Is the Cauchy sequence of natural numbers countable?

Question

Consider the set consisting of all Cauchy sequences $a_n$ with $a_n \in \mathbb{N}$ for all $n$. Is the set countable?

My idea:

It is straight forward to prove that any such cauchy sequence conatining only natural number can have only finite number of distinct terms i.e. Cauchy sequence consisting entirely of natural numbers, then all of the entries beyond some index are the same natural number.

Now if your consider all the sequences that end with $1$, it is clear that cardinality of this will be more than the cardinality of power set of natural numbers. Over all I can conclude that such a set has to be uncountable.

Can some one please validate the proof especially the second part where I try to prove the cardinality?

Answer 1

The first part of your analysis is fine, but the second is wrong. Consider the set of sequences that are eventually $1$. Let $S_n$ be the set of these sequences that are $1$ for all but at most the first $n$ terms. There is an easy bijection between $S_n$ and $\Bbb N^n$: just throw away all of the ones after the $n$-th term. But $\Bbb N^n$ is countable for each $n\in\Bbb Z^+$, so each $S_n$ is countable, and therefore $\bigcup_{n\in\Bbb Z^+}S_n$, being the union of countably many countable sets, is also countable.

Now do this for each natural number, not just $1$; there are only countably many of them, so you still end up with only countably many Cauchy sequences.

Answer 2

Let $S$ be the set of Cauchy sequences of natural numbers. As you pointed out, any such sequence is eventually constant, hence we can write $$ S=\bigcup_{n=1}^{\infty}S_n $$ where $S_n$ is the set of sequences $\{a_k\}$ of natural numbers such that $a_k=a_n$ for all $k\geq n$.

Now there is a bijection $\mathbb{N}^n\to S_n$, hence $S_n$ is countable. So $S$ is a countable union of countable sets, hence is countable.