1. ## Show Compactness

Let (X,d) have the property that every open cover of X has a finite subcover.
I want to show X is compact.

My start so far:

If X is not compact there exist a sequence $(x^{(n)})^{\infty}_{n=1}$ with no limit points. Then for every $x \in X$ there exists a ball $B(x,\epsilon)$ containing x which contains at most finitely many elements of this sequence..

2. Originally Posted by bram kierkels
Let (X,d) have the property that every open cover of X has a finite subcover.
I want to show X is compact.

My start so far:

If X is not compact there exist a sequence $(x^{(n)})^{\infty}_{n=1}$ with no limit points. Then for every $x \in X$ there exists a ball $B(x,\epsilon)$ containing x which contains at most finitely many elements of this sequence..
Great! So this is an open cover of X; and now what? - By our assumption there exists a finite subcover, i.e. finitely many of those balls that cover X, each of which contains only finitely many elements of the infinite sequence $x_n$. - Isn't this a contradiction? - Methinks it is.