# Sequence in $R$ with no convergent subsequence.

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 $$s_{n_1} = s_2, s_{n_2} = s_1, s_{n_3} = s_4, s_{n_4} = s_3, \dots$$

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…
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 $$n$$.