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Let me rephrase it. You mean take a bounded subset of real numbers, S, and find a subset of all the upperbounds of S, say D, such that sup S is not in D?

If I get you right, then yes.

Take D := {a : a = sup s + n, n is natural and n < 4} so the first element is sup s + 1 > sup s, and the remaining two terms are even larger than the first one.

But I think I got you wrong, go through the Completeness Axiom.

That is for any two set A and B such that for all a in A, b in B, a <= b, and they share a maximum of one element, then there exist at least one number x such that a <= x <= b

In particular, A have a supremum, sup A <= x and B have an infimum inf B > = x

sup A <= inf B if they are equal then they must be x.

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