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

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Let <math>A</math> and <math>B</math> be bounded subsets of the real line. Suppose that for every <math>a\in A</math> and <math>b\in B</math> we have <math>a\geq b</math>. Then <math>\inf(A)\geq \sup(B)</math>.
Let <math>A</math> and <math>B</math> be bounded subsets of the real line. Suppose that for every <math>a\in A</math> and <math>b\in B</math> we have <math>a\geq b</math>. Then <math>\inf(A)\geq \sup(B)</math>.
Actually, do <math>A</math> and <math>B</math> have to be bounded? I think they can even be empty!


==Proof==
==Proof==

Revision as of 22:36, 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 aA and bB we have ab. Then inf(A)sup(B).

Actually, do A and B have to be bounded? I think they can even be empty!

Proof

Let aA and bB be arbitrary. We have by hypothesis ab. Since b is arbitrary, we have that a is an upper bound of the set B, so taking the superemum over b we have asup(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