calculus - Showing that an unbounded set has an unbounded sequence

Suppose I have an unbounded set $S$ of Real numbers.
I want to show that I can find a sequence $\{a_n\}$ in $S$ such that $\lim_{n \rightarrow \infty}a_n=\infty$.

Here is what I have so far:

Since $S$ is unbounded, either $S$ is unbounded above or $S$ is unbounded below.

Case 1: $S$ is not bounded above

If there is no sequence $\{a_n\}$ in $S$ such that $\lim_{n \rightarrow \infty}a_n=\infty$ then there must be a number $M$ such that $\forall n$ in any sequence $\{a_n\}$ in $S$, $a_n\leq M$. This implies that $S$ is bounded by $M$, a contradiction.

Case 2: $S$ is not bounded below

The same argument as Case 1.

It seems obvious to me that I can always find such sequence but then I am not satisfied with my explanation.

Any ideas? Thank you!

By the way, I'm trying to prove that a subset of the Real numbers is compact if and only if every sequence in subset has a sub-sequence converging to point in the subset.

I'm in the last part of my proof. I want to show that S is bounded and I am going to use the answer to this question to finish the proof. Thanks!

Answer

If you know that there is no sequence $\{a_n\}$ such that $\lim_{n\to\infty} a_n = +\infty$, how do you conclude that there is a number $M$ such that all sequences are bounded by the same number $M$? Doesn't that assume what you are trying to prove?

That said, what you are trying to prove is as follows : Suppose $S$ is not bounded above, then for any natural number $n \in \mathbb{N}$,

$$S\cap[n, \infty) \neq \emptyset$$
Hence, we may choose any number $a_n \in S\cap[n,\infty)$, and consider the sequence $\{a_n\}$. Can you show that this sequence must be going to $\infty$?