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In mathematics, the supremum of an ordered set S is the least element that is greater than or equal to each element of S. Consequently, it is also referred to as the least upper bound (also lub and LUB). The supremum may, or may not, belong to the set S. If S contains a greatest element, then that element is the supremum; and if not, then the supremum does not belong to the set.

Suprema are often considered for subsets of real numbers, rational numbers, or any other well-known mathematical structures for which it is immediately clear what it means for an element to be "greater-or-equal" than another element. Nonetheless, the definition generalizes easily to the more abstract setting of order theory where one considers arbitrary partially ordered sets.

In any case, suprema must not be confused with minimal upper bounds, or with maximal or greatest elements. Some notes on these issues follow below.

    Supremum
        Supremum of a set of real numbers
            Approximation property
            Additive property
            Comparison property
        Suprema within partially ordered sets
            Greatest elements
            Maximal elements
            Minimal upper bounds
        Least-upper-bound property
        See also

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Supremum of a set of real numbers

In analysis the supremum or least upper bound of a set S of real numbers is denoted by sup(S) and is defined to be the smallest real number that is greater than or equal to every number in S. An important property of the real numbers is its completeness: every nonempty set of real numbers that is bounded above has a supremum. If, in addition, we define sup(S) = −∞ when S is empty and sup(S) = +∞ when S is not bounded above, then every set of real numbers has a supremum (see extended real number line).

Examples:

$sup = 3,$

$sup = sup = 1,$

$sup = sqrt,$

$sup left = 1$

$sup mathbb = infty,$

$sup = sup(A) + sup(B),$

$sup varnothing = -infty,$

The supremum of S may or may not belong to S. In particular, note the third example where the supremum of a set of rationals is irrational (which means that the rationals are incomplete). However, if the supremum value belongs to the set then it is the greatest element in the set. The term maximal element is also synonymous as long as one deals with real numbers or any other totally ordered set.

Since sup(S) is the least upper bound, to show that sup(S) ≤ a, one only has to show that a itself is an upper bound for S, i.e. one only has to show that x ≤ a for all x in S. Showing that sup(S) ≥ a is a bit harder: for any b < a, we must find an x in S with x ≥ b.

In functional analysis, one often considers the supremum norm (also sometimes referred to as the uniform norm) of a bounded function f

X -> R (or C); it is defined as

$|f|_=mbox$

and gives rise to several important Banach spaces.

See also: infimum or greatest lower bound, limit superior.

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Approximation property

Let S be a nonempty set of real numbers with a supremum,
say b = sup S. Then for every a < b there is some x in S such that

$a$

Proof:

First of all,

xleq b

for all x in S.
If we had

x leq a

for every x in S, then

a

would be an upper bound for S smaller than the least upper bound. Therefore

x >a

for at least one x in S.

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Additive property

Given nonempty subsets A and B of R, let C denote the set

$C=.$

If each of A and B has a supremum, then C has a supremum and
sup C = sup A + sup B.

proof:

Let a = sup A, b = sup B. If

zin C

then z = x + y, where

xin A

yin B

,so

z=x+yleq a+b

.
Hence a + b is an upper bound for C, so C has a supremum, say c = sup C, and

cleq a+b

. We show next that

a+bleq c

.
Choose any z > 0. By the approximation property, there is an x in A and a y in B such that a − z < x and b − z < y. Adding these inequalities we find

a+b-2z . Thus,

a + b < c + 2z for every z > 0 so

a+bleq c

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Comparison property

Given nonempty subsets S and T of R such that

sleq t

for every s in S and t in T. If T has a supremum then S has a supremum and

sup Sleq sup T

.

proof:

Let c = sup T. For

sleq tleq c

for every s in S and t in T, S is bounded above, thus S has a supremum. Let d = sup S.
By the approximation property, there is an s in S such that d − z < s for any z > 0. Therefore d − z <

sleq tleq c

. Because this holds for all z > 0, this implies that

dleq c

.

Lemma: Given real numbers a and b such that a < b + z for every z > 0. Then

aleq b

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Suprema within partially ordered sets

Least upper bounds are important concepts in order theory, where they are also called joins (especially in lattice theory). As in the special case treated above, a supremum of a given set is just the least element of the set of its upper bounds, provided that such an element exists.

Formally, we have: For subsets S of arbitrary partially ordered sets (P, ≤), a supremum or least upper bound of S is an element u in P such that

x ≤ u for all x in S, and

for any v in P such that x ≤ v for all x in S it holds that u ≤ v.

It can easily be shown that, if S has a supremum, then the supremum is unique: if u₁ and u₂ are both suprema of S then it follows that u₁ ≤ u₂ and u₂ ≤ u₁, and since ≤ is antisymmetric, one finds that u₁ = u₂. The dual concept of supremum, the greatest lower bound, is called infimum and is also known as meet.

If the supremum of a set S exists, it can be denoted as sup(S) or, which is more common in order theory, by

vee

S. Likewise, infima are denoted by inf(S) or

wedge

S.

Subsets of a partially ordered set may well fail to have a supremum, even if they have upper bounds. Some discussion on this is provided in the sections below, where the difference between suprema, maximal elements, and minimal upper bounds is stressed. As a consequence of the possible absence of suprema, classes of partially ordered sets for which certain types of subsets are guaranteed to have least upper bound become especially interesting. This leads to the consideration of so-called completeness properties and to numerous definitions of special partially ordered sets.

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Greatest elements

The difference between the supremum of a set and the greatest element of a set may not be immediately obvious. The difference is exemplified by the set of negative real numbers. Since 0 is not a negative number, this set has no greatest element: for every element of the set, there is another, larger element. For instance, for any negative real number x, there is a negative real number x/2, which is greater. On the other hand, the upper bounds of the set of negative reals as a subset of the real numbers obviously constitute of all real numbers greater than or equal to 0. Hence, 0 is the least upper bound of the negative reals.

In general, this situation occurs for all subsets that do not contain a greatest element. In contrast, if a set does contain a greatest element, then it also has a supremum given by the greatest element.

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Maximal elements

For an example where there are no greatest but still some maximal elements, consider the set of all subsets of the set of natural numbers (the powerset). We take the usual subset inclusion as an ordering, i.e. a set is greater than another set if it contains all elements of the other set. Now consider the set S of all sets that contain at most ten natural numbers. The set S has many maximal elements, i.e. elements for which there is no greater element. In fact, all sets with ten elements are maximal. However, the supremum of S is the (only and therefore least) set which contains all natural numbers. One can compute least upper bounds of an element of a powerset (i.e. a set of sets) by just taking the union of its elements.

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Minimal upper bounds

Finally, a set may have many minimal upper bounds without having a least upper bound. Minimal upper bounds are those upper bounds for which there is no strictly smaller element that also is an upper bound. This does not say that each minimal upper bound is smaller than all other upper bounds, it merely is not greater. Of course this is only possible when the given order is not a total one (like the real numbers above).

As an example, let S be the set of all finite subsets of natural numbers and consider the partially ordered set obtained by taking all sets from S together with the set of integers Z and the set of positive real numbers R+, ordered by subset inclusion as above. Then clearly both Z and R+ are greater than all finite sets of natural numbers. Yet, neither is R+ smaller than Z nor is the converse true: both sets are minimal upper bounds but none is a supremum.

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Least-upper-bound property

The least-upper-bound property is an example of the aforementioned completeness properties which is typical for the set of real numbers.

If an ordered set S has the property that every nonempty subset of S having an upper bound also has a least upper bound, then S is said to have the least-upper-bound property. As noted above, the set R of all real numbers has the least-upper-bound property. Similarly, the set Z of integers has the least-upper-bound property; if S is a nonempty subset of Z and there is some number n such that every element s of S is less than or equal to n, then there is a least upper bound u for S, an integer that is an upper bound for S and is less than or equal to every other upper bound for S.

An example of a set that lacks the least-upper-bound property is Q, the set of rational numbers. Let S be the set of all rational numbers q such that q² < 2. Then S has an upper bound (1000, for example, or 6) but no least upper bound in Q. For suppose p ∈ Q is an upper bound for S, so p² > 2. Then q = (2p+2)/(p + 2) is also an upper bound for S, and q < p. (To see this, note that q = p − (p² − 2)/(p + 2), and that p² − 2 is positive.)

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-upper-bound of the set.

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