Consider the set $S$ consisting of all cauchy sequences $\{a_n\}_{n \in \mathbb{N}}$ with $ a_n \in \mathbb{N}$ for all $n$. Is the set $S$ countable? Justify your Answer.
Attempt
Cauchy $\rightarrow$ Convergent.Let sequence $\{ a_n \}$ converge to $N$
Thus for any $\epsilon >0$ there exist $N_0 \in \mathbb{N}$ such that $|a_n-N|$ for all $n>N_0$. (DEFINITION 1)
Choose $\epsilon=0.001$ Because any two natural numbers differ by atleast $1$ We must have all $a_n=N$ for all $n>N_0$.
Define $D_{K}=\{$Sequences where $N_0=K \}$ (REFER DEFINITION 1 for $N_0$
$S=D_1 \cup D_2....... \cup D_K......$
Further note that each set $D_K$ where $N_0=K$ has cardinality$ \mathbb{N} \times \mathbb{N} \times \mathbb{N} ...... \times \mathbb{N} $
No. Of $\mathbb{N} $ is K+1.
Each set $D_K$ is countable because its product of finitely many $\mathbb{N} $.
Hence countable sum of countable set is countable hence S is countable.
The idea of this proof is very good. The proof itself is quite close to correct. The one point where I'd be more careful is the choice of $N_0$: as you've written it, the choice of $N_0$ is not unique. Perhaps try defining $D_K$ as the set of sequences for which $K$ is the least integer satisfying $|a_n-N|<0.001$ for all $n>K$