TITLE: There is no Baire bijection between $\mathbb R$ and the set of functions $\mathbb Z\to\mathbb R$ modulo shifts
QUESTION [5 upvotes]: Let $X$ denote the set $\mathbb{R}^\mathbb{Z}$ (the set of all functions from integers to reals), 
and $\sim$ the equivalence relation on $X$ defined by:
$f \sim g$  iff there is a $z \in \mathbb{Z}$ such that for all $z' \in \mathbb{Z}: f(z'+z) = g(z')$.
Consider the quotient set $X/\sim$.
Some years ago, Mike Oliver made the remark that no bijection between $X/\sim$ and $\mathbb{R}$ could possibly be a Baire function.
I could do with a hint (or two) on how to prove that.

REPLY [2 votes]: Forget about $\mathbb{R}^\mathbb{Z}$; just look at $2^\mathbb{Z}$.
Hint 1: any set $A\subseteq 2^\mathbb{Z}$ which has the Baire property and which is $\sim$-saturated (in the sense that $x\in A$, $x\sim y$ implies $y\in A$) must be either meager or comeager.
Hint 2: if $f : 2^\mathbb{Z}\to \mathbb{R}$ is Baire-measurable and constant on every $\sim$-equivalence class, think about what $f^{-1}(U)$ could be, where $U$ ranges over open sets.  Maybe do something with a countable basis for $\mathbb{R}$...
I hope this is enough of a hint.