# Let $\{a_n\}_n$ be a sequence and suppose $\{a_n\}_n$ isn't bounded above. Prove that there's a subsequence $\{a_{n_k} \}_k$ such that $a_{n_k} → ∞.$

I just wanted to see if my proof worked or not, or if there was any way to improve it.

Proof: In the case that $$\{a_n\}_n$$ is bounded below, we have $$a_n\to\infty$$ and so for any subsequence $$\{a_{n_k}\}_k$$ of $$\{a_n\}_n$$, we have $$a_{n_k}\to\infty$$. If the sequence is not bounded below, then take $$\{a_{n_k}\}_k$$ to be the sequence of positive, increasing terms of $$\{a_n\}_n$$. This sequence must exist, since $$\{a_n\}_n$$ is not bounded above, and since $$\{a_{n_k}\}_k$$ is positive and increasing, we have $$a_{n_k}\to\infty.$$ In either case, such subsequence exists.

If $${a_n}$$ is bounded below then $$a_n \to \infty$$? How do you get this?
Correct proof: For each $$k$$ choose $$n_k$$ such that $$n_k >n_{k-1}$$ and $$a_{n_k}>k$$. It is possible to find such integers inductively. This gives $$a_{n_k} \to \infty$$.
No, it is not correct. It is false that if $$(a_n)_{n\in\mathbb N}$$ is bounded below, then $$\lim_{n\to\infty}a_n=\infty$$. Take, for instance, $$a_n=n^{(-1)^n}$$.
Justa take, for each $$k\in\mathbb N$$ a $$n_k$$ such that $$a_{n_k}\geqslant k$$ and that $$n_k>n_{k-1}$$.