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!
 A: 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 $\infty$?
A: I would say that you're argument does not constitute a proof, but is just a reformulation of the claim.
For a proof, you'd have to construct a sequence $\{ a_n \}$ with $\lim_{n \to \infty} a_n = \infty$. Let's assume we're in case 1: $S$ is unbounded above. Since $S$ is unbounded, there is an $a_1 \in S$ with $a_1 > 1$. Since $S$ is unbounded, there is an $a_2 \in S$ with $a_2 > a_1 + 1$. Since $S$ is unbounded, there is an $a_3 \in S$ with $a_3 > a_2 + 1$. ... This gives a sequence $\{ a_n \}$ with the property that $a_n > n$ for all $n$.
Even shorter, you could simply pick each $a_n \in S$ such that $a_n > n$ to start with; also possible because $S$ is unbounded.
A: Case 2 does not seem to be correct. If we have a set unbounded below, it does not mean you have a sequence going to $\infty$. For example $S = \{ x\in\mathbb{R}\ |\ x < 0\}$.
A: Suppose a set $S$ is unbounded above. This means for all $M>0$, there exists some $x \in S$ such that $x > M$.
So, choose $n \in \mathbb{N}$, and let $x_n \in S$ be a number satisfying $x_n > n$. Then we must have $\lim_n x_n = \infty$.
