentioned I don't have it on hand so I'll regurgitate the reference that google gave me.)


Beer, Gerald. A Polish topology for the closed subsets of a Polish space. Proc. Amer. Math. Soc. 113 (1991), no. 4, 1123–1133. 


*Edit: but Theo Buehler has given a relevant reference to Kechris. See his comment.*


A variation of the earlier argument in the compact case should work in this context.


*Edit again: I just noticed that the definition of the topology I gave makes sense for nonempty closed subsets of $X$. This is not a serious problem and is in fact addressed in Beer's paper.*


Finally, it is hopeless to expect this to work for arbitrary Polish spaces $X$ and $Y$. As you noticed, for small spaces there are cardinality issues. When the spaces are large, you can also fiddle around with compactness/noncompactness, and other topological notions. There are just too many wild Polish spaces.


