Equivalence relation class of $X=\{0,1\}^{\mathbb{N}}$ Consider $X=\{0,1\}^{\mathbb{N}}$, i.e. the set of functions $\Bbb N \to \{0,1\}$.
Let $R$ be a relation on $X$ such that for every $f, g \in X$, $fRg$ iff the set $\{n\in\mathbb{N}|f(n)\neq g(n)\}$ is finite. 
I proved that the relation $R$ is an equivalence relation. Now I am trying to find the cardinality of $X/R$, the set of equivalence classes of $X$ modulo $R$. But I have some trouble to understand how to do so.
 A: consider $f : n \mapsto 0$. Let $\overline f$ be the set of functions that are related to $f$. Then, there is an obvious bijection between $\overline f$ and the set of all finite subsets of $\Bbb N$, which means that $\overline f$ is countable.
Every equivalence class in $X/R$ has the same cardinality $\aleph_0$ because of symmetry. Let $|X/R| = \kappa$.
Then, $\kappa \aleph_0 = 2^{\aleph_0}$, which means that $\kappa = 2^{\aleph_0}$.
A: Here is an explicit injection $X\to X/R$, which together with the surjection $X\to X/R$, the axiom of choice, and the Cantor-Bernstein theorem, allow you to conclude that $|X/R|=|X|$.
Constructing this injection is essentially finding $|X|$ subsets of $\mathbb{N}$ that are pairwise almost disjoint (rather than "essentially", I should say it is sufficient, because actually it is not necessary).
Find your preferred bijection $f: \mathbb{N} \to 2^{<\mathbb{N}}$ (set of finite sequences of $0$'s and $1$'s, it is clearly countable). This is just now a matter of finding $|X|$ pairwise almost-disjoint subsets of $2^{<\mathbb{N}}$.But this is really easy, for this one can think of $2^{\mathbb{N}}$ as sort of the "limit points" of $2^{<\mathbb{N}}$.
For each $u\in X$, put $A(u) = \{u_{\mid n} \mid n\in \mathbb{N}\}$ ($u_{\mid n}$ is the restriction of $u$ to $n=\{0,...,n-1\}$). Then if $A(u)\cap A(v)$ is infinite, it means there are arbitrarily high integers such that $u_{\mid n} = v_{\mid n}$, which implies $u=v$, so if $u\neq v$, $A(u)$ and $A(v)$ have finite intersection.
How does this help ? Well going back up this yields a family $B(u)$, $u\in 2^{\mathbb{N}}$ of subsets of $\mathbb{N}$ that have a pairwise finite intersection (put $B(u) = \{f^{-1}(x) : x\in A(u)\}$, check that this works), and then send $u\mapsto [B(u)]$ where $[B(u)]$ is the $R$-class of $B(u)$ ($B(u) \subset \mathbb{N}$ and so can be seen as a function $\mathbb{N}\to 2$). Accoridng to what we just proved, this is injective. 
Of course Kenny Lau's argument using cardinal arithmetic is fine as well, but this provides an explicit injection, and the injection itself doesn't need the axiom of choice, so that even in ZF you have $|X|\leq |X/R|$ and $|X/R| \leq^* |X|$, which is a subtler result
