Thread: Could someone help me prove this.

Here's the problem

Show that S, subset R, is compact iff every infinite subset of S contains at least one accumulation point.

I've proven the => part:

Suppose that S was compact. THen, let T be an infinite subset of S. Then, since every subset of a compact set is compact, T is compact. Hence, T is bounded. Thus by the Bolzano-Weierstrass theorem, T has an accumulation point.

But, I'm having a good deal of difficulty proving the converse.

every subset of a compact set is compact

This is not true, but I'll leave the counterexample to you (it's not difficult).

Originally Posted by Chris11

Show that S, subset R, is compact iff every infinite subset of S contains at least one accumulation point.

For the idea is to assume there exists a sequence with no accumulation points in , then there exist intervals with center at that contain (from ) only (why?). Clearly is closed (why?) so it is compact (why?), but can the cover have a finite subcover? (I picked a sequence because it's easier to visualize, but T need not be countable)

For just use Heine Borel.