# Compactness; closed and bounded. Heine-Borel

An extract (an excerpt really) from http://pirate.shu.edu/~wachsmut/ira/topo/proofs/cpctbdd.html

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

Proof

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 $M$?

-
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

## 2 Answers

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 $[-n,n]$.

-
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

Let me TRY:

Let $\{a_k: k \in N\}$ is unbounded. Then for each $n \in N$, $n$ cannot be the bound of the sequence, i.e, there exists $k(n) \in N$, such that $|a_{k(n)}| > n$. Then we get a subsequece $\{a_{k(n)}\}$ which satisfies the condition.

Hope it be helful for you.

-
That's my question. I am asking why does $|a_n| > n$ mean it is unbounded? – Hawk Apr 13 '13 at 3:37