# Infimums and Darboux sums

1. Feb 18, 2010

### Gavins

1. The problem statement, all variables and given/known data
If b>a, find the infimum of b - a for arbritrary b,a.

2. Relevant equations
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3. 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.

2. Feb 18, 2010

### Dick

Re: Infimum

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?

3. Feb 18, 2010

### Gavins

Re: Infimum

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 $$\Sigma$$(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.

4. Feb 18, 2010

### Dick

Re: Infimum

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.

5. Feb 22, 2010

### Gavins

Re: Infimum

Thanks for that. Got it all sorted out now.