Assume that there exists a lower upper bound \sigma < s of A. Since (I_n)_{n\in\mathbb{N}} is a sequence of nested intervals, the interval lengths get arbitrarily small; in particular, there exists an interval with a length smaller than s-\sigma. But from s\in I_n one gets s-a_n and therefore a_n>\sigma. Following the rules of this construction, a_n would have to be an upper bound of A, contradicting property 2 of all sequences of nested intervals.
