# A subset in R^n is bounded if and only if it is totally bounded.

1. Jan 7, 2014

### bobby2k

1. The problem statement, all variables and given/known data
Prove that a subset in R^n, where n is a finite number, is bounded if and only if it is totally bounded.

2. Relevant equations
If A is the subset, A is bounded if there is a point b in R^n such that d(x,b)<= K, for a every x in A.

A is totally bounded if for every e> 0, there is a finite number of ball with radius e, that covers A.

3. The attempt at a solution

totally bounded -> bounded

Chose a number e, there is a finite number P of open balls that cover A, each with radius e.
Chose an element b in A. s=sup{d(b,l): l is the center of the balls} must exist since we only have finite amount of balls. Then the max distance from b to another point in A is s+e, and hence A is bounded.
Is this proof correct for this implication?

bounded -> totally bounded
There is an element in R^n such that for any element x in A d(x,b) <= K, for a real number K.
I do not see how to proceed here?
Since Q^n is dense I guess that if we make balls with radius epsilon around every element in R^n that is rational we have a countable number of balls that cover A?, then the problem is to reduce it to a finite number of balls?

Last edited: Jan 7, 2014
2. Jan 7, 2014

### Mandelbroth

Since $A$ is bounded, can we make $A\subseteq B_K$, where $B_K=\{x\in\mathbb{R}^n:d(b,x)\leq K\}$?