supremum n infimum

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Supremum and infimum 1 Introduction We have already seen two equivalent forms of the completenes completenesss axiom for the reals:: the montone convergence theorem and the statement that every Cauchy reals sequence has a limit. 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 ℝ but applies to arbitrary subsets of ℝ 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 ⊆ ℝ . We say that A is bounded above if there is b ∈ ℝ 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 ∈ ℝ 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 ⊆ ℝ .

We say that b ∈ ℝ 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 ∈ ℝ 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 ∈ ℚ : 0 ≤ x ≤ 2 doesn't have a maximum element, though it does have a minimum element, 0 . On the other hand B = x ∈ ℚ : 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 ∈ 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 ∈ 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 and a lower bound for A . So A does not have lea st upper bound or greatest lower bound.

∈ℝ

is both an upper bound

Let A = ℕ . 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 ∈ ℕ then inf ℕ = 0 . Otherwise you will have inf ℕ = 1 . Let A = ℤ . 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 ⊆ ℝ 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 ∈ ℚ 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 ∈ ℕ 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 ∈ ℝ and we need only take K ∈ ℕ 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 ∈ ℝ . 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 ∈ 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 ⊆ ℝ be non-empty and bounded below. Then there is a greatest lower bound of A , inf A . Proof

Let c ∈ ℝ 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 ∈ 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

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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 author contents glossary pdf previous next exercises This web page is available in xhtml, html and pdf. It is copyright and is one of Richard Kaye's Sequences and Series Web Pages. It may be copied under the terms of the Gnu Free Documentation Licence (http://www.gnu.org/copyleft/fdl.html). There is no warranty. Web page design and creation are by GLOSS.

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