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More generally, one may define upper bound and least upper bound for any subset of a partially ordered set X, with “real number” replaced by “element of X ”. In this case, we say that X has the least-upper-bound property if every non-empty subset of X with an upper bound has a least upper bound in X.
For example, if the domain is the set of all real numbers, one can assert in first-order logic the existence of an additive inverse of each real number by writing ∀x ∃y (x + y = 0) but one needs second-order logic to assert the least-upper-bound property for sets of real numbers, which states that every bounded, nonempty set of 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 ...
That is, the least upper bound (sup or supremum) of the empty set is negative infinity, while the greatest lower bound (inf or infimum) is positive infinity. By analogy with the above, in the domain of the extended reals, negative infinity is the identity element for the maximum and supremum operators, while positive infinity is the identity ...
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.
The notion of complete lattice generalizes the least-upper-bound property of the reals. One completion of S is the set of its downwardly closed subsets, ordered by inclusion . A related completion that preserves all existing sups and infs of S is obtained by the following construction: For each subset A of S , let A u denote the set of upper ...
A totally ordered set is said to be complete if every nonempty subset that has an upper bound, has a least upper bound. For example, the set of real numbers R is complete but the set of rational numbers Q is not. In other words, the various concepts of completeness (not to be confused with being "total") do not carry over to restrictions.
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 bound for f.