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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 .
Similarly, the set of integers has the least-upper-bound property; if is a nonempty subset of and there is some number such that every element of is less than or equal to , then there is a least upper bound for , an integer that is an upper bound for and is less than or equal to every other upper bound for .
The theorem states that if you have an infinite matrix of non-negative real numbers , such that the rows are weakly increasing and each is bounded , where the bounds are summable < then, for each column, the non decreasing column sums , are bounded hence convergent, and the limit of the column sums is equal to the sum of the "limit column ...
There may be elements, besides the least element, that have no predecessor (see § Natural numbers below for an example). A well-ordered set S contains for every subset T with an upper bound a least upper bound, namely the least element of the subset of all upper bounds of T in S. If ≤ is a non-strict well ordering, then < is a strict well ...
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 ...
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 ...
The rational number line Q does not have the least upper bound property. An example is the subset of rational numbers = {<}. This set has an upper bound. However, this set has no least upper bound in Q: the least upper bound as a subset of the reals would be √2, but it does not exist in Q.