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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!

Thanks!

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- 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!

Thanks!

- #2

tiny-tim

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(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

Is this part of some longer proof?

- #3

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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

Science Advisor

Homework Helper

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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 …

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

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! Thank you so so much!

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