Sequence in $R$ with no convergent subsequence.

Question

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 $

Answer 1

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…

Answer 2

Note that in the first line of the proof, it should be that for every $n$ there exists an $N\in{\mathbb{N}}$ s.t. $|s_N|>n$.

e.g. take the unbounded sequence $s_n=\frac{n}{2}$. clearly $s_n$ is unbounded but $|s_n|=\frac{n}{2}\not>n$ for all $n$.