11
$\begingroup$

(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.

$\endgroup$
2
  • $\begingroup$ The original question: math.stackexchange.com/questions/36634 $\endgroup$
    – Asaf Karagila
    May 10, 2011 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. $\endgroup$ May 11, 2011 at 8:20

3 Answers 3

10
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ And of course Juris' example is precisely what you get when you look at the Wijsman topology associated with the discrete metric on $\omega$. $\endgroup$ May 10, 2011 at 11:04
  • 1
    $\begingroup$ 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. $\endgroup$ May 10, 2011 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, 2011 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, 2011 at 20:09
4
$\begingroup$

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 $n\in A$.

$\endgroup$
4
  • 1
    $\begingroup$ I'm sorry, but I don't get the point you're making. Could you be so kind as to elaborate a little? $\endgroup$ May 10, 2011 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. $\endgroup$ May 10, 2011 at 10:54
  • $\begingroup$ This should have been a comment rather than an answer. $\endgroup$ May 10, 2011 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. $\endgroup$ May 10, 2011 at 10:59
4
$\begingroup$

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?
$\endgroup$
1
  • $\begingroup$ That is awesome. Thanks for posting this! $\endgroup$
    – Asaf Karagila
    Sep 23, 2012 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.