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 $a_{n_k} → ∞.$

Question

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.

Answer 1

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 $a_{n_k} \to \infty$.

Answer 2

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 $n_k>n_{k-1}$.