Limit of a sequence bounded below but has no cluster points

Question

Question is this: Let $a_n$ be a sequence of real numbers. Prove that if $a_n$ is bounded below and has no cluster points then $a_n$ → ∞.

I could not really find a way to prove it. Could you give me a hand?

Regards, Amadeus

Answer 1

HINT: If $a_n\not\to\infty$, then the sequence has an infinite subsequence that is bounded above. (Why?) That subsequence is actually a bounded sequence. What do you know about cluster points of bounded sequences?

Added: Suppose that $a_n\not\to\infty$. Then there is some $M$ such that for all $m\in\Bbb N$ there is a $k\ge m$ with $a_k<M$. Let $A=\{k\in\Bbb N:a_k<M\}$ clearly $A$ is infinite, so we can list it as $A=\{n_k:k\in\Bbb N\}$ in such a way that $n_0<n_1<n_2<\ldots~$. We now have a subsequence $\sigma=\langle a_{n_k}:k\in\Bbb N\rangle$ of the original sequence.

The original sequence is bounded below by hypothesis, so $\sigma$ is also bounded below, and by construction $\sigma$ is bounded above by $M$. Being a bounded sequence in $\Bbb R$, $\sigma$ has a convergent subsequence $\sigma'=\langle a_{n_{k_i}}:i\in\Bbb N\rangle$; let $x$ be the limit of $\sigma'$.

Can each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|x-x_{n_{k_i}}|<\epsilon$ whenever $i\ge m_\epsilon$; that’s just the definition of convergence. But that means that for each $\epsilon>0$, infinitely many terms of the original sequence are in the interval $(x-\epsilon,x+\epsilon)$, i.e., $x$ is a cluster point of the original sequence.

We’ve shown that if $a_n\not\to\infty$, then the sequence has a cluster point; it follows immediately that if it has no cluster point, then $a_n\to\infty$, by taking the contrapositive.