Suppose B is a compact set in Real numbers and A is a subset of B. Prove that if A is an infinite set, then A has a cluster point.

Question

So far I have that Proof: Suppose $B$ is a compact set in real numbers and $A\subseteq B$. I will do proof by contradiction. Assume $A$ is an infinite set and $A$ does not have a cluster point in K.

I’m not sure the next step I know I want to show a contradiction but I’m not sure where it will come from yet. Is it sufficient to use proof by contradiction.

Answer 1

If $A$ has no cluster point, then, for each $b\in B$, there is some $\varepsilon_b>0$ such that $(b-\varepsilon_b,b+\varepsilon_b)$ either contains no point of $A$ or, if it does, then that point is $b$ itself. Then$$B\subset\bigcup_{b\in B}(b-\varepsilon_b,b+\varepsilon_b),$$and therefore $\{(b-\varepsilon_b,b+\varepsilon_b)\mid b\in B\}$ is an open cover of $B$. Since $B$ is compact, it has a finite subcover. But then, since each $(b-\varepsilon_b,b+\varepsilon_b)$ contains, at most, one element of $A$, $A$ would be finite.

Answer 2

I think this statement is not true. For example $B=[0,1]$ but $A=\{1,1/2,...,1/n,...\}$.