Equivalence classes of real sequences, an interesting concept of closeness Consider $\mathbb R^{\mathbb N}$, the set of infinite sequences of reals. Two such sequences are equivalent if and only if they eventualy coincide. That is, if $x_1,x_2,\dots$ is one of the sequences, and $y_1,y_2,\dots$ is the other one, then we say that they are equivalent if and only if there is some $N$ such that for all $n\ge N$ we have $x_n=y_n$. 
We want to pick a representative from each class, which can be done using the axiom of choice but, in addition, we require that the representatives we choose must have following 
"Nice property" :

If the sequence $Y=y_1,\dots,y_n,\dots$ is the chosen representative of some class, then the sequence $K_1,K_2,K_3,\dots$ must also be the chosen representative of its class, where 
  $K_i= y_{2i-1}$ for all $i$ and, similarly, so should be the sequence $Z_1,Z_2,Z_3,\dots$, where $Z_i= y_{2i}$ for all $i$, i.e., the odd sequence and the even sequence from $Y$ are both representatives for some class.

How can I prove this family of representatives exists?

Some ideas:
Consider the set following set of pickings:
Some representatives are picked from some classes that respect the "Nice property":
 i.e If we picked  $y_1,....y_n ...$ as a representative of a class, we also picked the $K_i$ and $Z_i$ as representatives of their own class .(odd and even)
We induce the order relation of inclusion: if a choice of pickings contains all the pickings of some other picking then we say it majorizes it. Since a chain is bounded by the union of all the pickings and that element is also a nice picking, i.e the union) We can apply ZORN's lemma to get a maximal picking.
Suppose this maximal picking doesn't have a representative choosen in the class of sequences which are eventualy equal to some $b_1,b_2....b_k...$
The only way we wouldn't be able to pick a sequence $b_1...b_n ... $and add it to our set of representatives would be if splitting and splitting it in$ Z_i $and $K_i $ sequences would produce a class with two distinct choosen representatives since we are bounded to respect the nice property.
For sequence $K$ and $Z$ define their concaternation as
$ K*Z = k_1 z_1 k_2 z_2 k_3 z_3 ...$
So, by above we can say that if we have two sequences $K_i$ and$ Z_i $choosen as representatives their concaternation will be safe to add $(K_i*Z_i)$ so by maximality we already have it, also if we have representatives$ Z_1, Z_2 ... $( those are sequences not terms of sequences) it is safe  to say we already have$ Z_1*(Z_2*(Z_3*.....))) $as a limit, since it doesn't produce a contradiction in the sense of the above remark. So our maximal picking is in some sense closed.
 A: Here is a possible way in which things can go wrong: Consider the sequence $a = \langle 1, 2, 3, 4, \dots , n, n + 1, \dots  \rangle$. Let $[a]$ denote the equivalence class of $a$. Construct a tree of sequences $\langle a_{\sigma} : \sigma \in 2^{< \omega} \rangle$ such that $a_{\phi} = a$ and for every $n \geq 1$, $a_{1^{n+1}}$ is the odd part of $a_{1^n}$ and $a_{1^{n} 0}$ is the even part of $a_{1^n}$ except that it starts with $0$. For example, $a_1 = 1, 3, 5, 7, \dots $, $a_{0} = 0, 4, 6, 8, \dots $, $a_{11} = 1, 5, 9, 13, \dots$, $a_{10} = 0, 7, 11, 15, \dots$. If $\sigma \neq \phi, 1^n$, let $a_{\sigma 0}, a_{\sigma 1}$ be the even and odd parts of $a_{\sigma}$. Let $C = \{a_{\sigma} : \sigma \in 2^{< \omega}, \sigma \notin \{\phi, 1^n : n \geq 1 \}\}$. Then $C$ is a partial nice set of representatives. But there is no way to choose a member from $[a]$ to extend $C$!
A: Pinky's remark actually also shows that this is impossible.
Let $a = \langle a_0, a_1, a_2, \dots \rangle$ where $a_n = 1$ if $n  = 2^k - 1$ and $a_n = 0$ otherwise. Suppose that $C$ is a nice complete set of representatives. Then the constant zero sequence $\overline{0} = \langle 0, 0, 0, \dots \rangle$ must be in $C$. But now it is impossible to have anything in $C$ equivalent to $a$.
