# Thread: Could someone help me prove this.

1. ## 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.

2. 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 $\displaystyle \Rightarrow$ the idea is to assume there exists a sequence $\displaystyle T=(x_k)_k$ with no accumulation points in $\displaystyle S$, then there exist intervals $\displaystyle U_{k}$ with center at $\displaystyle x_k$ that contain (from $\displaystyle T$) only $\displaystyle x_k$ (why?). Clearly $\displaystyle T$ is closed (why?) so it is compact (why?), but can the cover $\displaystyle \{ U_k \}$ have a finite subcover? (I picked a sequence because it's easier to visualize, but T need not be countable)

For $\displaystyle \Leftarrow$ just use Heine Borel.