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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.

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.

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter1: Fundamentals
Section1.7: Relations
Problem 18E: Let (A) be the power set of the nonempty set A, and let C denote a fixed subset of A. Define R on...
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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.
Transcribed Image Text: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
Transcribed Image Text: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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