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Upper bound proof

  1. Sep 19, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Prove or disapprove, for non-empty, bounded sets S and T in ℝ :

    sup(SUT) = max{sup(S), sup(T)}


    2. Relevant equations

    The least upper bound axiom of course.


    3. The attempt at a solution

    Since we know S and T are non-empty and bounded in the reals, each of them contains a supremum by the least upper bound axiom. Let : L1 = sup(S) ^ L2 = sup(T) be these least upper bounds for S and T respectively.

    Since SUT is also a bounded non-empty set, it also contains a supremum by the axiom. Let L = sup(SUT) denote SUT's least upper bound.

    We want to show that L = max{L1, L2}

    Not quite sure how to proceed from here.
     
  2. jcsd
  3. Sep 19, 2012 #2
    Prove that it can't be less and can't be greater.
     
  4. Sep 19, 2012 #3

    Zondrina

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    Uhm, well since S is a subset of SUT and T is a subset of SUT, we know that SUT which is comprised of all of the points of S and T must also contain the least upper bounds of both sets. That is L1, L2 are in SUT.

    Is this the right direction?
     
  5. Sep 19, 2012 #4
    This is not true. The least upper bound need not be contained in its set. For example, (0, 1) does not contain its least upper bound, nor does (2, 3), nor will (0, 1) U (2, 3).
     
  6. Sep 19, 2012 #5

    Zondrina

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    Ah yes I see, so from what you told me before, I must somehow show :

    L1, L2 < L < L1, L2 ?
     
  7. Sep 19, 2012 #6

    Zondrina

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    So for L to be the sup(SUT), it must be an upper bound, that is x ≤ L for all x in SUT.

    It also has to be the least upper bound, that is for any upper bound of SUT, say M, L ≤ M.

    Are these points relevant? If so I believe I know how to do this.
     
  8. Sep 19, 2012 #7

    jbunniii

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    Well, that's the definition of the least upper bound, so it's certainly relevant.

    Using your notation, you need to show that both of the following are impossible:

    [tex]L < \max(L_1, L_2)[/tex]
    [tex]L > \max(L_1, L_2)[/tex]

    So start by assuming one of these and finding a contradiction.
     
  9. Sep 19, 2012 #8
    Well, these are the definition of the least upper bound. Yes, these could be use to prove that sup SUT is not less and is not greater than max {sup S, sup T}.
     
  10. Sep 19, 2012 #9

    jbunniii

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    Hint: consider what the two inequalities mean.

    [itex]L < \max(L_1, L_2)[/itex] means that either [itex]L < L_1[/itex] or [itex]L < L_2[/itex] (or both).

    [itex]L > \max(L_1, L_2)[/itex] means that [itex]L > L_1[/itex] and [itex]L > L_2[/itex]
     
  11. Sep 19, 2012 #10

    Zondrina

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    So I'll write out my whole proof below here :

    Since we know S and T are non-empty and bounded in the reals, each of them contains a supremum by the least upper bound axiom. Let sup(S) ^ sup(T) be these least upper bounds for S and T respectively.

    Since SUT is also a bounded non-empty set, it also contains a supremum by the axiom. Let sup(SUT) denote SUT's least upper bound.

    We want to show that sup(SUT) = max{sup(S), sup(T)}. So to do this we want to show sup(SUT) ≤ max{sup(S), sup(T)} and sup(SUT) ≥ max{sup(S), sup(T)}.

    To show sup(SUT) ≤ max{sup(S), sup(T)} consider the following :

    Suppose x is in S, then we know x ≤ sup(S) ≤ max{sup(S), sup(T)}.
    Suppose y is in T, then we know y ≤ sup(T) ≤ max{sup(S), sup(T)}.

    Putting these together we know that sup(SUT) ≤ max{sup(S), sup(T)}.

    To show sup(SUT) ≥ max{sup(S), sup(T)} let M be an upper bound for SUT. Then consider that since M is an upper bound for the union of S and T, this tells us that either M is an upper bound for S or M is an upper bound for T.

    If M is an upper bound for S, then M ≥ sup(S). If M is an upper bound for T, then M ≥ sup(T). So it follows now that sup(SUT) ≥ max{sup(S), sup(T)}.

    Now since sup(SUT) ≤ max{sup(S), sup(T)} and sup(SUT) ≥ max{sup(S), sup(T)}, we only have one option left, it must be that sup(SUT) = max{sup(S), sup(T)} as desired.
     
  12. Sep 19, 2012 #11
    I think it would be more straightforward to say that for any x (or y) from S(or T), x (y) is no greater than M; and because sup S and sup T are the least upper bounds, they are also necessarily no greater than M. Essentially, just like the first part.
     
  13. Sep 19, 2012 #12

    Zondrina

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    Yes I see what you're saying.

    Thanks for the help guys :)
     
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