- #1

i_hate_math

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Claim: Let A be a non-empty subset of R

a. Prove the claim.

b. Does the claim still hold if we replace R+ with R? Explain briefly.

So I have spent the past hours trying to prove this claim using the theorem:

a = sup(A) ⇔ ∀ ε>0 (( ∀ x∈A x ≤ a + ε )∧( ∃ x∈A x > a - ε ))

I aim to prove true for: ∀ ε>0 (( ∀ x∈A x

As for the proof:

Firstly, it is given that A is bounded above, this means sup(A) exists.

Letting a = sup(A) gives: ∀ ε>0 (( ∀ x∈A x ≤ a + ε )∧( ∃ x∈A x > a - ε )) statement 1

I was able to use the definition of supremum to get a≥x and since a,x>0 , a

But the this result is not as helpful when proving the second part (namely, ∀ ε>0 ( ∃ x∈A x > a - ε )), neither is statement 1 (tried squaring both sides and messing with the inequalities but didn't work).

What I really need to prove is just there exist an x such that x

Please help.

^{+}= {x ∈ R : x > 0} which is bounded above, and let B = {x^{2}: x ∈ A}. Then sup(B) = sup(A)^{2}.a. Prove the claim.

b. Does the claim still hold if we replace R+ with R? Explain briefly.

So I have spent the past hours trying to prove this claim using the theorem:

a = sup(A) ⇔ ∀ ε>0 (( ∀ x∈A x ≤ a + ε )∧( ∃ x∈A x > a - ε ))

I aim to prove true for: ∀ ε>0 (( ∀ x∈A x

^{2}≤ a^{2}+ ε )∧( ∃ x∈A x^{2}> a^{2}- ε )), which is equivalent to proving ∀ ε>0 (( ∀ x∈B x ≤ b + ε )∧( ∃ x∈B x > b - ε )), given how set B is related to A.As for the proof:

Firstly, it is given that A is bounded above, this means sup(A) exists.

Letting a = sup(A) gives: ∀ ε>0 (( ∀ x∈A x ≤ a + ε )∧( ∃ x∈A x > a - ε )) statement 1

I was able to use the definition of supremum to get a≥x and since a,x>0 , a

^{2}≥x^{2}. With this result, it is easy to get the first part of the definition for B: ∀ ε>0 ( ∀ x∈A x^{2}≤ a^{2}+ ε ).But the this result is not as helpful when proving the second part (namely, ∀ ε>0 ( ∃ x∈A x > a - ε )), neither is statement 1 (tried squaring both sides and messing with the inequalities but didn't work).

What I really need to prove is just there exist an x such that x

^{2}>^{a}- ε, this seems almost to be the definition of supremum. I took a further step, attempting to prove by cases with case 1 being a∈A, this gives the desired result easily; case 2 is a∉A, again, the difficulties arise.Please help.

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