1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Supremum of a set in ℝ

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Let T be a set such that:
    [tex]T=\{t\in\mathbb{R}/t^{2}<2\}[/tex]


    2. Relevant equations

    a) Justify the existence of a real number a such that a=Sup(T)

    b) Prove that the proposition [tex]a^{2}<2[/tex] is false.

    c) Suppose that [tex]a^{2}>2[/tex]. Prove that we can find a contradiction with a=Sup(T).

    d) Conclude that finally [tex]a^{2}=2[/tex]



    3. The attempt at a solution

    a) Ok we have [tex]t^{2}<2[/tex] which implies t<√2. There for [tex]t\in]-\infty,\sqrt{2}[\subset\mathbb{R}[/tex] , and since T is a non empty set bounded from the top. Hence by the completeness axiom there exists an a in ℝ such that a=Sup(T).

    b) I have a feeling that I will have to use a proof by contradiction, but I don't know how to start.

    c) I started this one with letting a and b be two upper bounds of T such that a≠b and let ε>0:

    Let ε'=a-b>0 . Therefore there exists an x in T such that: a-ε'<x. Therefore, after substituting ε' with its value we get that there exists an x in T such that b<x. Hence we reach a contradiction with b being an upper bound of T therefore b<a. And finally a=Sup(T)

    d) Well from b) we have that a^2<2 is a false statement and from c) we found that a^2>2 contradicts with a=Sup(T), which leaves us with a^2=2 and therefore a=Sup(T).

    Thank you for your help in advanced.
     
  2. jcsd
  3. Oct 17, 2013 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    OK, although you could have worded it better. Your claim is that ##T## is a nonempty set bounded from above. So give an example of an element in ##T##, and give an explicit upper bound. You provided ##\sqrt{2}## as an upper bound, and while it's "obvious" that this is correct, a line or two to prove it would be a good idea.

    If ##a^2 < 2##, can you find an element of ##T## which is greater than ##a##?

    I don't see how ##a-b > 0## follows from your assumption.

    The conclusion is that ##a^2 = 2##. You already defined ##a = \sup(T)## in part (a), so there's no "therefore" about it.
     
  4. Oct 17, 2013 #3

    pasmith

    User Avatar
    Homework Helper

    This is putting the cart before the horse: the object of the question is to prove that there actually exists a positive real number [itex]a[/itex] such that [itex]a^2 = 2[/itex].
     
  5. Oct 17, 2013 #4
    B) No there is no element in T that's greater than a since a is the sup(T). Is that all?
     
  6. Oct 17, 2013 #5

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Let me rephrase the question. Suppose ##a^2 < 2##. Can you find a number ##t \in T## such that ##t > a##, and therefore ##a## cannot be the supremum of ##T##?
     
  7. Oct 18, 2013 #6
    Supposing that a^2<0 and that b is an upper bound of T. We want to prove that a<b.

    So by proof by contradiction we suppose that b<a and let ε=a-b>0. So by definition there exists an x in T such that a-ε<x. By substituting ε with it's value we get that b<x , and that contradicts with our assumption of a<b (b is an upper bound). Finally a^2<2 is false and a=sup(T). Is this ok now?
     
  8. Oct 18, 2013 #7

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think you mean ##a^2 < 2##.

    I don't see where you used the assumption that ##a^2 < 2##.

    I don't think you want to introduce ##b## as an upper bound of ##T##. Try this instead: Suppose that ##a## is any real number such that ##a^2 < 2##. Then I claim that there is some ##t \in T## (not necessarily an upper bound of ##T##) such that ##a < t##. Therefore ##a## is not even an upper bound of ##T##, so it cannot be the supremum of ##T##. All you need to do is prove the claim.
     
  9. Oct 18, 2013 #8
    Ok so I let there be two elements of T (a and t) such that a≠t, and we want to prove that a<t<2.

    So with a proof by contradiction we'll say that t<a<2 and let ε=t-a>0. When we plug in ε to the previous inequality we get that ε+a<a⇔ ε<0 which contradicts with it being greater than 0. I'm pretty sure this isn't it though.
     
  10. Oct 18, 2013 #9

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, first of all, you want ##a < t < \sqrt{2}##, not ##a < t < 2##. And you can't "let there be" two such elements. Start with an element ##a \in T## (i.e. ##a^2 < 2##). You want to prove that there EXISTS an element ##t\in T## such that ##a < t##.
     
  11. Oct 18, 2013 #10

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hint: if ##a^2 < 2##, then there exists a rational number of the form ##p^2/q^2## such that ##a^2 < p^2/q^2 < 2##. (Why?)
     
  12. Oct 18, 2013 #11
    Well like you said and if we have p and q two co-prime integers then when we square root them we get: a<p/q<√2 and multiply by q and we get aq<p<q*√2. I was thinking that's going to say that p isn't an integer but you can't say from what I've reached.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Supremum of a set in ℝ
  1. Supremum of sets (Replies: 3)

Loading...