This question rose up when i was reading a problem the author used to argue against the axiom of choice.
Consider the set of all (infinite) sequences of 0's and 1's. Q1) How many such sequences are there? (Countably infinite or uncountably infinite (I am currently trying to prove it is uncountably infinite, which intuitively seems correct)).
Q2) Define equivalence classes such that $(a)R(b) <=> (a)$ and $(b)$ differ in only finitely many positions. How many equivalence classes are there? (The article i read seemed to implicitly claim countably infinitely many equivalence classes upon which they applied axiom of choice)
It’s a standard result that the set of infinite sequence of zeroes and ones is uncountable: there’s a bijection between it and $\wp(\Bbb N)$. Each equivalence class of $R$, however, is countable: given a sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ of zeroes and ones, there are only countably many finite $F\subseteq\Bbb N$ and hence only countably many sequences differing from $\sigma$ in only finitely many places. Assuming the axiom of choice, a countable union of countable sets is countable, so there must be uncountably many equivalence clases.