5
$\begingroup$

Let $X$ be a Polish space, $\mathcal B$ the sigma-algebra of Borel sets. Let $E$ be an aperiodic countable Borel equivalence relation on $X \times X$ (this means that every class of equivalence is countably infinite). A set $C\in \mathcal B$ is called a complete section for $E$, if $\forall x \in X$ $\exists y \in C$ such that $(x, y) \in E$.

$\textit{Question:}$ Let $f : X \to \mathbb R$ be a Borel real-valued function. Is there a complete section $C$ such the restriction of $f$ on the set $C$, $f|_C$, is a continuous function.

$\endgroup$
3
$\begingroup$

The answer is No.

A suitable counterexample can be constructed as follows.

On the real line $\mathbb R$ consider the equivalence relation $E=\{(x,y)\in\mathbb R\times \mathbb R:x-y\in\mathbb Q\}$.

Fix a countable base $\{U_n\}_{n\in\omega}$ of the topology on the real line. Take a countable set $X=\{x_n\}_{n\in\omega}\subset\mathbb R$ such that $x_n-x_m\notin \mathbb Q$ for any distinct $n,m\in\omega$.

In the real line $\mathbb R$ consider the $G_\delta$-set $$G:=\mathbb R\setminus\bigcup_{n\in\omega}(\mathbb R\setminus U_n)\cap(x_n+\mathbb Q)=(\mathbb R\setminus(X+\mathbb Q))\cup\bigcup_{n\in\omega}U_n\cap(x_n+\mathbb Q)$$ and the Borel function $f:G\to\mathbb R$ $$f(x)=\begin{cases} n&\mbox{if $x\in U_n\cap (x_n+\mathbb Q)$ for some $n\in\omega$};\\ 0&\mbox{otherwise}. \end{cases} $$ It is easy to see that the equivalence relation $E_G:=E\cap(G\times G)$ on $G$ has no complete section $C$ with continuous restriction $f|C$; moreover, for any complete section $C$ of $E_G$, the restriction $f|C$ has no continuity points.


There are (a bit more involved) counterexamples even for very good equivalence relations.

To construct a suitable example, take the convergent sequence $S:=\{0\}\cup\{2^{-n}:n\ge0\}\subset\mathbb R$ and on the compact zero-dimensional space $X:=S^\omega\times S$ consider the equivalence relation $$E=\{((x,y),(x',y'))\in X\times X:x=x'\}.$$ Observe that the quotient map $q:X\to X/E=S^\omega$ is open and closed.

Now take any bijective map $p:S\to D$ to a countable space $D$. Its countable power $P:S^\omega\to D^\omega$, $P:(x_n)_{n\in\omega}\mapsto (p(x_n))_{n\in\omega}$, is known as the Pawlikowski function and is a standard example of a Borel function of the first Baire class, which is not $\sigma$-continuous. More precisely, the Pawlikowski function has the property that a subset $C\subset S^\omega$ is nowhere dense in $S^\omega$ if the restriction $P|C$ is continuous.

We claim that the Borel function $f:=P\circ q:X\to D^\omega$ and the equivalence relation $E$ yield a counterexample to the question posed by Shrey. Indeed, assume that the equivalence relation $E$ has a complete section $C$ with continuous restriction $f|C$. For every $s\in S$ let $C_s:=\{x\in S^\omega:(x,s)\in C\}$ and observe that the continuity of the restriction $f|C$ implies the continuity of the map $P|C_s$ and hence nowhere density of $C_s$ in $S^\omega$. Since the space $S^\omega$ is compact and hence Baire, we conclude that $\bigcup_{s\in S}C_s\ne S^\omega$, which contradicts the choice of $C$ as a complete section of $E$.

$\endgroup$
  • $\begingroup$ Thank you for your reply. Edit: Let me edit this problem to ask the following : Is there a complete section $C$ such the restriction of $f$ on the set $C$, $f|_C$, is bounded. $\endgroup$ – Shrey Jul 26 '18 at 17:32
  • $\begingroup$ @Shrey It seems that the first example has the unboundedness property you required. $\endgroup$ – Taras Banakh Jul 26 '18 at 19:52
  • $\begingroup$ Thank you. If you allow me to ask further. would it be possible to do so if I restrict my equivalence class to aperiodic non-smooth hyperfinite Borel equivalence relation. Or is there any other case where it is possible. $\endgroup$ – Shrey Jul 26 '18 at 20:26
  • $\begingroup$ @Shrey Sorry, this is too complicated for me. I do not know what is hyperfinite and non-smooth equivalence relation. $\endgroup$ – Taras Banakh Jul 26 '18 at 20:32
  • $\begingroup$ You don't need to be sorry. Thank you for all you help. I really appreciate it. $\endgroup$ – Shrey Jul 26 '18 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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