Supremum

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In mathematics, a supremum is a largest possible quantity (subject to given conditions), if such a thing exists; or otherwise, within a larger set of quantities, it is a minimal larger choice (if such exists). For example, under different tax systems there might be a 'largest' percentage tax rate that anyone would have to pay; and this could mean that someone actually paid at that rate, or it might refer instead to the top rate that limits the percentage a very high earner might pay.

Table of contents

1 Supremum of a set of real numbers
2 Supremum of a poset
3 Comparison with maximum
4 Least-upper-bound property

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 { 1, 2, 3 } = 3

sup { x in R : 0 < x < 1 } = sup { x in R : 0 ≤ x ≤ 1 } = 1

sup { x in Q : x² < 2 } = √2

sup { (-1)ⁿ − 1/n : n = 1, 2, 3, ...} = 1

sup Z = +∞

Note that the supremum of S may or may not belong to S. If the supremum value belongs to the set then we can say there is a largest (or maximum) element in the 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 ε > 0, we must find an x in S with x ≥ a − ε.

In functional analysis, one often considers the supremum norm of a bounded function f : X -> R (or C); it is defined as

and gives rise to several important Banach spaces.

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

Supremum of a poset

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 it follows that u₁ = u₂.

In an arbitrary partially ordered set, there may exist subsets which don't have a supremum. In a lattice every nonempty finite subset has a supremum, and in a complete lattice every subset has a supremum.

Comparison with maximum

The difference between the supremum of a set and the maximum element of a set may not be immediately obvious. The difference is exemplified by the set of negative real numbers. This set has no maximum element; for every element of the set, there is another, larger element. For example, for any negative real number x, there is a negative real number x/2, which is greater. But, although the negative real numbers has no maximum, it does -- like all bounded non-empty sets of real numbers -- have a supremum; namely, 0.

If, on the other hand, a set does contain a maximum element, then this maximum element is also the supremum of the set.

Least-upper-bound property

If an ordered set S has the property that every nonempty subset of S has 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 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 q² − 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.