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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!

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  • $\begingroup$ What you said there is the usual definition of a compact set... $\endgroup$ – Karolis Juodelė Sep 27 '13 at 15:53
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    $\begingroup$ @KarolisJuodelė : There are many definitions of a compact set, and the one stated above is "sequential compactness", which is not always the same thing (it is in $\mathbb{R}$ however) $\endgroup$ – Prahlad Vaidyanathan Sep 27 '13 at 15:54
  • $\begingroup$ Yes, ''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'' is a definition of a compact set. But we must be able to show too that this definition is reasonable. $\endgroup$ – chowching Sep 27 '13 at 16:06
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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 $\infty$?

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  • $\begingroup$ Thanks for giving better answer than others.+1 $\endgroup$ – Silent Sep 27 '13 at 15:54
  • $\begingroup$ Okay I got it now. Thank you! $\endgroup$ – chowching Sep 27 '13 at 16:13
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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.

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  • $\begingroup$ I have this in my mind, thank you for putting it into words. Thanks! $\endgroup$ – chowching Sep 27 '13 at 16:11
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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 $S = \{ x\in\mathbb{R}\ |\ x < 0\}$.

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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 $\lim_n x_n = \infty$.

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