What uniquely characterizes equivalence classes of eventually equal binary sequences?

Question

Let $X$ be the set of all infinite binary sequences. (Or we can think of them as subsets of $\mathbb{N}$ or real numbers between $0$ and $1$.) Let us define an equivalence relation $\sim$ on $X$ by saying that $(a_n)\sim(b_n)$ if they’re eventually equal, i.e. if there exists a natural number $N$ such that $a_n=b_n$ for all $n\geq N$. And let $Y$ be the set of equivalence classes of elements of $X$ under $\sim$.

My question is, characterizes a given equivalence class in $Y$? We can’t say, e.g. “having a 1 in the 100th place”, because for every sequence that does have 1 in the 100th place, there are also sequences in the same equivalence class that don’t. And in general for any sequence $(a_n)$ and any naturap number $k$, there exists a sequence $(b_n)\sim(a_n)$ where $a_k\neq b_k$.

So what is the minimum information needed to unambiguously specify a given element of $Y$? By the way, this is somewhat similar to the notion of germs, which I discuss here.

Answer 1

The relation $\sim$ is more commonly denoted $E_0$, and I'll refer to it that way below.

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One piece of information which certainly suffices is to provide an element of the given equivalence class. Understanding how difficult this is leads to the question:

How complicated must a transversal for $E_0$ be?

(A transversal for an equivalence relation is simply a choice of representative for each equivalence class.) It turns out that in a precise sense, $E_0$ is quite complicated in this regard: there is no Borel transversal. That is, if $A\subseteq\mathbb{N}^\mathbb{N}$ is Borel, then there is some $a\in\mathbb{N}^\mathbb{N}$ which is $E_0$-related to no element of $A$ or is $E_0$-related to more than one element of $A$.

• Incidentally, this can be pushed further under set-theoretic assumptions, e.g. assuming appropriate large cardinals there is no transversal for $E_0$ in the (very large) inner model $L(\mathbb{R})$ - see this Mathoverflow answer (it begins by talking about the Vitali relation $x-y\in\mathbb{Q}$, but then moves to $E_0$, and indeed the two are appropriately equivalent).

In fact, this property characterizes $E_0$ uniquely within a fairly broad context: the Harrington-Kechris-Louveau theorem says that whenever $E$ is a Borel equivalence relation on $\mathbb{N}^\mathbb{N}$, then either $E$ has a Borel transversal or there is a continuous reduction of $E_0$ to $E$ ("$E$ is at least as complicated as $E_0$"). The general study of Borel reducibility of (not-too-high-complexity) equivalence relations is quite well-studied.

All this says that there is no easy way to pick a "canonical" representative from an $E_0$-class; this is a good indication that there isn't a "simple" way to describe an $E_0$-class in general. (Of course, that's not definitive, but I think your question is not precise enough at present to admit a truly definitive answer.)