# Proving every bounded infinite set has a limit point without using Bolzano-Weierstrass.

I am trying to prove that every bounded, infinite set has at least one limit point through the concept of open covers and without using Bolzano Weierstrass. I have no idea where to start. So far I've only assumed that a set A is bounded and infinite, so for some real number M, the set A is entirely contained in the closed and bounded set [-M, M]. By Heine Bornel I know that each open cover for [-M, M] has a finite subcover, but honestly I don't know where to go from there. Am I in the right direction?

• What you are trying to prove is Bolzano-Weiertrass. And yes, your starting point is correct : assume that there is no limit point, and try to construct an open cover which cannot have a finite subcover. Mar 29, 2016 at 14:18

Suppose $S$ is an infinite set contained in the interval $I=[-M,M]$. If $S$ has no limit points in interval $I$, then for every $x\in [-M,M]$ there is $r_x>0$ such that the ball $B(x,r_x)$ contains only finitely many points of $S$.
Notice that the family $\{B(x,r_x):x\in [-M,M]\}$ is an open cover of $[-M,M]$, so there are points $x_1,\ldots,x_n$ such that
$$[-M,M]\subseteq \bigcup_{i=1}^n B(x_i,r_{x_i}).$$
Now, by piggeonhole principle, $S\cap B(x_i,r_{x_i})$ must be infinite for some $i=1,\ldots,n$. This contradicts the choice of $r_{x_i}$.
You can suppose the interval is $[0,1]$. If $A$ has no limit point, then for each $x\in[0,1]$ you can find its open neighborhood $U_x$ such that $U_x\cap A = \{x\}$. These from an open cover, so from compactness there are $U_{x_1}, \ldots, U_{x_n}$ covering $[0,1]$. This means $A = \{x_0, \ldots, x_n\}$, contradiction.