Basic Epsilon Question

  • Thread starter Seacow1988
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  • #1

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

Thanks!
 

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  • #2
tiny-tim
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Hi Seacow1988! :smile:

(have an epsilon: ε :wink:)
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. :confused:

Is this part of some longer proof?
 
  • #3
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!
 
  • #4
tiny-tim
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Hi Seacow1988! :smile:
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 :frown:
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.
 
  • #5
! Thank you so so much!
 

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