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A real number x is the least upper bound (or supremum) for S if x is an upper bound for S and x ≤ y for every upper bound y of S. The least-upper-bound property states that any non-empty set of real numbers that has an upper bound must have a least upper bound in real numbers.
13934 and other numbers x such that x ≥ 13934 would be an upper bound for S. The set S = {42} has 42 as both an upper bound and a lower bound; all other numbers are either an upper bound or a lower bound for that S. Every subset of the natural numbers has a lower bound since the natural numbers have a least element (0 or 1, depending on ...
In its simplest form, it says that a non-decreasing bounded-above sequence of real numbers ... converges to its smallest upper bound, its supremum. Likewise, a non-increasing bounded-below sequence converges to its largest lower bound, its infimum. In particular, infinite sums of non-negative numbers converge to the supremum of the partial sums ...
This defines two Cauchy sequences of rationals, and so the real numbers l = (l n) and u = (u n). It is easy to prove, by induction on n that u n is an upper bound for S for all n and l n is never an upper bound for S for any n. Thus u is an upper bound for S. To see that it is a least upper bound, notice that the limit of (u n − l n) is 0 ...
In this poset, 60 is an upper bound (though not a least upper bound) of the subset {,,,}, which does not have any lower bound (since 1 is not in the poset); on the other hand 2 is a lower bound of the subset of powers of 2, which does not have any upper bound. If the number 0 is included, this will be the greatest element, since this is a ...
In mathematical analysis, limit superior and limit inferior are important tools for studying sequences of real numbers.Since the supremum and infimum of an unbounded set of real numbers may not exist (the reals are not a complete lattice), it is convenient to consider sequences in the affinely extended real number system: we add the positive and negative infinities to the real line to give the ...
Following the rules of this construction, would have to be an upper bound of , contradicting property 2 of all sequences of nested intervals. In two steps, it has been shown that s {\displaystyle s} is an upper bound of A {\displaystyle A} and that a lower upper bound cannot exist.
If T is the empty set, then {v} is an upper bound for T in P. Suppose then that T is non-empty. We need to show that T has an upper bound, that is, there exists a linearly independent subset B of V containing all the members of T. Take B to be the union of all the sets in T. We wish to show that B is an upper bound for T in P.