# Compactness with accumulation points

1. Oct 12, 2011

### sazanda

1. The problem statement, all variables and given/known data

Let K be a subset of R. Prove that if every sequence in K has an accumulation point, then K must be compact.

2. Relevant equations

I tried to proof it below. Am I on the right track?

3. The attempt at a solution

My intuition;

Let x_n be sequence in K whose accumulation point is x, then there is a sub-sequence.
x_n_k converges to x.
Since sub-sequence x_n_k converges to x, x_n_k is a Cauchy sequence.
We know that Cauchy sequences are bounded. Thus x_n is bounded. so K is bounded.

Since all accumulation points are in K so K is closed.

By Heine-Borel Thm, K is compact.

2. Oct 12, 2011

### PAllen

Not quite. I think you're better off showing the equivalent (each implies the other):

if K is not compact, there exists a sequence without an accumulation point.

3. Oct 12, 2011

### Fredrik

Staff Emeritus
Here you appear to be saying that if a sequence has a bounded subsequence, the sequence is bounded. Consider 1/2, 2, 1/3, 3, 1/4, 4, ...