Can you help me prove A is compact?

Question

Show that a subset A of R^n is compact if, and only if, every continuous function from A to R is bounded.

If A is compact, then for every continuous function f from A to R, the set f(A) is compact and, therefore, bounded. So the conclusion is trivial. But the converse seems to be kinda more difficult to prove.

Thank you

Answer 1

OK, let's prove the converse. Let's use Heine Borel Theorem and a contra-positive argument.

Suppose A is not compact. Then, by Heine Borel Theorem, A is not bounded or is not closed.

Suppose A is not bounded. Then, for every M >0, there exists x in A such that ||x|| > M. Define f:A --> R by f(x) = ||x||. Then, f is clearly continuous and unbounded.

Now, suppose A is not closed. Then, A has an accumulation point a that does not belong to A. Define f:A-->R by f(x) = 1/||x - a||. Then, f is continuous on A. Since a is an accumulation point of A, for every M>0 we can find an x in A with ||x -a|| <1/M, so that f(x) = 1/||x - a|| > M. Since M is arbitrary, it follows f is unbounded.

We conclude that, if A is not compact, then there exists a continuous function f:A --> R which is unbounded. By contra-position, the assertion is proved.

Answer 2

If there is a sequence x_i from A converging to X,
and X does not belong to A, then

F(x) = 1/ [distance from x to X]

is continuous, but unbounded, because F(x_i) is unbounded.