Ex 4:4.) Let S be a non empty bounded set in IR.aj Let azo, and let S := {as: SES}. Prove thatinf (as) - a inf 5, sup (as) = a sups.b. Let beo and let bS={bs: SES}. Prove thatinf (b5) = b supS, sup (bs) =binf S. //

Let S be any -empty bounded set in a) Let , and let Since, is a bounded subset of , Sup and Inf…

Answered: Ex 4.) Let Sbe a non empty bounded set…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Ex 4:
4.) Let S be a non empty bounded set in IR.
aj Let azo, and let S := {as: SES}. Prove that
inf (as) - a inf 5, sup (as) = a sups.
b. Let beo and let bS={bs: SES}. Prove that
inf (b5) = b supS, sup (bs) =binf S. //

Expert Solution

Step 1: "Introduction to the solution"

Let S be any -empty bounded set in $straight real numbers.$

a) Let $a greater than 0$ , and let $a S space colon equals open curly brackets a s colon space s element of S close curly brackets$

Since, $S$ is a bounded subset of $straight real numbers$ , Sup $open parentheses S close parentheses$ and Inf $open parentheses S close parentheses$ exists.

Let $S u p left parenthesis S right parenthesis equals M comma space I n f left parenthesis S right parenthesis equals m$

Since, Sup(S) $equals M$ , it follows that $open parentheses 1 close parentheses$ $s less or equal than M comma for all s element of S....... left parenthesis 1 right parenthesis$

and $open parentheses 2 close parentheses$ $for all epsilon greater than 0 comma space$ there exists an element $y element of S$ such that $y greater than left parenthesis M minus epsilon over a right parenthesis.......... left parenthesis 2 right parenthesis$

Since,Inf(S)=M, it follows that (3) $m less or equal than s comma space for all s element of S........ left parenthesis 3 right parenthesis$

and $open parentheses 4 close parentheses$ $for all epsilon greater than 0 comma$ there exists an element $z element of S$ such that $z less than left parenthesis m plus epsilon over a right parenthesis........... left parenthesis 4 right parenthesis$