How is the following proof is correct?
Claim: If $S$ is a non-empty unbounded subset of R, then there exists a sequence $(s_n)$ with
values in $S$ which has no convergent subsequences.
Proof: Since $S$ is unbounded, for all $n$, there exists $s_n \in S$ so that $|s_n| > n$ (since, if not $S$ would be bounded above by $n$ and bounded below by $-n$, hence bounded).
Suppose that $(s_{n_k})$ is a convergent subsequence of $(s_n)$. Then $(s_{n_k})$ is
bounded. So, there exists $M$ such that $|s_{n_k}
| \leq M$ for all $k \in N$. However, if we choose $k \in N$
so that $k \geq M$, then $|s_{n_k}
| > n_k ≥ k ≥ M$, so $|s_{n_k}| > M$. Contradiction.
This seems to me erronenous. Suppose my subsequence were that $s_{n_1} = s_2, s_{n_2} = s_1, s_{n_3} = s_4, s_{n_4} = s_3, \dots $
Best Answer
That proof is correct. Don't forget that, in order that $(s_{n_k})_{k\in\Bbb N}$ is a subsequence, the sequence $(n_k)_{k\in\Bbb N}$ must be a strictly increasing sequence of natural numbers. Therefore, yes, $n_1\geqslant1$, $n_2\geqslant 2$, and so on…