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Proving Sup(A)

  1. Jul 7, 2013 #1
    1. The problem statement, all variables and given/known data
    A=(1,2) prove that sup(A)=2

    2. Relevant equations
    this is how it was proved by the master

    2≥x for all x in R
    ∴ 2 is an upper bound of A
    let u be any upper bound of A

    suppose u<2
    therefore there exists r in R s.t. u<r<2
    1.5 ε R --> 1.5 ≤ u

    now 1.5≤u<r<2 --> 1<r<2 * :confused:
    r ε A with u<r --- contradiction

    ∴ 2≤u
    ∴ 2 = sup(A) //

    here i can't understand the line marked with *
    how can 1.5≤u<r<2 imply 1<r<2 ? shouldn't it be corrected as 1.5<r<2?

    is there any other way to prove this?
  2. jcsd
  3. Jul 7, 2013 #2


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    It's actually a weaker statement: if r is between 1.5 and 2 it is definitely between 1 and 2. You could also replace 1.5 by 1.

    The proof is fairly standard, however I'm not too thrilled about the apparently arbitrary occurrence of 1.5.

    The shortest proof you would like to give is something like:
    "Suppose u < 2. Let ##\epsilon = 2 - u## and ##r = u + \epsilon / 2 \in \mathbb R##. Then ##u < r < 2## so ##r \in A##, therefore u is not an upper bound".

    However, you can easily see that this proof will not work if u < 1 (e.g. take u = 0, the above construction would give you r = 1 which is not in A). This is where your teacher introduces the constraint ##1.5 \le u##, because we know that ##1.5 \in A## so any u smaller than that is definitely not an upper bound. You could fix this by replacing the first line of the proof from "Suppose u < 2" to "Clearly ##u \le 1## is not an upper bound for A. Suppose 1 < u < 2".
  4. Jul 7, 2013 #3


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    You mean "for all x in A".

    This is nonsense. Saying that u<2 does not necessarily mean that u is larger than 1.5.

    That's the part you are confused about? 1< 1.5, certainly so if 1.5< u then 1< u follows immediately. Technically it is the "transitive" property of <. If 1< 1.5 and 1.5< u then 1< u.
    It's the statement above, that "1.5 ε R --> 1.5 ≤ u" that should confuse you. That's non-sense.
    What is true is that, since u is an upper bound on the set and 2 is in the set, we have [itex]2\le u[/itex] and certainly 1.5< 2 so [itex]1.5< 2\le u[/itex] which gives 1.5< u.

    But there is no reason in the world to introduce "1.5". That has nothing at all to do with the problem. Even if we were working in the set of integers, where there is NO "1.5", 2 would still be the least upper bound and exactly the same proof would work.

    What you can say is that if u is an upper bound for {1, 2}, then we must, by definition of "upper bound", have [itex]2\le u[/itex] so that we cannot have u< 2.
  5. Jul 7, 2013 #4


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    Halls, I was confused about A as well, I think it is the open interval (1, 2) instead of the two-element set {1, 2}.
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