If S⊂R is a set, then x∈R is a cluster point if for every ϵ>0, the set (x−ϵ,x+ϵ)∩S∖{x} is not empty. That is, if there are points of S arbitrarily close to x. For example, S:={1/n:n∈N} has a unique (only one) cluster point 0, but 0∈/S. Prove the following version of the Bolzano--Weierstrass theorem:
Theorem. Let S⊂R be a bounded infinite set, then there exists at least one cluster point of S.
Hint: If S is infinite, then S contains a countably infinite subset. That is, there is a sequence {xn}n=1∞ of distinct numbers in S.