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Today 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 $Y$?


I gave it some thought, and had some insights that I thought would be worth mentioning,

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.

Despite that, not all hope is lost - my teacher keeps mentioning the analogy to recursive sets, and he said that in that context there can be such universal set - so there might still be some hope.

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I cross posted this on MO: – Asaf Karagila May 10 '11 at 8:19
your question remind me my problem. Could you please take a look here:…? – Ilya Jun 8 '11 at 15:11
up vote 1 down vote accepted

The question was answered on MathOverflow by Clinton Conley.

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