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

Revision as of 22:30, 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 $a \in A$ and $b \in B$ we have $a \geq b$ . Then $inf (A) \geq sup (B)$ .

Proof

Let $a \in A$ and $b \in B$ be arbitrary. We have by hypothesis $a \geq b$ . Thus, $a$ is an upper bound of the set $B$ , so taking the superemum over $b$ we have $a \geq sup (B)$ (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 $inf (A) \geq sup (B)$ , as required.

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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>.
+==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.
 ==Applications==