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.


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 $E$.

  • $\begingroup$ 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

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