(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.

  • $\begingroup$ The original question: math.stackexchange.com/questions/36634 $\endgroup$ – Asaf Karagila May 10 '11 at 8:19
  • $\begingroup$ It might be worth pointing out that another way to phrase your question is along the lines of "what sort of Polish topology can I put on the set of open subsets of $X$ that makes the membership relation open in the product?" Then, for Polish $X$, the answer would be that there's always at least one Polish topology that works, but it's hopeless to expect in general that you can meet any homeomorphism class with such a topology. $\endgroup$ – Clinton Conley May 11 '11 at 8:20

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.

  • $\begingroup$ And of course Juris' example is precisely what you get when you look at the Wijsman topology associated with the discrete metric on $\omega$. $\endgroup$ – Clinton Conley May 10 '11 at 11:04
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    $\begingroup$ Kechris only mentions Beer's result and cites the exact same paper. However, in proving that the Effros Borel space is standard, he embeds $X$ into some compactification $\overline{X}$ and shows that the closed subsets of $X$ are a $G_{\delta}$ in the closed subsets of $\overline{X}$. By Kuratowski's theorem then, $\operatorname{CL}(X)$ is Polish, which seems good enough for your answer. This can be found in section 12.C on page 75. $\endgroup$ – Theo Buehler May 10 '11 at 11:15
  • $\begingroup$ @Clinton: Sorry for the delayed answer. I had to catch my teacher so we could sit and go over your answer (as it was slightly over my head). Firstly, I noted that much would have been simpler had I required $Y$ perfect, regardless I was told that it is not a problem, as Polish spaces tend to have injective continuous (but not necessarily homeomorphism) functions from the Baire space, which is perfect. So indeed you have answered my question "For all $X$ there is a uniquely-universal set in the Baire space, for $\Pi^0_1$ sets of $X$". (cont...) $\endgroup$ – Asaf Karagila May 18 '11 at 20:06
  • $\begingroup$ (...cont) So did I understand that correctly? A natural question now would be whether or not this extends to the rest of the Borel hierarchy of $X$? And what would be if we also require $X$ to be perfect? Many many thanks! $\endgroup$ – Asaf Karagila May 18 '11 at 20:09

If $X =\omega$ with the discrete topology and $Y= \mathcal{P}(\omega)$ with the Cantor set topology let $G$ be the set of all $(A,n)$ such that $n\in A$.

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    $\begingroup$ I'm sorry, but I don't get the point you're making. Could you be so kind as to elaborate a little? $\endgroup$ – Theo Buehler May 10 '11 at 10:38
  • $\begingroup$ I am simply pointing out a case in which there is a universal set that provides a unique slice for each open set. In the case of $\omega$ with the discrete topology every set ia open. Each such set is also a unique element of the Cantor space providing the unique $y$ you asked for. Of course, this does not answer the general question you asked. $\endgroup$ – Juris Steprans May 10 '11 at 10:54
  • $\begingroup$ This should have been a comment rather than an answer. $\endgroup$ – Juris Steprans May 10 '11 at 10:55
  • $\begingroup$ Thanks for the clarification, that's what I suspected, but I was afraid I missed an important point towards the general answer. I didn't ask the question, I was just curious about the answer to Asaf's question. $\endgroup$ – Theo Buehler May 10 '11 at 10:59

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:

  1. If $Y$ is a locally compact zero dimensional Polish space which is not compact, then $2^\omega\times Y$ has UU.

  2. If $Y$ is Polish, then $\omega^\omega \times Y$ has UU iff $Y$ is not compact.

  3. 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?
  • $\begingroup$ That is awesome. Thanks for posting this! $\endgroup$ – Asaf Karagila Sep 23 '12 at 13:32

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