Prove that a compact set in the real numbers is a bounded set.

Question

In order to prove this I can see that I should use the definition of compact. In order to do a direct proof. So far I have: Proof: Let x be an arbitrary point in the set and consider the collection of open balls G = {B_n(x): n = 1,2,3,4...}

Answer 1

I’ll include both the standard short indirect proof, as well as a more instructive direct proof for your intuition.

A direct instructive proof, following your lead considering balls:

Let $x\in\mathbb{R}$ (in fact this works for any metric space), denote the compact set $S\in\mathbb{R}$ and define $B(x,r)$ to be the open balls with centre $x$ and radius $r\in\mathbb{N}.$ These balls cover all of $\mathbb{R}$ (just take $r\to\infty$, same for general metric space with union of balls), but the compact set $S\in\mathbb{R}$ is inside $\mathbb{R}$ so it is covered by the balls as well. An equivalent definition for compactness of a set that is if each cover has a finite subcover for the set. This means that in particular a finite subset of the balls cover $S$. Take your bounding set to be the largest such ball.

The standard indirect proof:

Let $S\subset\mathbb{R}$ be compact and suppose for a contradiction that $S$ is not bounded. Then $$\forall\ n\in\mathbb{N}\ \exists\ a_n\in S\ \text{st } |a_n|>n.$$ But then no subsequence $\left(a_{n_i}\right)\subset \left( a_n\right)$ can converge, so by definition $S$ is not compact. This is the contradiction we require to conclude that $$S\subset\mathbb{R}\ \text{compact}\implies\ S\ \text{bounded.}$$

Stay safe!

Answer 2

First, some intuition. Let's think of a non bounded set, such as $\mathbb R$ with the standard Euclidean metric. Why is this not compact? well using the definition, if we consider the collection of open balls $\{B(x, 1) : x\in \mathbb R\}$, then clearly this collection is an open cover, but any finite subset of this open cover definitely wont cover all real numbers. This is the intuition, that for unbounded sets, there should be some infinite collection of open sets that cover the set, but these sets are too small individually to cover the set with just a finite number of them. (Note that not any open cover will do, we have to find the right open cover).

Ok, so now let's give a "direct" proof. Let $X$ be an unbounded set in the metric space $(M, \rho)$. If you use the open cover I described in the previous paragraph, the proof is more or less the same. I'll provide a different open cover.

Fix some $x\in X$. Let $G = \{B(x,n): n\in\mathbb N\}$. This collection of open balls is just balls of ever-increasing radii centered at $x$. $G$ is an open cover of $X$, since for any $y\in X$, one has $y\in B(x, N)$ where $N$ is a natural number greater than $\rho(x,y)$.

Let $H = \{B(x,n_1), B(x,n_2), \dots, B(x, n_k)\} \subseteq G$ be a finite subset of $G$. Let $M=\max \{n_1, n_2, \dots, n_k\}$. Then since $X$ is unbounded, there is a $z\in X$ with $\rho (x,z) > M$ and therefore $z\notin B(x,n_i), 1 \leq i \leq k$. Thus we have our result since any finite subset of $G$ cannot cover $X$