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>. | 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 and be bounded subsets of the real line. Suppose that for every and we have . Then .
Proof
Let and be arbitrary. We have by hypothesis . Since is arbitrary, we have that is an upper bound of the set , so taking the superemum over we have (remember, is the least upper bound, whereas is just another upper bound). Since was arbitrary, we see that is a lower bound of the set . Taking the infimum over , we have , as required.