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Homework Help: Having a difficult time with supremum proof

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose a = sup(A) and b = sup(B). Let A + B = [itex]\left\{[/itex]x + y;x[itex]\in[/itex]A; y[itex]\in[/itex]B[itex]\right\}[/itex]. Show that a + b = sup(A + B).

    2. Relevant equations

    3. The attempt at a solution

    I'm honestly not sure where to start. Any help guys?
  2. jcsd
  3. Sep 3, 2011 #2
    Well, a=sup(A), and b=sup(B). What does that mean for you??

    You need to show a+b=sup(A+B). What does that mean?? What do you need to prove?
  4. Sep 3, 2011 #3
    I suppose saying I have no clue where to start was a bit of an overstatement.

    a is not only the upperbound of A but the smallest possible upperbound. Same for b/B. This is the definition of a supremum.

    When the sets are added together, the supremum of that new set, call it C, is the same as the supremum's of sets A and B individually added together. That is, c=sup(A+B)=sup(C)=a+b.

    Here is my intuition: if you take the largest values from set A and set B and add them together, this is the greatest number in (A+B). Call this value z. The supremum of (A+B) lies at the next incremental value greater than z. Does this make sense?
  5. Sep 3, 2011 #4


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    Yes, this is what we want to show.

    Ah, but you don't know A or B has a largest value! This is only true if max(A) = sup(A), and max(B) = sup(B).

    Go through your notes/book. What is the mathematical definition of a supremum? Least upper bound is not enough.
  6. Sep 3, 2011 #5
    If S is a nonempty subset of numbers, then M=sup(S) if and only if M is an upperbound for S AND for any z>0, M-z is not an upperbound for S.

    Is this a solid definition?
  7. Sep 3, 2011 #6


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    Almost. You need to say mathematically what it means for M-z to not be an upperbound for S:

    M=sup(S) if and only if M is an upperbound for S AND for any z>0, there exists an s in S such that s > M-z.

    I wish I could draw a number line to illustrate this, but this basically says that if M is a supremum, we can choose any z>0 and we'll always find a value between M-z and M that's in S.
  8. Sep 3, 2011 #7
    That makes perfect sense, actually. Where should I go from here?
  9. Sep 3, 2011 #8


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    Well, we need to prove two things:

    1) a+b is an upperbound.
    2) For a fixed z>0, there exists s in A+B such that s > a+b-z

    For 1), How do you show a value is an upperbound of a set?
  10. Sep 3, 2011 #9
    Let A[itex]_{w}[/itex]=set of whole parts of elements of A+B
    A[itex]_{w}[/itex] is bounded from above because A+B is bounded. Let a[itex]_{w}[/itex]=max(A[itex]_{w}[/itex]. And let A[itex]_{1}[/itex] [itex]\subseteq[/itex] A+B be the subset of A+B containing all elements that begin a[itex]_{w}[/itex]. ...(decimal expansion). Assume a[itex]_{w}[/itex] [itex]\geq[/itex] 0. Let a[itex]_{1}[/itex]=max (1/10)'s digit of all elements in A+B. Let A[itex]_{2}[/itex] [itex]\subseteq[/itex] A[itex]_{1}[/itex] be the subset of all elements in A[itex]_{1}[/itex] beginning a[itex]_{w}[/itex].a[itex]_{1}[/itex]...
    Continue in this manner...
    Let C=a[itex]_{w}[/itex].a[itex]_{1}[/itex]a[itex]_{2}[/itex]a[itex]_{3}[/itex]...
    Claim: C=sup(A+B)
    given any x[itex]\in[/itex]A+B, a[itex]_{w}[/itex] [itex]\geq[/itex] the whole part of x. If they're equal, a[itex]_{1}[/itex] [itex]\geq[/itex] the (1/10)'s digit of x. If these are equal, a[itex]_{2}[/itex] [itex]\geq[/itex] (1/100)'s digit of x...
    For any n[itex]\geq[/itex]1 there is an element of A[itex]\ni[/itex]X[itex]_{n}[/itex] such that X[itex]_{n}[/itex]=a[itex]_{w}[/itex].a[itex]_{1}[/itex]a[itex]_{2}[/itex]a[itex]_{3}[/itex]...a[itex]_{n}[/itex]
    [itex]\left|[/itex]C-X[itex]_{n}[/itex][itex]\right|[/itex][itex]\leq[/itex][itex]\frac{1}{10^{n}}[/itex], so any number less than s-[itex]\frac{1}{10^{n}}[/itex] is not an upperbound. Since this is true for all n, no number below C is an upperbound. Thus, C=sup(A+B)=a+b.

    Am I way off here?
  11. Sep 3, 2011 #10
    The part that says \left and \right is supposed to be absolute value. I'm not sure how to input that.
  12. Sep 3, 2011 #11
    Can anyone tell me if I'm on the right track here?
  13. Sep 3, 2011 #12
    I don't see what your post 9 actually has to do with proving that sup(A+B)=a+b...
    You're making it way more complicated than it needs to be.

    First, you need to prove that a+b is an upper bound of A+B. For that, take x+y in A+B. Can you prove that [itex]x+y\leq a+b[/itex]?
  14. Sep 3, 2011 #13
    Well, since x[itex]\in[/itex]A, it is by definition less than a, which is the upperbound of A. Similarly, y is less than b. Therefore, x+y[itex]\leq[/itex]a+b. I'm not sure how to formalize this intuition, though.
  15. Sep 3, 2011 #14
    Yes, that's ok. This proves that a+b is an upper bound of A+B. Now prove that it is the least upper bound...
  16. Sep 3, 2011 #15
    I'm a little stuck in thinking about this. Intuitively, I know a is the least upperbound of A by definition, and b is the least upperbound of B, similarly. Having a least upperbound of A+B greater than a+b implies that either a or b was not a least upperbound to begin with. I'm not sure where to go from here...
  17. Sep 3, 2011 #16
    You could prove that there exists a sequence in A+B that converges to a+b.

    In general:
    If z is an upper bound of a set Z, and if there exists a sequence [itex](z_n)_n[/itex] in Z that converges to z, then z is the supremum of Z.

    Now, try to prove that "in general"-statement, and try to see why such a sequence should exist.
  18. Sep 3, 2011 #17
    I think I was following well until this point, but now I'm totally lost. Would you mind explaining this? By the way, I really appreciate the help.
  19. Sep 3, 2011 #18
    You did see sequences yet, did you??

    Well, assume that z is an upper bound of Z, and let's say that there are element [itex]z_n\in Z[/itex] such that [itex]z_n\rightarrow z[/itex]. Then I claim that z is the least upper bound.

    Let's try to prove this.
    Let's assume that w is another upper bound of Z. We will want to prove that [itex]z\leq w[/itex]. Do you follow until here?? Can you prove this?
  20. Sep 3, 2011 #19
    Yes, I follow until there. I'm not sure how I prove that, though...
  21. Sep 3, 2011 #20
    [itex]z_n\leq w[/itex]. Now take the limit...
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