What uniquely characterizes equivalence classes of eventually equal binary sequences? 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.
 A: The relation $\sim$ is more commonly denoted $E_0$, and I'll refer to it that way below.

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.)
