Understanding Epsilon: Why a+e<b for all e>0 implies a≤b

[Seacow1988]

Why is it true that: if a+e<b for all e>0 then a≤b? What is the meaning of epsilon here?

Thanks!

 

[tiny-tim]

Hi Seacow1988!

(have an epsilon: ε )

Why is it true that: if a+e<b for all e>0 then a≤b?

uhh? that would only apply if b was infinite.

Is this part of some longer proof?

 

[Seacow1988]

Thanks for the reply! It's a concept taken from a larger proof:

Let A and B be nonempty bounded subsets of R. Let
S = A + B = {a + b : a in A, b in B}.

We want to show that sup S= Sup A + Sup B

Let α = supA, β = supB, and γ = sup(A + B).

Part 1 of the proof is:

Let e > 0 be given. Since α−e/2 < α = supA, we can find a in A such
that α−e/2 < a. Similarly, we can find b in B such that β−e/2 < b.
Let c = a + b. Then (α + β) − e = (α − e/2) + (β − e/2) < a + b = c
and c belongs to A+B. It follows that (α+β)−e < sup(A+B) = γ.
Since this holds for all e > 0, if follows that α + β ≤ γ.

I don't understand the last sentence.

Thanks!

 

[tiny-tim]

Hi Seacow1988!

Let e > 0 be given. Since α−e/2 < α = supA, we can find a in A such
that α−e/2 < a. Similarly, we can find b in B such that β−e/2 < b.
Let c = a + b. Then (α + β) − e = (α − e/2) + (β − e/2) < a + b = c
and c belongs to A+B. It follows that (α+β)−e < sup(A+B) = γ.
Since this holds for all e > 0, if follows that α + β ≤ γ.

This not what you originally wrote …

Why is it true that: if a+e<b for all e>0 then a≤b?

… this is: if a-e<b for all e>0 then a≤b.

It's because if a > b, then there's an e (= (a-b)/2 for example) with a - e > b.

 

[Seacow1988]

! Thank you so so much!