In mathematics, the least-upper-bound property (sometimes called completeness, supremum property or l.u.b. property)[1] is a fundamental property of the real numbers. More generally, a partially ordered setX has the least-upper-bound property if every non-empty subset of X with an upper bound has a least upper bound (supremum) in X. Not every (partially) ordered set has the least upper bound property. For example, the set of all rational numbers with its natural order does not have the least upper bound property.
More generally, one may define upper bound and least upper bound for any subset of a partially ordered setX, 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, the set Q of rational numbers does not have the least-upper-bound property under the usual order. For instance, the set
has an upper bound in Q, but does not have a least upper bound in Q (since the square root of two is irrational). The construction of the real numbers using Dedekind cuts takes advantage of this failure by defining the irrational numbers as the least upper bounds of certain subsets of the rationals.
It is possible to prove the least-upper-bound property using the assumption that every Cauchy sequence of real numbers converges. Let S be a nonempty set of real numbers. If S has exactly one element, then its only element is a least upper bound. So consider S with more than one element, and suppose that S has an upper bound B1. Since S is nonempty and has more than one element, there exists a real number A1 that is not an upper bound for S. Define sequences A1, A2, A3, ... and B1, B2, B3, ... recursively as follows:
Check whether (An + Bn) ⁄ 2 is an upper bound for S.
If it is, let An+1 = An and let Bn+1 = (An + Bn) ⁄ 2.
Otherwise there must be an element s in S so that s>(An + Bn) ⁄ 2. Let An+1 = s and let Bn+1 = Bn.
Then A1 ≤ A2 ≤ A3 ≤ ⋯ ≤ B3 ≤ B2 ≤ B1 and |An − Bn| → 0 as n → ∞. It follows that both sequences are Cauchy and have the same limit L, which must be the least upper bound for S.
Applications
The least-upper-bound property of R can be used to prove many of the main foundational theorems in real analysis.
Intermediate value theorem
Let f : [a, b] → R be a continuous function, and suppose that f (a) < 0 and f (b) > 0. In this case, the intermediate value theorem states that f must have a root in the interval [a, b]. This theorem can be proved by considering the set
S = {s ∈ [a, b] : f (x) < 0 for all x ≤ s} .
That is, S is the initial segment of [a, b] that takes negative values under f. Then b is an upper bound for S, and the least upper bound must be a root of f.
Bolzano–Weierstrass theorem
The Bolzano–Weierstrass theorem for R states that every sequencexn of real numbers in a closed interval [a, b] must have a convergent subsequence. This theorem can be proved by considering the set
S = {s ∈ [a, b] : s ≤ xn for infinitely many n}
Clearly,
, and S is not empty.
In addition, b is an upper bound for S, so S has a least upper bound c.
Then c must be a limit point of the sequence xn, and it follows that xn has a subsequence that converges to c.
Extreme value theorem
Let f : [a, b] → R be a continuous function and let M = sup f ([a, b]), where M = ∞ if f ([a, b]) has no upper bound. The extreme value theorem states that M is finite and f (c) = M for some c ∈ [a, b]. This can be proved by considering the set
S = {s ∈ [a, b] : sup f ([s, b]) = M} .
By definition of M, a ∈ S, and by its own definition, S is bounded by b.
If c is the least upper bound of S, then it follows from continuity that f (c) = M.
Heine–Borel theorem
Let [a, b] be a closed interval in R, and let {Uα} be a collection of open sets that covers[a, b]. Then the Heine–Borel theorem states that some finite subcollection of {Uα} covers [a, b] as well. This statement can be proved by considering the set
S = {s ∈ [a, b] : [a, s] can be covered by finitely many Uα} .
The set S obviously contains a, and is bounded by b by construction.
By the least-upper-bound property, S has a least upper bound c ∈ [a, b].
Hence, c is itself an element of some open set Uα, and it follows for c < b that [a, c + δ] can be covered by finitely many Uα for some sufficiently small δ > 0.
This proves that c + δ ∈ S and c is not an upper bound for S.
Consequently, c = b.
History
The importance of the least-upper-bound property was first recognized by Bernard Bolzano in his 1817 paper Rein analytischer Beweis des Lehrsatzes dass zwischen je zwey Werthen, die ein entgegengesetztes Resultat gewähren, wenigstens eine reelle Wurzel der Gleichung liege.[3]
^Bartle and Sherbert (2011) define the "completeness property" and say that it is also called the "supremum property". (p. 39)
^Willard says that an ordered space "X is Dedekind complete if every subset of X having an upper bound has a least upper bound." (pp. 124-5, Problem 17E.)