Can you please provide a bit more details to the first part of the proof where L=SupR. How do you know that sup R is a lower bound for S?How do you know that it is greater than or equal to every other lower bound? Suppose that SCR is a non-empty set which is bounded below. Definea set R byR= {bER| b is a lower bound for S}.Prove that sup R = inf S. Given that,SCIRis a Non-empty set which is bounded below.a set R by :{bEIR | b is a lower bound for sint s.We have to Prove, Sup RSince,and defineThenandS is Bounded below, so I bo E IRbo is a lower bound for s.R isNon emptybo & RMore oven9andR =XERi.e.This= R isEveryhasThus, By Supremum Property of IR gaR hasandx ≤ 3is True for all XERBoundedaboveb ≤ L ;L = Sup RandNon empty bonded above Subset of IRSupremum in IR.a supremum.VBER(11) since,everyandL = Sup RThis, It shows that,L =Hence, by (*) & (**)SE SSESWex is lower bound for 5.getSup R = inf 5SaySneh thatL is a lower bound of SL any lower bound of S.inf S(Proved)Las Lisот uрреrfor R.is an upper bound of RL≤8 for everyse s.*bound

Answered: Image /qna-images/answer/394da6ee-6537-489d-b65e-e80301752192.jpg

Answered: Suppose that SCR is a non-empty set…

Suppose that SCR is a non-empty set which is bounded below. Define a set R by R= {bER| b is a lower bound for S}. Prove that sup R = inf S.

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

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Can you please provide a bit more details to the first part of the proof where L=SupR.

How do you know that sup R is a lower bound for S?

How do you know that it is greater than or equal to every other lower bound?

Suppose that SCR is a non-empty set which is bounded below. Define
a set R by
R= {bER| b is a lower bound for S}.
Prove that sup R = inf S.

Given that,
SCIR
is a Non-empty set which is bounded below.
a set R by :
{
bEIR | b is a lower bound for s
int s.
We have to Prove, Sup R
Since,
and define
Then
and
S is Bounded below, so I bo E IR
bo is a lower bound for s.
R is
Non empty
bo & R
More oven
9
and
R =
XER
i.e.
This
= R is
Every
has
Thus, By Supremum Property of IR g
a
R has
and
x ≤ 3
is True for all XER
Bounded
above
b ≤ L ;
L = Sup R
and
Non empty bonded above Subset of IR
Supremum in IR.
a supremum.
VBER
(11) since,
every
and
L = Sup R
This, It shows that,
L =
Hence, by (*) & (**)
SE S
SES
We
x is lower bound for 5.
get
Sup R = inf 5
Say
Sneh that
L is a lower bound of S
L any lower bound of S.
inf S
(Proved)
L
as Lis
от uрреr
for R.
is an upper bound of R
L≤8 for every
se s.
*
bound

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