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Is it possible to have in a subset S from the set of Real numbers a finite number of upperbounds and yet have no least upperbound?
Jiminikiz
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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