Supremum and infimum

1 Introduction

We have already seen two equivalent forms of the completeness axiom for the reals: the montone convergence theorem and the statement that every Cauchy sequence has a limit. The second of these is useful as it doesn't mention the order relation < and so applies to the complex numbers for instance. (It turns out that the set of complex numbers is also complete in the Cauchy sense.) This web page describes a third approach due to Dedekind. This uses the order relation < of but applies to arbitrary subsets of rather than sequences. There are a number of places (including results about the radius of convergence for power series, and several results in more advanced analysis) where Dedekind's approach is particularly helpful.

2 Supremum and infimum

We start with a straightforward definition similar to many others in this course. Read the definitions carefully, and note the use of and here rather than < and > .

Definition 2.1

Let A .

We say that A is bounded above if there is b such that x A   ( x b ) . The number b is called an upper bound for A .

We say that A is bounded below if there is c such that x A   ( x c ) . The number c is called an lower bound for A .

If we are interested in the best upper bound or best lower bound we need to consider the following.

Definition 2.2

Let A .

We say that b is a least upper bound for A if b is an upper bound for A and no c < b is also an upper bound. In other words if x A   ( x b ) and c < b   x A   ( x > c ) . The least upper bound of a set A may not exist, but if it does it is unique, because if we have two distince upper bounds c , d then one of these must be larger and so it cannot be a least upper bound. When it exists, the least upper bound of a set A is called the supremum of A and denoted sup A .

We say that c is a greatest lower bound for A if c is a lower bound for A and no d > c is also a lower bound. In other words if x A   ( x c ) and d > c   x A   ( x < d ) . The greatest lower bound of a set A may not exist, but if it does it is unique, and is called the infimum of A and denoted inf A .

Supremums and infimums are a bit like maximums and minimums for infinite sets A .

The problem with maximums and minimums is that they are only guaranteed to exist for finite sets A . For example, A = x : 0 x 2 doesn't have a maximum element, though it does have a minimum element, 0 . On the other hand B = x : 0 x 4 has both a maximum (2) and a minumum (0).

In fact, for a nonempty set A , the maximum element of A exists if and only if sup A exists and is an element of A . Similarly, the minimum element of A exists if and only if inf A exists and is an element of A .

Examples 2.1

Let A = 1 - 1 n : n . Then A has no largest element, i.e., max A doesn't exist, but sup A = 1 since 1 is an upper bound and any c < 1 is less than some 1 - 1 n by the Archimedean property. Note that sup A is not an element of A .

Let A = 1 n : n . Then A has no smallest element, i.e., min A doesn't exist, but inf A = 0 since 0 is a lower bound and any d > 0 is greater than some 1 n by the Archimedean property. Note that inf A is not an element of A .

Let A = [ 2 , 3 ] . In this case A does have largest and smallest elements, and sup A = 3 and inf A = 2 .

Let A be the empty set. Then by convention every b is both an upper bound and a lower bound for A . So A does not have least upper bound or greatest lower bound.

Let A = . Then A does not have any upper bound, by the Archimedean property. But A does have a lower bound, such as -1 . The greatest lower bound is determined by your convention on the set of natural numbers. If you prefer 0 then inf = 0 . Otherwise you will have inf = 1 .

Let A = . Then A does not have any upper bound nor any lower bound, by the Archimedean property again.

Theorem 2.1 (Completeness of reals, supremum form)

Let A be non-empty and bounded above. Then there is a least upper bound of A , sup A .

Proof

This is proved by a variation of a proof of an earlier result that every real number has a monotonic sequence of rationals converging to it.

We start by defining a sequence ( a n ) using induction on n . Start with any a 0 with some x A such that a 0 < x . This just uses the assumption that A is nonempty. Now inductively assume that a n is defined and a n < x for some x A , i.e., a n is not an upper bound for A . If a n + 1 n is an upper bound for A then let a n + 1 = a n . Otherwise, let a n + 1 = a n + k n where k = K 0 - 1 and K 0 is chosen to be the least natural number such that a n + K 0 n x for all x A . Such a K exists since A is bounded above by some b and we need only take K so that a n + K n b , using the Archimedean Property. So since there is some such K there must be a least such number, K 0 .

By construction, ( a n ) is a nondecreasing sequence of rationals and bounded above by b , an upper bound for A . It follows that ( a n ) converges to some l . We shall show that this l is the least upper bound of A .

Subproof

Suppose l is not an upper bound of A .

Subproof

Then there is x A such that l < x . But this gives a contradiction, for if n is such that x - l > 1 n we consider the n th stage of the construction, where we chose a n + 1 = a n + k n with k greatest so that there is some y A with a n + k n < y . But a n + k n l < x - 1 n so a n + k + 1 n l + 1 n < x contradicting this choice of k .

Thus l is after all an upper bound. To see it is the least upper bound, we again suppose otherwise.

Subproof

Suppose l is not the least upper bound of A .

Subproof

Then there is some m < l such that every x A has x m . This again is impossible. Since a n l , there must be some a n with m < a n l . (To see this, put ε = l - m and a n with a n - l < ε .) But by construction of a n there is always some x A with a n < x . Therefore m < x and m is not after all an upper bound for A .

This completes the proof.

Theorem 2.2 (Completeness of reals, infimum form)

Let A be non-empty and bounded below. Then there is a greatest lower bound of A , inf A .

Proof

Let c be a lower bound for A and let B = - x : x A . Then b = - c is an upper bound for B and hence by the previous result B has a least upper bound, l . It follows easily that - l is a greatest lower bound for A , for if x A then l - x as - x B so - l x , and if m > - l then - m < l so - m is not an upper bound for B , so there is x B with x > - m hence - x < m and clearly - x A .

These results are equivalent to the monotone convergence theorem. To see this, suppose ( a n ) is a bounded nondecreasing sequence. Then let A = a n : n . By the fact that the sequence is bounded and by completeness theorem above, l = sup A exists, and it is a nice exercise to show that a n l as n .

3 Summary

You have seen another form of the completeness axiom for the reals, which is useful in that it doesn't involve any sequences and may be applied to subsets of .

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