Proof
Consider yet again the set . Set . We are going to prove that . We'll rule out that and . The idea is for the former to show that it's opposite violates being the least upper bound, and for the latter that isn't an upper bound:
(): Assume the opposite. In search of some element in that doesn't bound, notice that if we choose some for some then:
But now assuming , allows us space, namely the part, to fit our new number and stay less than . Specifically, choose large enough such that:
Which implies that (we constructed this starting from this inequality and working backwards) so consequently:
Thus contradicting that is an upper bound for . Thus .