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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.
There is a corresponding greatest-lower-bound property; an ordered set possesses the greatest-lower-bound property if and only if it also possesses the least-upper-bound property; the least-upper-bound of the set of lower bounds of a set is the greatest-lower-bound, and the greatest-lower-bound of the set of upper bounds of a set is the least ...
If (,) is a partially ordered set, such that each pair of elements in has a meet, then indeed = if and only if , since in the latter case indeed is a lower bound of , and since is the greatest lower bound if and only if it is a lower bound. Thus, the partial order defined by the meet in the universal algebra approach coincides with the original ...
Thus, the infimum or meet of a collection of subsets is the greatest lower bound while the supremum or join is the least upper bound. In this context, the inner limit, lim inf X n, is the largest meeting of tails of the sequence, and the outer limit, lim sup X n, is the smallest joining of tails of the sequence. The following makes this precise.
Part of what this argument shows is that there is a least upper bound of the sequence 0.9, 0.99, 0.999, etc.: the smallest number that is greater than all of the terms of the sequence. One of the axioms of the real number system is the completeness axiom, which states that every bounded sequence has a least upper bound.
For example, the set of natural numbers has no maximum, though it has a minimum. If an infinite chain S is bounded, then the closure Cl(S) of the set occasionally has a minimum and a maximum, in which case they are called the greatest lower bound and the least upper bound of the set S, respectively.
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 ...
Proof of the Extreme Value Theorem. By the boundedness theorem, f is bounded from above, hence, by the Dedekind-completeness of the real numbers, the least upper bound (supremum) M of f exists. It is necessary to find a point d in [a, b] such that M = f(d). Let n be a natural number. As M is the least upper bound, M – 1/n is not an upper ...