What is the infimum of b - a for b>a?

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Homework Help Overview

The problem involves finding the infimum of the expression b - a under the condition that b is greater than a, with a focus on the context of Darboux sums in analysis.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of the infimum, with one attempting to prove that 0 is the greatest lower bound. Another participant questions whether the proof holds for all values of a and b, suggesting specific values to test the bounds.

Discussion Status

The discussion is active, with participants exploring different approaches to proving the infimum. There is acknowledgment of the challenges in establishing the greatest lower bound, and some guidance has been provided regarding the selection of values for a and b.

Contextual Notes

Participants are working under the assumption that b must always be greater than a, and there is a focus on the implications of this condition for the infimum of the difference b - a.

Gavins
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Homework Statement


If b>a, find the infimum of b - a for arbritrary b,a.


Homework Equations


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The Attempt at a Solution


So this is for a darboux sum but I'm stuck on proving that 0 is the inf. It seems really easy but I can't seem to think straight.

First thing is 0 is a lower bound of b - a.
b - a > 0
b > a

Next is to prove it's the greatest lower bound. I tried to do this by supposing x > 0 is the greatest lower bound and find some contradiction but I can't figure it out.
 
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Ok, go ahead. Suppose x>0 is a lower bound. Pick b=x/2 and a=0. Is x a lower bound for b-a?
 


Does that not only prove it then when a = 0? I guess I might as well say what I'm doing. Basically, I'm trying to get the darboux sum for a function where f(x) = 1 if x = 0 and f(x) = 0 otherwise. I got the lower darboux sum and then when I try to do the upper darboux sum, I get [tex]\Sigma[/tex](xj+1 - xj)sup f. So if I expand this sum, I get 0 for every interval except the interval that contains 0 ie [xk,xk+1]. The upper darboux sum is xk+1 - xk. Clearly, the inf is 0 but I need to prove this.
 


Gavins said:
Does that not only prove it then when a = 0? I guess I might as well say what I'm doing. Basically, I'm trying to get the darboux sum for a function where f(x) = 1 if x = 0 and f(x) = 0 otherwise. I got the lower darboux sum and then when I try to do the upper darboux sum, I get [tex]\Sigma[/tex](xj+1 - xj)sup f. So if I expand this sum, I get 0 for every interval except the interval that contains 0 ie [xk,xk+1]. The upper darboux sum is xk+1 - xk. Clearly, the inf is 0 but I need to prove this.

Well, no. The assumption that was that x>0 was a lower bound of b-a for ALL b>a. If it's not a lower bound for SOME b>a, then x>0 is NOT a lower bound. But you are thinking about this way too generally. Of course you can make the upper sum as small as you want by picking x_(j+1)-x_j as small as you want. You might make the interval containing 0 be [-1/n,1/n]. So sure, you can make that as close to zero as you want by picking n large enough.
 


Thanks for that. Got it all sorted out now.
 

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