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I Upper bound and supremum problem

  1. Apr 12, 2017 #1
    Claim: Let A be a non-empty subset of R+ = {x ∈ R : x > 0} which is bounded above, and let B = {x2 : 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 x2 ≤ a2 + ε )∧( ∃ x∈A x2 > a2 - 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 , a2≥x2. With this result, it is easy to get the first part of the definition for B: ∀ ε>0 ( ∀ x∈A x2 ≤ a2 + 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 x2 > 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.
     
    Last edited: Apr 12, 2017
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  3. Apr 12, 2017 #2

    PeroK

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    If you can assume (or prove) that for ##x, y > 0, \ x \le y \ \Leftrightarrow x^2 \le y^2##, then I would suggest a different approach using the definition of supremum as the least upper bound.
     
    Last edited: Apr 12, 2017
  4. Apr 12, 2017 #3

    andrewkirk

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    The second part is readily proved by contradiction. Assume it is false, ie ##\neg(\forall\varepsilon>0,\ \exists x\in B:\ x>a^2-\varepsilon)##, where ##a\triangleq \sup(A)##. That is equivalent to ##\exists \varepsilon>0,\ \forall x\in B:\ x\leq a^2-\varepsilon## which, because the square function is a monotonic bijection between ##A\subseteq\mathbb R^+## and ##B\subseteq\mathbb R^+##, is equivalent to ##\exists \varepsilon>0,\ \forall y\in A:\ y^2\leq a^2-\varepsilon##.

    Try completing the square of ##a^2-\varepsilon## in order to find an upper bound for ##A## that is less than ##a##, which will give us the contradiction we need.
     
  5. Apr 12, 2017 #4
    I don't quite get what you're suggesting. Could you be a bit more specific about using the definition of supremum?
     
  6. Apr 12, 2017 #5
    Thank you very much for your quick response, very thorough explanation. I'll work on it!
     
  7. Apr 12, 2017 #6

    PeroK

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    The supremum is the least upper bound. ##a = sup(A)## means:

    1) ##a## is an upper bound for A (##\forall x \in A, x \le a##)

    2) If ##b## is an upper bound for ##A## then ##a \le b##

    That's what a supremum is, by definition.

    To help you a little, if ##a = sup(A)## you could show thta ##a^2 = sup(B)## by showing that:

    1) ##\forall x \in B, x \le a^2##

    2) If ##b## is an upper bound for ##B## then ##a^2 \le b##
     
  8. Apr 12, 2017 #7
    I am able to prove (1) as it is equivalent to ∀x∈A x2 ≤ a2, which can be derived from ∀x∈A x ≤ a (for positive x and a).
    But I am having trouble with the second part, because the only thing we know about b so far is ∀x∈b x ≤ b, can also be written as ∀x∈A x2 ≤ b. This does not , though, give the desired a2≤b. Could you enlighten me a bit further?
     
  9. Apr 12, 2017 #8
    I've tried your suggested method, and it worked well and I am well convinced. However, there are still a few things I am not certain:
    1. Is it sound to get from ∀ε>0 ∃x∈A, x≤a that ∀ε>0 ∃x∈A, x2≤a2 ?
    2. For when I am rewriting the logical notations of these statements, do I need to assign different parameters x, y .etc for elements in different sets? e.g. x and y in your reply

    Thanks heaps.
     
  10. Apr 12, 2017 #9

    andrewkirk

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    Yes, it is sound because
    $$x\geq 0\to\left(x\leq a\ \leftrightarrow x^2\leq a^2\right)$$
    It would be worthwhile to prove that formally as an exercise.

    It is the absence of that antecedent condition ##x\geq 0## that prevents part (b) of the question in the OP from being true.
    It is not strictly necessary, because the scope of a variable is restricted to the smallest containing formula in which it is quantified. However I think it is good practice to use different variables for members of different sets, because it makes it easier for the reader to follow.
     
  11. Apr 13, 2017 #10

    PeroK

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    You need to use the fact that ##a = Sup(A)##. Hint: use ##\forall x \in A, x \le \sqrt{b}##
     
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