Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

An extract (an excerpt really) from

Theorem: $S \subset \mathbb{R}^n$ is compact $\iff $ closed and bounded


Suppose $S$ unbounded. Then $\forall n, \exists a_n \in S$ with $|a_n| > n$ with no subsequence of $(a_n)$ converging. So $S$ is not compact, a contradiction. So $S$ is bounded.

Can someone explain this step for me?

It says suppose the sequence is unbounded, then for each $n$, we have $|a_n| > n$.

Why are we bounding it away by $n$? Why the indices of the sequence must be equal to $n$? Why can't I have something like $|a_2| > 5$? Moreover, why $n$? Why not say $|a_n| > M$ for some big $M$?

share|cite|improve this question
To avoid future link rot, please copy down as much of the relevant information as possible into your question (and present link rot for that matter - I can't get the link to work currently...). At minimum, take a screenshot and include it as an image in your question. – Zev Chonoles Apr 13 '13 at 3:01

A set is called "$bounded$" if it isn't contained in a finite interval.Otherwise unbounded.

So a set may be unbounded with being bounded below like $[1,\infty)$ or bounded above like (-$\infty,1]$.

If for some n , there exist no $a_n\in S $ with |$a_n|>n$ that would mean S is bounded and

contained in the finite interval $[-n,n]$.

share|cite|improve this answer
Sorry, don't you mean $|a_n| \leq n$? – Hawk Jun 28 '13 at 2:33
No,--but it says : there exists " no " ... – Halil Duru Jul 2 '13 at 11:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.