TITLE: Doubt about the construction of $ \mathbb R$ by Cauchy sequences.
QUESTION [5 upvotes]: The exercise says:

Let $S$ be the set of Cauchy sequences of rational numbers. Define the
relation $ \sim $ in $S \times S$ as $(x_n)\sim (y_n)$ if
$\lim_{n\to\infty} x_n - y_n =0$. Prove that this is an equivalence
relation. What is the size of each equivalence class? How many
equivalence classes are?

I did the prove that the relation is an equivalence relation and solved how many equivalence classes there are, but I'm not clear about the size of each equivalence class. I think there is the same than the cardinality of $ \mathbb R$ by the construction of $\mathbb R$ by Cauchy sequences, but I'm not sure.

REPLY [5 votes]: Suppose $\langle x_n\rangle$ is a Cauchy sequence and that $\langle s_n\mid n\in\Bbb N\rangle$ is a sequece with $s_n=0$ or $s_n=1$ for each $n$. Then check that $\langle \frac{s_n}{n}+x_n\rangle$ is also a Cauchy sequence and that $\lim_{n\to\infty} (\frac{s_n}{n}+x_n)=\lim_{n\to\infty}x_n$ as well. Hence, $\langle x_n\rangle\sim\langle \frac{s_n}{n}+ x_n\rangle$.
So, the number of Cauchy sequences equivalent to $\langle x_n\rangle$ is at least as large as the number of sequences $\langle s_n\mid n\in\Bbb N\rangle$ such that $s_n=0$ or $s_n=1$ for each $n\in\Bbb N$. The cardinality of the set of all such sequences is $2^{\aleph_0}$.
Furthermore, each Cauchy sequence is a countable sequence of rationals, and there are countably many rationals, so there are at most $\aleph_0^{\aleph_0}=2^{\aleph_0}$ many Cauchy sequences that could be a member of the equivalence class to begin with.
Therefore, indeed, the equivalence class of a Cauchy sequence has cardinality $2^{\aleph_0}$, which is the cardinality of $\Bbb R$.