% LaTex2e % Uniquely Universal Sets % A. Miller May 2011 \def\lastrevised{January 12, 2012.} \documentclass[12pt]{article} \usepackage{amssymb,latexsym} \pagestyle{myheadings} \def\header{Uniquely Universal Sets \hfill} %\def\header{\today} \markboth\header\header \def\cutcrap{\addressarn\end{document}} %\def\cutcrap{} \def\al{\alpha} \def\bb{{\mathcal B}} \def\be{\beta} \def\bspc{{\om^\om}} \def\calc{{\mathcal C}} \def\caln{{\mathcal N}} \def\cc{{\mathcal C}} \def\cl(#1){{ cl}(#1)} \def\concat(#1){\hat{\phantom{a}}\la #1\ra} \def\cont{{\mathfrak c}} \def\cspc{{2^\om}} \def\de{\delta} \def\dom(#1){dom(#1)} \def\eps{\epsilon} \def\ga{\gamma} \def\gg{{\mathcal G}} \def\hrusak{Hru\v{s}\'ak } \def\ii{{\mathcal I}} \def\isolated{\oplus 1} \def\ka{\kappa} \def\la{\langle} \def\nicetree{{\rm NT}} \def\om{\omega} \def\pair(#1,#2){\la #1, #2 \ra} \def\pf{\par\bigskip\noindent proof:\par} \def\poset{{\mathbb P}} \def\power{{\mathcal P}} \def\pp{\mathfrak p} \def\proof{\pf} \def\pr{\prime} \def\qed{\par\noindent QED\par\bigskip} \def\qq{{\mathbb Q}} \def\ra{\rangle} \def\res{{\upharpoonright}} \def\rmand{\mbox{ and }} \def\rmforall{\mbox{ for all }} \def\rmiff{\mbox{ iff }} \def\rmif{\mbox{ if }} \def\rmor{\mbox{ or }} \def\root{{\la\ra}} \def\rr{{\mathbb R}} \def\si{\sigma} \def\sm{\setminus} \def\st{\;:\;} % such that \def\su{\subseteq} \def\symdiff{{\Delta}} \def\uu{{\mathcal U}} \def\zerovec{{\vec{0}}} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{define}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{prop}[theorem]{Proposition} \newtheorem{ques}[theorem]{Question} \newtheorem{cor}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \newenvironment{enum} {\begin{list}{(\alph{enumi})} {\usecounter{enumi}\setlength{\rightmargin}{\leftmargin}}} {\end{list}} \def\addressarn{\begin{flushleft} Arnold W. Miller \\ miller@math.wisc.edu \\ http://www.math.wisc.edu/$\sim$miller\\ University of Wisconsin-Madison \\ Department of Mathematics, Van Vleck Hall \\ 480 Lincoln Drive \\ Madison, Wisconsin 53706-1388 \\ \end{flushleft}} \begin{document} \begin{center} {\large Uniquely Universal Sets} \end{center} \begin{flushright} Arnold W. Miller\\ \end{flushright} \begin{center} Abstract\footnote{ MSC2010: 03E15 \par Keywords: Universal set, Unique parameterization, Polish spaces, Cantor space, Baire space. \par Last revised \lastrevised} \end{center} \begin{quote} 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\u{s}\'ak 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: \par 1. If $Y$ is a locally compact zero dimensional Polish space which is not compact, then $2^\omega\times Y$ has UU. \par 2. If $Y$ is Polish, then $\omega^\omega \times Y$ has UU iff $Y$ is not compact. \par 3. If Y is a $\sigma$-compact subset of a Polish space which is not compact, then $\omega^\omega \times Y$ has UU. \end{quote} For any space $Y$ with a countable basis there exists an open set $U\su 2^\om\times Y$ which is universal for open subsets of $Y$, i.e., $W\su Y$ is open iff there exists $x\in 2^\om$ with $$U_x=^{{\rm def}}\{y\in Y\st (x,y)\in U\}=W.$$ To see this let $\{B_n\st n<\om\}$ be a basis for $Y$. Define $$(x,y)\in U \rmiff \exists n\; (x(n)=1 \rmand y\in B_n).$$ More generally if $X$ contains a homeomorphic copy of $\cspc$ then $X\times Y$ will have a universal open set. In 1995 Michael \hrusak mentioned the following problem to us. Most of the results in this note were proved in June and July of 2001. \bigskip\bigskip \begin{center} \hrusak's problem. \end{center} \begin{quote} Let $X$, $Y$ be topological spaces, call $X$ the parameter space, and $Y$ the base space. When does there exists $U\subseteq X\times Y$ which is uniquely universal for the open subsets of $Y$? This means the $U$ is open and for every open set $W\su Y$ there is a unique $x\in X$ such that $U_x=W$. \end{quote} Let us say that $X\times Y$ satisfies UU (uniquely universal property) if there exists such an open set $U\su X\times Y$ which uniquely parameterizes the open subsets of $Y$. Note that the complement of $U$ is a closed set which uniquely parameterizes the closed subsets of $Y$. \begin{prop} (\hrusak) $2^\om\times 2^\om$ does not satisfy UU. \end{prop} \proof The problem is the empty set. Suppose $U$ is uniquely universal for the closed subsets of $2^\om$. Then there is an $x_0$ such that $U_{x_0}=\emptyset$ but all other cross sections are nonempty. Take $x_n\to x_0$ but distinct from it. Since all other cross sections are non-empty we can choose $y_n\in U_{x_n}$. But then $y_n$ has a convergent subsequence, say to $y_0$, but then $y_0\in U_{x_0}$. \qed More generally: \begin{prop}\label{hrus2} (\hrusak) Suppose $X\times Y$ has UU and $Y$ is compact. Then $X$ must have an isolated point. \end{prop} \pf Suppose $U\subseteq X\times Y$ witnesses UU for closed subsets of $Y$ and $U_{x_0}=\emptyset$. For every $y\in Y$ there exists $U_y\times V_y$ open containing $(x_0,y)$ and missing $U$. By compactness of $Y$ finitely many $V_y$ cover $Y$. The intersection of the corresponding $U_y$ isolates $x_0$. \qed Hence, for example, $2^\om\times (\om+1)$, $\bspc\times(\om+1)$, and $\bspc\times \cspc$ cannot have UU. \begin{prop} Let $\cspc\isolated$ be obtained by attaching an isolated point to $\cspc$. Then $(\cspc\isolated)\times \cspc$ has UU. \end{prop} \pf Define $T\su 2^{<\om}$ to be a nice tree iff \begin{enum} \item $s\su t\in T$ implies $s\in T$ and \item if $s\in T$, then either $s\concat(0)$ or $s\concat(1)$ in $T$. \end{enum} Let $\nicetree\su\power(2^{<\om})$ be the set of nice trees. Define the universal set $U$ by $$ U=\{(T,x)\in \nicetree\times \cspc \st \forall n \;\; x\res n\in T\}.$$ Note that the empty tree $T$ is nice and parameterizes the empty set. Also $\nicetree$ is a closed subset of $\power(2^{<\om})$ with exactly one isolated point (the empty tree), and hence it is homeomorphic to $\cspc\isolated$. \qed \begin{ques} Does $(\cspc \isolated)\times [0,1]$ have UU? \end{ques} \begin{remark}\label{powerset} $2^\om\times \om$ has the UU property. Just let $(x,n)\in U$ iff $x(n)=1$. \end{remark} \begin{ques} Does either $\rr\times \om $ or $[0,1]\times \om$ have UU? Or more generally, is there any example of UU for a connected parameter space? \end{ques} Recall that a topological space is Polish iff it is completely metrizable and has a countable dense subset. A set is $G_\de$ iff it is the countable intersection of open sets. The countable product of Polish spaces is Polish. A $G_\de$ subset of a Polish space is Polish (Alexandrov). % kur p.208 A space is zero-dimensional iff it has a basis of clopen sets. All compact zero-dimensional Polish spaces without isolated points are homeomorphic to $\cspc$ (Brouwer). A zero-dimensional Polish space is homeomorphic to $\bspc$ iff compact subsets have no interior (Alexandrov-Urysohn). For proofs of these facts see Kechris \cite{kec} p.13-39. \begin{prop} \label{locallycompact} Suppose $Y$ is a zero dimensional Polish space. If $Y$ is locally compact but not compact, then $\cspc\times Y$ has UU. So, for example, $\cspc\times (\om\times 2^\om)$ has UU. \end{prop} \pf Let $\bb$ be a countable base for $Y$ consisting of clopen compact sets. Define $G\su\bb$ is good iff $$G=\{b\in\bb \st b\su \bigcup G\}$$ Let $\gg\su\power(\bb)$ be the family of good subsets of $\bb$. We give the $\power(\bb)$ the topology from identifying it with $2^{\bb}$. Since $\bb$ is an infinite countable set $\power(\bb)$ is homeomorphic to $\cspc$. A sequence $G_n$ for $n<\om$ converges to $G$ iff for each finite $F\su\bb$ we have that $G_n\cap F=G\cap F$ for all but finitely many $n$. Hence $\gg$ is a closed subset of $\power(\bb)$ since by compactness $b\su \bigcup G$ iff $b\su \bigcup F$ for some finite $F\su G$. There is a one-to-one correspondence between good families and open subsets of $Y$: Given any open set $U\su Y$ define $$G_U=\{b\in\bb\st b \su U\}$$ and given any good $G$ define $U_G=\bigcup G$. (Note that the empty set is good.) We claim that no $G_0\in \gg$ is an isolated point. Suppose for contradiction it is. Then there must be a basic open set $N$ with $\{G_0\}=\gg\cap N$. A basic neighborhood has the following form $$N=N(F_0,F_1)=\{G\su \bb\st F_0\su G\rmand F_1\cap G=\emptyset\}$$ where $F_0, F_1\su \bb$ are finite. For each $b\in F_1$ since $G_0$ is good, $b$ is not a subset of $\bigcup G_0$, and since $\bigcup F_0\su \bigcup G_0$, we can choose a point $z_b\in b\sm\bigcup F_0$. Since $Y$ is not compact, $Y\sm (\bigcup (F_0\cup F_1))$ is nonempty. Fix $z\in Y\sm (\bigcup (F_0\cup F_1))$. Now let $U_1=Y\sm \{z_b\st b\in F_1\}$ and let $U_2=U_1\sm\{z\}$. Then $G_{U_1},G_{U_2}$ are distinct elements of $N\cap\gg$. Hence $\gg$ is a compact zero-dimensional metric space without isolated points and therefore it is homeomorphic to $\cspc$. To get a uniquely universal open set $U\su \gg\times Y$ define: $$(G,y)\in U\rmiff \exists b\in G\;\; y\in b.$$ \qed Example \ref{examplepolish} is a countable Polish space $Z$ such that $\cspc\times Z$ has UU, but $Z$ is not locally compact. \begin{lemma}\label{onto} Suppose $f:X\to Y$ is a continuous bijection and $Y\times Z$ has UU. Then $X\times Z$ has UU. \end{lemma} \proof Given $V\su Y\times Z$ witnessing UU, let $$U=\{(x,y)\st (f(x),y)\in V\}.$$ \qed Many uncountable standard Borel sets\footnote{A standard Borel set is a Borel subset of a Polish space.} are the bijective continuous image of the Baire space $\om^\om$. According to the footnote on page 447 of Kuratowski \cite{kur} Sierpinski proved in a 1929 paper that any standard Borel set in which every point is a condensation point is the bijective continuous image of $\bspc$. We weren't able to find Sierpinski's paper but we give a proof of his result in Lemma \ref{standardborel}. We first need a special case for which we give a proof. \begin{lemma}\label{bspcontocspc} There is a continuous bijection $f:\bspc\to\cspc$. \end{lemma} \pf Let $\pi:\om\to\om+1$ be a bijection. It is automatically continuous. It induces a continuous bijection $\pi:\bspc\to (\om+1)^\om$. But $(\om+1)^\om$ is a compact Polish space without isolated points, hence it is homeomorphic to $\cspc$. \qed Remark. If $C\su\cspc\times\bspc$ is the graph of $f^{-1}$, then $C$ is a closed set which uniquely parameterizes the family of singletons of $\bspc$. \begin{cor} \label{corparam} If $\cspc\times Y$ has UU, then $\bspc\times Y$ has UU. \end{cor} \begin{ques} Is the converse of Corollary \ref{corparam} false? That is: Does there exist $Y$ such that $\bspc\times Y$ has UU but $\cspc\times Y$ does not have UU? \end{ques} \begin{lemma}\label{bspconto} Suppose $X$ is a zero-dimensional Polish space without isolated points. Then there exists a continuous bijection $f:\bspc\to X$. \end{lemma} \pf Construct a subtree $T\su \om^{<\om}$ and $(C_s\su X\st s\in T)$ nonempty clopen sets such that: \begin{enumerate} \item $C_{\root}=X$, \item if $s\in T$ is a terminal node, then $C_s$ is compact, and \item if $s\in T$ is not terminal, then $s\concat(n)\in T$ for every $n\in\om$ and $C_s$ is partitioned by $(C_{s\concat(n)}\st n<\om)$ into nonempty clopen sets each of diameter\footnote{This is with respect to a fixed complete metric on $X$.} less than $\frac1{|s|+1}$. \end{enumerate} For each terminal node $s\in T$ choose a continuous bijection $f_s:[s]\to C_s$ given by Lemma \ref{bspcontocspc}. Define $f:\bspc\to X$ by $f(x)=f_{x\res n}(x)$ if there exists $n$ such that $x\res n$ is a terminal node of $T$ and otherwise determine $f(x)$ by the formula: $$\{f(x)\}=\bigcap_{n<\om} C_{x\res n}$$ Checking that $f$ is a continuous bijection is left to the reader. \qed Remark. An easy modification of the above argument shows that any zero-dimensional Polish space is homeomorphic to a closed subspace of $\bspc$. It also gives the classical result that if no clopen sets are compact, then $X$ is homeomorphic to $\bspc$. A different proof of Lemma \ref{bspconto} is given in Moschovakis \cite{mos} p. 12. \begin{define} We use $\cl(X)$ to denote the closure of $X$. \end{define} \begin{prop} \label{polish} Suppose $Y$ is Polish but not compact. Then $\bspc\times Y$ has the UU. So for example, $\bspc\times\bspc$ and $\bspc\times\rr$ both have UU. \end{prop} \pf We assume that the metric on $Y$ is complete and bounded. Let $\bb$ be a countable basis for $Y$ of nonempty open sets which has the property that no finite subset of $\bb$ covers $Y$. For $s,t\in\bb$ define $t\lhd s$ iff $\cl(t)\subseteq s$ and diam$(t)\leq\frac12 $ diam$(s)$. \begin{lemma}\label{lemprop} Suppose $ G\subseteq \bb$ has the following properties: \begin{enumerate} \item[(1)] for all $t,s\in\bb$ if $t\subseteq s\in G$, then $t\in G$ and \item[(2)] $\forall s\in\bb\;$ if ($\forall t\lhd s\;\; t\in G$), then $s\in G$, \end{enumerate} then for any $s\in\bb$ if $s\subseteq\bigcup G$, then $s\in G$. \end{lemma} \pf Suppose (1) and (2) hold but for some $s\subseteq\bigcup G$ we have $s\notin G$. Note that there cannot be a sequence $(s_n:n\in\om)$ starting with $s_0=s$, and with $s_{n+1}\lhd s_n$ and $s_n\notin G$ for each $n$. This is because if $\{x\}=\bigcap_{n\in\om}s_n$, then $x\in s\subseteq\bigcup G$ and so for some $t\in G$ we have $x\in t$. But then for some sufficiently large $n$ we have that $s_n\subseteq t$ putting $s_n\in G$ by (1). Hence there must be some $t\unlhd s$ with $t\notin G$ but for all $r\lhd t$ we have $r\in G$. This is a contradiction to (2). \qed Let $\gg\su\power(\bb)$ be the set of all $G$ which satisfy the hypothesis of Lemma \ref{lemprop}. Then we have a $G_\delta$ subset $\gg$ of $2^{\bb}$, which uniquely parameterizes the open sets. Hence the set $U$ witnesses the unique universal property: $$U=\{(G,x)\in\gg\times Y\st x\in \bigcup G\}.$$ To finish the proof it is enough to see that no point in $\gg$ is isolated. If $\gg$ has an isolated point, there must be $s_1,\ldots, s_n$ and $t_1,\ldots,t_m$ from $\bb$ such that $$N= \{G\in \gg\st s_1\in G,\ldots,s_n\in G,\;t_1\notin G,\ldots, t_m\notin G\} $$ is a singleton. Let $W=s_1\cup\cdots\cup s_n$. Since $N$ contains the point $$G=\{s\in\bb\st s\su W\}$$ it must be that $N=\{G\}$. For each $j$ choose $x_j\in t_j\sm W$. Since the union of the $s_i$ and $t_j$ does not cover $Y$, we can choose $$z\in Y\sm (s_1\cup\cdots \cup s_n\cup t_1\cup\cdots\cup t_m).$$ Take $r\in\bb$ with $z\in r$ but $x_j\notin r$ for each $j=1,\ldots,m$. Let $$G^\pr=\{s\in\bb\st s\su r\cup W\}.$$ Then $G^\pr\in N$ but $G^\pr\neq G$ which shows that $N$ is not a singleton. It follows from Lemma \ref{onto} and \ref{bspconto} that $\bspc\times Y$ has the UU. \qed Proposition \ref{hrus2} and \ref{polish} show that: \begin{cor} For $Y$ Polish: \centerline{ $\bspc\times Y$ has UU iff $Y$ is not compact.} \end{cor} \begin{ques}\label{ques17} Does $\cspc\times \bspc$ have UU? \end{ques} \begin{ques} Does $\cspc\times \rr$ have UU? \end{ques} Our next result follows from Proposition \ref{sigmacompact} but has a simpler proof so we give it first. \begin{prop}\label{ctbl} If $X$ is a countable metric space which is not compact, then $\bspc\times X$ has UU. So, for example, $\bspc\times\qq$ has UU. \end{prop} \proof We produce a uniquely universal set for the open subsets of $X$. First note that there exists a countable basis $\bb$ for $X$ with the property that it is closed under finite unions and $X\setminus B$ is infinite for every $B\in \bb$. To see this fix $\{x_n\st n<\om\}\su X$ an infinite set without a limit point, i.e., an infinite closed discrete set. Given a countable basis $\bb$ replace it with finite unions of sets from $$\{B\sm\{x_m:m>n\}\st n\in\om\rmand B\in\bb\}.$$ We may assume also that $\bb$ includes the empty set. Next let $$\poset=\{(B,F): B\in\bb,\;\;F\in[X]^{<\om},\rmand B\cap F=\emptyset\}.$$ Then $\poset$ is a partial order determined by $p\leq q$ iff $B_q\su B_p$ and $F_q\su F_p$. For $p\in\poset$ we write $p=(B_p,F_p)$. For $p,q\in\poset$ we write $p\perp q$ to stand for $p$ and $q$ are incompatible, i.e., there does not exist $r\in \poset$ with $r\leq p$ and $r\leq q$. We will code open subsets of $X$ by good filters on $\poset$. Define the family $\gg$ of good filters on $\poset$ to be the set of all $G\su\poset$ such that \begin{enumerate} \item $p\leq q$ and $p\in G$ implies $q\in G$, \item $\forall p,q\in G$ exists $r\in G$ with $r\leq p$ and $r\leq q$, \item $\forall x\in X\;\exists p\in G \;\; x\in B_p\cup F_p$, and \item $\forall p\in\poset$ either $p\in G$ or $\exists q\in G\;\; p\perp q$. \end{enumerate} Since the poset $\poset$ is countable we can identify $\power(\poset)$ with $\power(\om)$ and hence $\cspc$. We give $\gg\su\power(\poset)$ the subspace topology. Note that $\gg$ is $G_\de$ in this topology. Note also that the sets $$[p]=\{G\in\gg\st p\in G\}$$ form a basis for $\gg$ (use conditions (2) and (4) to deal with finitely many $p_i$ in $G$ and finitely many $q_j$ not in $G$). Note that since $X\sm B_p$ is always infinite, for any $p\in\poset$ there exists $r,q\leq p$ with $r\perp q$. Namely, for some $x\in X\sm (B_p\cup F_p)$ put $x$ into $B_r\cap F_q$. It follows that no element of $\gg$ is isolated. So $\gg$ is a zero-dimensional Polish space without isolated points. Hence by Lemma \ref{bspconto} there is a continuous bijection $f:\bspc\to\gg$. For $G\in \gg$, let $$U_G=\bigcup \{B_p\st p\in G\}.$$ For any $U\su X$ open, define $$G_U=\{p\in\poset\st B_p\su U \rmand F_p\cap U=\emptyset\}.$$ The maps $G\to U_G$ and $U\to G_U$ show that there is a one-to-one correspondence between $\gg$ and the open subsets of $X$. Finally define $\uu\su \gg\times X$ by $$(G,x)\in \uu \rmiff \exists p\in G \;\; x\in B_p.$$ This witnesses the UU property for $\gg\times X$ and so by Lemma \ref{onto}, we have UU for $\bspc\times X$. \qed \begin{ques} Does $\cspc\times\qq$ have UU? \end{ques} Our next result Proposition \ref{sigmacompact} implies Proposition \ref{ctbl} but needs the following Lemma: \begin{lemma}\label{standardborel}(Sierpinski) Suppose $B$ is a Borel set in a Polish space for which every point is a condensation point. Then there exists a continuous bijection from $\bspc$ to $B$. \end{lemma} \pf We use that every Borel set is the bijective image of a closed subset of $\bspc$. This is due to Lusin-Souslin see Kechris \cite{kec} p.83 or Kuratowski-Mostowski \cite{kurmos} p.426. Using the fact that every uncountable Borel set contains a perfect subset it is easy to construct $K_n$ for $n<\om$ satisfying: \begin{enumerate} \item $K_n\su B$ are pairwise disjoint, \item $K_n$ are homeomorphic copies of $\cspc$ which are nowhere dense in $B$, and \item every nonempty open subset of $B$ contains infinitely many $K_n$. \end{enumerate} Let $B_0=B\sm\bigcup_{n<\om}K_n$. We may assume $B_0$ is nonempty, otherwise just split $K_0$ into two pieces. Since it is a Borel set, there exists $C\su\bspc$ closed and a continuous bijection $f:C\to B_0$. Define $$\Gamma=\{s\in \om^{<\om}\st [s]\cap C=\emptyset \rmand [s^*]\cap C\neq \emptyset\}$$ where $s^*$ is the unique $t\su s$ with $|t|=|s|-1$. Without loss we may assume that $C$ is nowhere dense and hence $\Gamma$ infinite. Let $\Gamma=\{s_n\st n<\om\}$ be a one-one enumeration. Note that $\{C\}\cup\{[s_n]\st n<\om\}$ is a partition of $\bspc$. Inductively choose $l_n>l_{n-1}$ with $K_{l_n}$ a subset of the ball of radius $\frac1{n+1}$ around $f(x_n)$ for some $x_n\in C\cap [s_n^*]$. For each $n<\om$ let $f_n:[s_n]\to K_{l_n}$ be a continuous bijection. \bigskip Then $g=f\cup\bigcup_{n<\om}f_n$ is a continuous bijection from $\bspc$ to $B_0\cup\bigcup_{n<\om}{K_{l_n}}$. \bigskip\noindent To see that it is continuous suppose for contradiction that $u_n\to u$ as $n\to\infty$ and $|g(u_n)-g(u)|>\eps>0$ all $n$. Since $C$ is closed if infinitely many $u_n$ are in $C$, so is $u$ and we contradict continuity of $f$. If $u\in [s_n]$, then we contradict the continuity of $f_n$, So, we may assume that all $u_n$ are not in $C$ but $u$ is in $C$. By the continuity of $f$ we may find $s\su u$ with $f([s]\cap C)$ inside a ball of radius $\frac\eps{3}$ around $f(u)$. Find $n$ with $\frac1{n+1}<\frac\eps3$ for which there is $m$ such that $u_m\in [s_n]$ and $s_n\supseteq s$. But then $g(u_m)=f_n(u_m)\in K_{l_n}$ and $K_{l_n}$ is in a ball of radius $\frac1{n+1}$ around some $f(x_n)$ with $x_n\in [s_n^*]\cap C$. This is a contradiction: $$d(g(u),g(u_m))\leq d(f(u),f(x_n))+d(f(x_n),f_n(u_m))\leq \frac23\eps.$$ Next let $I=\om\sm\{l_n\st n<\om\}$. Then there exists continuous bijection $$h:I\times\bspc\to \bigcup_{i\in I}K_i.$$ Finally $g\cup h$ is a continuous bijection from $\bspc \oplus (I\times\bspc)$ to $B_0\cup \bigcup_{n<\om}K_n=B$. Since $\bspc \oplus (I\times\bspc)$ is a homeomorphic copy of $\bspc$ we are done. \qed \begin{prop} \label{sigmacompact} $\bspc\times Y$ has UU for any $\si$-compact but not compact subspace $Y$ of a Polish space. So for example, $\bspc\times (\qq\times \cspc)$ has UU. \end{prop} \pf Let $Y=\bigcup_{n<\om}K_n$ where each $K_n$ is compact. Since $Y$ is not compact it contains an infinite closed discrete set $D$. Choose a countable basis $\bb$ for $Y$ such that for any $b\in\bb$ the closure of $b$ contains at most finitely many points of $D$. Define $G\su\bb$ to be good iff for every $b\in\bb$ if $\cl(b)\su\bigcup G$ then $b\in G$. Let $\gg\su\power(\bb)$ be the family of good sets. Note that $\gg$ is a ${\bf \Pi}^0_3$ set: $$G\in\gg \rmiff \forall b\in\bb \;\;(\forall n \;\;\cl(b)\cap K_n\su \bigcup G)\to b\in G$$ Note that $\cl(b)\cap K_n\su\bigcup G$ iff there is a finite $F\su G$ with $\cl(b)\cap K_n\su\bigcup F$. To finish the proof it is necessary to see that basic open sets in $\gg$ are uncountable. Given $b_i,c_j\in\bb$ for $i0$ let $$K_n=\{(0^{n} \concat(\star,x_0,x_1,\ldots),y)\in\cspc\times Y\st ((n,x),y)\in C\}$$ By $0^{n}\concat(\star,x_0,x_1,\ldots)$ we mean a sequence of $n$ zeros followed by the special symbol $\star$ and then the (binary) digits of $x$. Note that the $K_n$ converge to $\zerovec$. Let $$K_0=\{(x,y)\st ((0,x),y)\in C\}$$ Let $B=\bigcup_{n<\om}K_n\su S\times Y$ where $$S=\cspc\cup\bigcup_{n>0}\{0^{n}\concat(\star,x_0,x_1,\ldots) \st x\in\cspc\}.$$ Note that $S$ is homeomorphic to $\cspc$ and there is a one-to-one correspondence between the cross sections of $B$ and cross sections of $C$. Note that $B$ is closed in $S\times Y$: If $(x_n,y_n)\in B$ for $n<\om$ is a sequence converging to $(x,y)\in S\times Y$ and $x$ is not the zero vector it is easy to see that $(x,y)\in B$. On the other hand if $x$ is the zero vector, then since $B_\zerovec=Y$, it is automatically true that $(x,y)\in B$. Hence UU holds for $\cspc\times Y$. For the opposite direction just note that $(\om+1)\times \cspc$ is homeomorphic to $\cspc$ and there is a continuous bijection from $\om\times\cspc$ onto $(\om+1)\times \cspc$. \qed Next we describe our counterexample to a converse of Proposition \ref{locallycompact}. Let $Z=(\om\times\om)\cup \{\infty\}$. Let each $D_n=\{n\}\times\om$ be an infinite closed discrete set and let the sequence of $D_n$ ``converge'' to $\infty$, i.e., each neighborhood of $\infty$ contains all but finitely many $D_n$. Equivalently $Z$ is homeomorphic to: $$Z^\pr=\{x\in\bspc \st |\{n\st x(n)\neq 0\}|\leq 1\}.$$ The point $\infty$ is the constant zero map, while $D_n$ are the points in $Z^\pr$ with $x(n) \neq 0$. Note that $Z^\pr$ is a closed subset of $\bspc$, hence it is Polish. This seems to be the simplest nonlocally compact Polish space. \begin{example} \label{examplepolish} $Z$ is a countable nonlocally compact Polish space such that $2^\om\times Z$ has the UU. \end{example} \pf $Z=\bigcup_{n<\om}D_n\cup\{\infty\}$. Note that $X\su Z$ is closed iff $\infty\in X$ or $X\su\bigcup_{i\leq k}D_i$ for some $k<\om$. By Lemma \ref{omegatimescspc} it is enough to see that $(\om\times 2^\om)\times Z$ has the UU. Let $P_n=\{n\}\times 2^\om$. Use $P_0$ to uniquely parameterize all subsets of $Z$ which contain the point at infinity, see Remark \ref{powerset}. Use $P_1$ to uniquely parameterize all $X\su D_0$, including the empty set. For $n=1+2^{k-1}(2l-1)$ with $k,l>0$ use $P_n$ to uniquely parameterize all $X\su \bigcup_{i\leq k}D_i$ such that $D_k$ meets $X$ and the minimal element of $D_k\cap X$ is the $l$-th element of $D_k$. \qed Our next two results have to do with Question \ref{ques17}. \bigskip \begin{prop} Existence of UU for $\cspc\times \bspc$ is equivalent to: \par There exist a $\cc\su \power(\om^{<\om})$ homeomorphic to $\cspc$ such that every $T\in \cc$ is a subtree of $\om^{<\om}$ (possibly with terminal nodes) and such that for every closed $C\su\bspc$ there exists a unique $T\in\cc$ with $C=[T]$. \end{prop} \pf Given $C\su \cspc\times\bspc$ witnessing UU for closed subsets of $\bspc$. Let $$[T]=\{(s,t):([s]\times [t])\cap C \neq \emptyset\}.$$ Define $f:\cspc\to \power(\om^{<\om})=2^{\om^{<\om}}$ by $f(x)(s)=1$ iff $(x\res n,s)\in T$ where $n=|s|$. Then $f$ is continuous, since $f(x)(s)$ depends only on $x\res n$ where $n=|s|$. If $T_x=f(x)$, then $[T_x]=C_x$. Hence $f$ is one-to-one, so its image $\cc$ is as described. \qed \bigskip \begin{prop} Suppose $\bspc\times Y$ has UU where $Y$ is any topological space in which open sets are $F_\si$. Then there exists $U\su \cspc\times Y$ an $F_\si$ set such that all cross sections $U_x$ are open and for every open $W\su Y$ there is a unique $x\in \cspc$ with $U_x=W$. \end{prop} \pf Let $\om\isolated$ denote the discrete space with one isolated point adjoined and let $\om+1$ denote the compact space consisting of a single convergent sequence. Then $(\om\isolated)^\om$ is homeomorphic to $\bspc$ and $(\om+1)^\om$ is homeomorphic to $\cspc$. Assume that $U\su (\om\isolated)^\om \times Y$ is an open set witnessing UU. Then $U$ is an $F_\si$ set in $(\om+1)^\om\times Y$. To see this note that a basic clopen set in $(\om\isolated)^\om$ could be defined by some $s\in (\om\isolated)^{<\om}$ by $$[s]=\{x\in (\om\isolated)^{\om}\st s\su x\}.$$ But it is easy to check that $[s]\su (\om+1)^\om$ is closed in the topology of $(\om+1)^\om$. Since $U$ is open in $(\om\isolated)^\om \times Y$ there exists $s_n$ and open sets $W_n\su Y$ such that $$U=\bigcup_{n<\om} [s_n]\times W_n$$ Hence $U$ is the countable union of $F_\si$ sets and so it is $F_\si$ in $(\om+1)^{<\om}$. \qed \bigskip Here $\oplus$ refers to the topological sum of disjoint copies or equivalently a clopen separated union. \begin{prop} Suppose $X_i\times Y_i$ has UU for $i\in I$. Then $$(\prod_{i\in I}X_i)\times (\bigoplus_{i\in I} Y_i)\mbox{ has UU.}$$ So, for example, if $\cspc\times Y$ has UU, then $\cspc\times (\om\times Y)$ has UU. \end{prop} \pf Define $$((x_i)_{i\in I},y)\in U \rmiff \exists i\in I\;\; (x_i,y)\in U_i$$ where the $U_i\su X_i\times Y_i$ witness UU. \qed \bigskip Except for Proposition \ref{hrus2} we have given no negative results. The following two propositions are the best we could do in that direction. \begin{prop} Suppose $U\su X\times Y$ is an open set universal for the open subsets of $Y$. If $X$ is second countable, then so is $Y$. \end{prop} \pf $U$ is the union of open rectangles of the form $B\times C$ with $B$ open in $X$ and $C$ open in $Y$. Clearly we may assume that $B$ is from a fixed countable basis for $X$. Since $\bigcup_i B\times C_i=B\times \bigcup_i C_i$ we may write $U$ as a countable union: $$U=\bigcup_{n<\om} B_n\times C_n$$ where the $B_n$ are basic open sets in $X$ and the $C_n$ are open subsets of $Y$. But this implies that $\{C_n\st n<\om\}$ is a basis for $Y$ since for each $x\in X$ $$U_x=\bigcup \{C_n\st x\in B_n\}$$ \qed \begin{prop} There exists a partition $X\cup Y=\cspc$ into Bernstein sets $X$ and $Y$ such that for every Polish space $Z$ neither $Z\times X$ nor $Z\times Y$ has UU. \end{prop} \pf Note that up to homeomorphism there are only continuum many Polish spaces. If there is a UU set for $Z\times X$, then there is an open $U\su Z\times\cspc$ such that $U\cap (Z\times X)$ is UU. Note that the cross sections of $U$ must be distinct open subsets of $\cspc$. Hence it suffices to prove the following: Given $\uu_\al$ for $\al<\cont$ such that each $\uu_\al$ is a family of open subsets of $\cspc$ either \begin{enum} \item there exists distinct $U,V\in\uu_\al$ with $U\cap X=V\cap X$ or \item there exists $U\su\cspc$ open such that $U\cap X\neq V\cap X$ for all $V\in\uu_\al$. \end{enum} And the same for $Y$ in place of $X$. Let $P_\al$ for $\al<\cont$ list all perfect subsets of $\cspc$ and let $\{z_\al\st\al<\cont\}=\cspc$. Construct $X_\al,Y_\al\su\cspc$ with \begin{enumerate} \item $X_\al\cap Y_\al=\emptyset$ \item $\al<\be$ implies $X_\al\su X_\be$ and $Y_\al\su Y_\be$ \item $|X_\al\cup Y_\al|=|\al|+\om$ \item If there exists distinct $U,V\in\uu_\al$ such that $U\symdiff V$ is a countable set disjoint from $X_{\al}$, then there exists such a pair with $U\symdiff V\su Y_{\al+1}$ \item If there exists distinct $U,V\in\uu_\al$ such that $U\symdiff V$ is a countable set disjoint from $Y_{\al+1}$, then there exists such a pair with $U\symdiff V\su X_{\al+1}$ \item $P_\al$ meets both $X_{\al+1}$ and $Y_{\al+1}$ \item $z_\al\in (X_{\al+1}\cup Y_{\al+1})$ \end{enumerate} First we do (4) then (5) and then take care of (6) and (7). Let $X=\bigcup_{\al<\cont}X_\al$ and $Y=\bigcup_{\al<\cont}Y_\al$. Fix $\al$ and let us verify (a) or (b) holds. Take any point $p\in X\sm X_{\al+1}$. If (b) fails there must be $U,V\in\uu_\al$ with $X=X\cap U$ and $X\sm\{p\}=X\cap V$. Then $(U\symdiff V)\cap X_{\al+1}=\emptyset$. Since $X$ is Bernstein and $(U\symdiff V)\cap X$ has only one point in it, it must be that $U\symdiff V$ is countable. Then by our construction we have chosen distinct $U,V\in\uu_\al$ with $U\symdiff V\su Y$ therefore $U\cap X=V\cap X$, so (a) holds. A similar argument goes through for $Y$. \qed Finally, and conveniently close to the bibliography, we note some papers in the literature which are related to the property UU. Friedberg \cite{friedberg} proved that there is one-to-one recursively enumerable listing of the recursively enumerable sets. This is the same as saying that there is a (light-face) $\Sigma_1^0$ subset $U\su\om\times\om$ which is uniquely universal for the $\Sigma_1^0$ subsets of $\om$. Brodhead and Cenzer \cite{brodhead} prove the analogous result for (light-face) $\Sigma_1^0$ subsets of $\cspc$. Becker \cite{becker} considers unique parameterizations of the family of countable sets by Borel or analytic sets. Gao, Jackson, Laczkovich, and Mauldin \cite{gao} consider several other problems of unique parameterization. \begin{thebibliography}{99} \bibitem{brodhead} Brodhead, Paul; Cenzer, Douglas; Effectively closed sets and enumerations. Arch. Math. Logic 46 (2008), no. 7-8, 565-582. \bibitem{becker} Becker, Howard; Borel and analytic one-one parametrizations of the countable sets of reals. Proc. Amer. Math. Soc. 103 (1988), no. 3, 929-932. \bibitem{friedberg} Friedberg, Richard M.; Three theorems on recursive enumeration. I. Decomposition. II. Maximal set. III. Enumeration without duplication. J. Symb. Logic 23 1958 309-316. \bibitem{gao} Gao, Su; Jackson, Steve; Laczkovich, Mikls; Mauldin, R. Daniel; On the unique representation of families of sets. Trans. Amer. Math. Soc. 360 (2008), no. 2, 939-958. \bibitem{kec} Kechris, Alexander S.; {\bf Classical descriptive set theory.} Graduate Texts in Mathematics, 156. Springer-Verlag, New York, 1995. xviii+402 pp. ISBN: 0-387-94374-9 \bibitem{kur} Kuratowski, K.; {\bf Topology.} Vol. I. New edition, revised and augmented. Translated from the French by J. Jaworowski Academic Press, New York-London; Pastwowe Wydawnictwo Naukowe, Warsaw 1966 xx+560 pp. \bibitem{kurmos} Kuratowski, Kazimierz; Mostowski, Andrzej; {\bf Set theory.} With an introduction to descriptive set theory. Translated from the 1966 Polish original. Second, completely revised edition. Studies in Logic and the Foundations of Mathematics, Vol. 86. North-Holland Publishing Co., Amsterdam-New York-Oxford; PWN---Polish Scientific Publishers, Warsaw, 1976. xiv+514 pp. \bibitem{mos} Moschovakis, Yiannis N.; {\bf Descriptive set theory.} Studies in Logic and the Foundations of Mathematics, 100. North-Holland Publishing Co., Amsterdam-New York, 1980. xii+637 pp. ISBN: 0-444-85305-7 \end{thebibliography} \cutcrap \newpage \begin{center} {\bf Miscellaneous Crapolo} \end{center} How about Unique Parameterization (UP) for other families of sets? It was remarked that if $f:\bspc\to\cspc$ is 1-1,onto, continuous, then $$U=\{(x,y): f(y)=x\}\subseteq \cspc\times\bspc$$ is UP for the singleton subsets of $\bspc$. How about doubleton s? Does there exist a closed set $C\subseteq\cspc\times\bspc$ such that $|C_x|=2$ for every $x$ and for every $F\subseteq\bspc$ of size 2 there exists an $x$ with $C_x=F$? (If so, can we get it to be UP?). \bigskip Note that we can do singletons (uniquely) and using that we can do $F$ of size 1 or 2 uniquely. For $F$ of size 3 can do but I don't know about uniquely. Can also use $\om\times\cspc$ or $\bspc$ as unique parameters for doubleton s. \bigskip What about other classes of closed sets. For example, the perfect subsets of $\cspc$? Or the family of finite closed sets. Or the family of infinite discrete subsets of $\bspc$? \bigskip \bigskip For $X$ a Polish space let $C(X)$ be the compact non-empty subsets of $X$ with the Hausdorff metric: $$d(K,L)=sup_{y\in L}dist(y,K)\;\;+\;\;sup_{x\in K}dist(x,L)$$ I think this is always a Polish space. If we define $Q\su C(X)\times X$ by $(K,x)\in Q$ iff $x\in K$ we get a uniquely universal set for the compact subsets of $X$. \bigskip Is being a uniquely universal or universal set open sets absolute for Polish products $U\su X\times Y$? \bigskip Can $U\subseteq X\times Y$ witness simultaneously the UU for each? Or even just universal without the unique? No it cannot. If $U_x=Y$ then $x$ is in every $U^y$. \bigskip Investigate UU for maps $f:Y\to Z$, i.e. $F:X\times Y\to Z$ giving every map as a unique cross section. \bigskip What about doubly universal uniqueness sets? Suppose $U\su X\times Y$ is a uniquely universal set. Define $V,W\su (X\times X)\times Y$ by $$V=\{((x,y),z): (x,z)\in U\}\rmand W=\{((x,y),z): (y,z)\in U\}$$ Then for any pair of open sets $A,B\su Y$ there is a unique $z=(x,y)$ with $V_z=A$ and $W_z=B$. \bigskip On question about $\cspc\times\bspc$: what kind of function is $T\to \hat{T}$ where $\hat{T}=\{s\in T: T_s$ has branch$\}$? \bigskip Its $\Sigma_1^1$. Note that it is one-to-one onto from a putative universal set to Trees without terminal nodes (except root). \bigskip\bigskip On the question: Does there exists a closed (perfect?) $\calc\subseteq P(\om^{<\om})$ of trees such that for every $C\subseteq\bspc$ closed there exists a unique $T\in\calc$ with $[T]=C$? Various thought on the matter: $$\hat{T}=\{s\in T:[s]\cap[T]\not=\emptyset\}$$ $$D(T)=\{s\in T:\exists n \; sn\in T\}$$ $\caln$ subtrees of $\om^{<\om}$ with no terminal nodes equivalently fixed points of $D$. Note that $\calc\to\caln$ via $T\to \hat{T}$ is 1-1 and onto. $\caln\to cl(\bspc)$ via $T\to[T]$ is one-one and onto. \addressarn \end{document}