Universal sets in metric spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T06:55:41Z http://mathoverflow.net/feeds/question/64477 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64477/universal-sets-in-metric-spaces Universal sets in metric spaces Asaf Karagila 2011-05-10T08:19:15Z 2012-09-23T12:45:36Z <p>(I am cross-posting this from math.SE as it seems to be slightly over the top for that site.)</p> <p>I saw in the class the theorem:</p> <p>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:</p> <p>For all $U\subseteq X$ open, there exists $y\in Y$ such that <code>$U = \{x\mid\langle x,y\rangle\in G\}$</code>.</p> <p>$G$ with this property is called <em>universal</em>.</p> <p>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.</p> <p>My question is whether or not this $G$ can be modified such that for every $U\subseteq X$ open there is a <em>unique</em> $y\in Y$ such that <code>$U = \{x\mid\langle x,y\rangle\in G\}$</code>? Perhaps we need to require more, or possibly even less, from $X$ and $Y$?</p> <p>Some thoughts:</p> <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/64477/universal-sets-in-metric-spaces/64487#64487 Answer by Juris Steprans for Universal sets in metric spaces Juris Steprans 2011-05-10T10:28:47Z 2011-05-10T10:28:47Z <p>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 $n\in A$.</p> http://mathoverflow.net/questions/64477/universal-sets-in-metric-spaces/64488#64488 Answer by Clinton Conley for Universal sets in metric spaces Clinton Conley 2011-05-10T10:58:34Z 2011-05-10T13:03:24Z <p>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.</p> <p>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 <code>$\{(x,A) \in X \times \mathcal{P}(X) : x \in A\}$</code> is "uniquely" universal closed.</p> <p>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 <em>Vietoris topology</em>, generated by sets of the form</p> <p><code>$\{K : K \subseteq U\}$</code> and <code>$\{K : K \cap U = \emptyset\}$</code>,</p> <p>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 <code>$G = \{(x,K) \in X \times \mathcal{K}(X) : x \in K\}$</code>. 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 <code>$U \times \{K : K \cap U = \emptyset\}$</code> is an open neighborhood of $(x_0, K_0)$ disjoint from $G$.</p> <p>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.</p> <p>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 <em>Wijsman topology</em>. 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.)</p> <p>Beer, Gerald. A Polish topology for the closed subsets of a Polish space. Proc. Amer. Math. Soc. 113 (1991), no. 4, 1123–1133. </p> <p><em>Edit: but Theo Buehler has given a relevant reference to Kechris. See his comment.</em></p> <p>A variation of the earlier argument in the compact case should work in this context.</p> <p><em>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.</em></p> <p>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.</p> http://mathoverflow.net/questions/64477/universal-sets-in-metric-spaces/107900#107900 Answer by Theo Buehler for Universal sets in metric spaces Theo Buehler 2012-09-23T12:45:36Z 2012-09-23T12:45:36Z <p>While idly browsing around I stumbled over the follwing paper and remembered this question:</p> <p>A.W. Miller, <em><a href="http://www.math.wisc.edu/~miller/res/hrus.pdf" rel="nofollow">Uniquely Universal Sets</a></em>, Topology and its Applications&nbsp;<strong>159</strong> (2012), pp. 3033&ndash;3041. It's available in various formats <a href="http://www.math.wisc.edu/~miller/res/index.html" rel="nofollow">here</a>.</p> <p>Let me quote the abstract (to avoid confusion: Miller's terminology reverses the r&ocirc;les of $X$ and $Y$ in your question):</p> <blockquote> <p>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 Hru&#353;&aacute;k 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:</p> <ol> <li><p>If $Y$ is a locally compact zero dimensional Polish space which is not compact, then $2^\omega\times Y$ has UU.</p></li> <li><p>If $Y$ is Polish, then $\omega^\omega \times Y$ has UU iff $Y$ is not compact.</p></li> <li><p>If $Y$ is a $\sigma$-compact subset of a Polish space which is not compact, then $\omega^\omega \times Y$ has UU.</p></li> </ol> </blockquote> <p>His results are mostly positive: &ldquo;a certain space or family of spaces has UU&rdquo; and various permanence properties. One nice &ldquo;negative&rdquo; result:</p> <blockquote> <p>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.</p> </blockquote> <p>He also raises a few questions, e.g.:</p> <ul> <li>Question&nbsp;4: Does $(2^\omega\oplus 1) \times [0,1]$ have UU?</li> <li>Question&nbsp;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?</li> <li>Question&nbsp;11: Is the converse of Corollary&nbsp;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?</li> </ul>