User:IssaRice/Taking inf and sup separately: Difference between revisions

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==Proof==
==Proof==


Let <math>a\in A</math> and <math>b\in B</math> be arbitrary. We have by hypothesis <math>a\geq b</math>. Thus, <math>a</math> is an upper bound of the set <math>B</math>, so taking the superemum over <math>b</math> we have <math>a \geq \sup(B)</math> (remember, <math>\sup(B)</math> is the ''least'' upper bound, whereas <math>a</math> is just another upper bound). Since <math>a</math> was arbitrary, we see that <math>\sup(B)</math> is a lower bound of the set <math>A</math>. Taking the infimum over <math>a</math>, we have <math>\inf(A) \geq \sup(B)</math>, as required.
Let <math>a\in A</math> and <math>b\in B</math> be arbitrary. We have by hypothesis <math>a\geq b</math>. Since <math>b</math> is arbitrary, we have that <math>a</math> is an upper bound of the set <math>B</math>, so taking the superemum over <math>b</math> we have <math>a \geq \sup(B)</math> (remember, <math>\sup(B)</math> is the ''least'' upper bound, whereas <math>a</math> is just another upper bound). Since <math>a</math> was arbitrary, we see that <math>\sup(B)</math> is a lower bound of the set <math>A</math>. Taking the infimum over <math>a</math>, we have <math>\inf(A) \geq \sup(B)</math>, as required.


==Applications==
==Applications==

Revision as of 22:33, 17 December 2018

This page describes a trick that is sometimes helpful in analysis.

Satement

Let A and B be bounded subsets of the real line. Suppose that for every a∈A and b∈B we have a≥b. Then inf(A)≥sup(B).

Proof

Let a∈A and b∈B be arbitrary. We have by hypothesis a≥b. Since b is arbitrary, we have that a is an upper bound of the set B, so taking the superemum over b we have a≥sup(B) (remember, sup(B) is the least upper bound, whereas a is just another upper bound). Since a was arbitrary, we see that sup(B) is a lower bound of the set A. Taking the infimum over a, we have inf(A)≥sup(B), as required.

Applications