Proving inf(A) is unique with an epsilon proof

[jaqueh]

Homework Statement

Let A be a subset of ℝ, now prove if inf(A) exists, then inf(A) is unique

The Attempt at a Solution

I am using an epsilon proof but i don't think i am going about it the right way, can someone nudge me in the right direction

 

[Ray Vickson]

Homework Statement

Let A be a subset of ℝ, now prove if inf(A) exists, then inf(A) is unique

The Attempt at a Solution

I am using an epsilon proof but i don't think i am going about it the right way, can someone nudge me in the right direction

How you do it depends very much on what *definition* you are using for "inf". Different (but equivalent) definitions would need different types of proofs.

RGV

 

[Robert1986]

Usually, for something like this, it is best to start of saying "Let x and y be inf(S)" and then show that x and y are, in fact, the same. Now, showing that |x - y| < epsilon might be a good strategy, but there are probably more efficient methods. But, as RGV said, it depends on how inf is defined for you. If you have defined it as a greatest lower bound, the method I mentioned works very well.

 

[jaqueh]

yes i think I am going to revise it as x and y are infimums for A then x<y and y<x so x=y

 

[scurty]

yes i think I am going to revise it as x and y are infimums for A then x<y and y<x so x=y

That's a contradiction the way you wrote it. Did you mean $x \leq y, \ \text{and} \ y \leq x,\ \text{so} \ x=y$?

 

[jaqueh]

yes that is what i meant and i did write it that way x≥y y≥x