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Sup(A) less than or equal to sup(B)

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Given A and B are sets of numbers, [tex]A \neq \left\{ \right\} [/tex], [tex]B[/tex] is bounded above, and [tex] A \subseteq B [/tex].

    Explain why sup(A) and sup(B) exist and why [tex]sup(A) \leq sup(B)[/tex].


    2. Relevant equations
    [tex] \exists r \in \mathbb R \: : \: r \geq a \: \forall a \in A [/tex]
    [tex] \exists r \in \mathbb R \: : \: A \subset \left[ - \infty , r \right] [/tex]


    3. The attempt at a solution
    If [tex] k \in B: k \geq s, \: \forall s \in S [/tex]. Then [tex] k = sup(B)[/tex].
     
    Last edited: Sep 21, 2010
  2. jcsd
  3. Sep 20, 2010 #2
    What is your attempted proof? Ie., where are you having trouble?
     
  4. Sep 20, 2010 #3
    Proving [tex] sup(a) \leq sup(B)[/tex]
     
  5. Sep 20, 2010 #4
    Didn't think that it would have been that easy.
     
  6. Sep 20, 2010 #5

    Char. Limit

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    Not too much of a problem. Just don't do it again and you'll be fine.
     
  7. Sep 20, 2010 #6
    So what you're saying is that I'm assuming there is an element in B, such that it is less than or equal to the sup(B), and then taking another element from A and because A is a subset of B, that the element in A is less than or equal to the sup(B)?
     
  8. Sep 20, 2010 #7
    Not quite (careful with how you phrase things). The supremum is the least upper bound on the set. But if any element of A is in B, and z=sup(B) is an upper bound on every element of B, then what can you say about the relationship between A and z?
     
  9. Sep 20, 2010 #8
    Without proving it, just explaining it:

    If some element of A is in B, and the sup(B) is the least upper bound of B, then sup(A) is less than or equal to the sup(B).
     
  10. Sep 20, 2010 #9
    Right, so you explained it correctly but that's not a proof because you just restated what you want to prove... instead you want to show the reasoning step by step, where each step follows logically from the other. You'll think the proof is simple once you get it yourself.
     
  11. Sep 20, 2010 #10
    Suppose [tex] b=sup(B), \forall x \in B : x \leq b [/tex] and [tex] a=sup(A), \forall y \in A : y \leq a [/tex].

    [tex] \exists m \in \mathbb R : m \geq b \ \forall b \in B, B \subset \left( - \infty , m \right] [/tex]

    Since [tex] A \subseteq B, \ \forall x \in A [/tex], then
    [tex] y \leq a \leq x \leq b [/tex], and since [tex] a=sup(A) [/tex] and [tex] b=sup(B) [/tex], then [tex] y, x \in A[/tex] and [tex] y \leq x [/tex], then [tex] a \leq b [/tex].

    Therefore: [tex] sup(A) \leq sup(B) [/tex]
     
  12. Sep 20, 2010 #11
    Hmmm, I think you can cut down on the number of things to define. Right now you have [itex]sup(A) \leq x[/itex] for all x in A, so that doesn't really make much sense.

    Keep the b = sup(B) so for all x in B, [itex]x\leq b[/itex] part. Now take an arbitrary p in A. What does the hypothesis about set containment tell us about p?
     
  13. Sep 20, 2010 #12
    Could I word it this way?

    Suppose [tex]\exists b \in B : \ b=sup(B), \ \forall x \in B : x \leq b [/tex].
    Then [tex] \exists a \in A: \ a=sup(A), \forall y \in A: \ y \leq a[/tex].

    [tex] \exists m \in \mathbb R : m \geq b \ \forall b \in B, B \subset \left( - \infty , m \right] [/tex]

    Since [tex] A \subseteq B, \ \forall x \in A [/tex] and [tex] \forall x \in B[/tex] , then
    since [tex] x \leq b [/tex] and [tex] x \geq a [/tex], therefore [tex] a \leq b [/tex].

    [tex] sup(A) \leq sup(B) [/tex]
     
  14. Sep 20, 2010 #13

    vela

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    I just wanted to bring up a technical point. It won't affect your proof much. Your definition of sup seems a bit strange. Are you sure that sup(A) has to be an element of A? With your definition, if A=(0,1), sup(A) does not exist because there's no element in (0,1) that's greater than or equal to every element in (0,1).
     
  15. Sep 21, 2010 #14
    Assume [tex] A = \left\{ 1- \frac{1}{n} : \ n \in \mathbb Z^{+} \right\} [/tex], prove sup(A) = 1.

    If 1 is the least upperbound such that [tex] \forall \epsilon > 0, 1 - \epsilon [/tex] is not an upperbound of A, then

    [tex] \exist a \in A: \ a \in \left[ 1 - \epsilon , 1 \right) [/tex], then
    [tex] a = 1 - \frac{1}{n_{0}} [/tex] for some [tex] n_{0} \in \mathbb Z^{+} [/tex]

    [tex] 1 - \frac{1}{n_{0}} \geq 1 - \epsilon [/tex]
    [tex] - \frac{1}{n_{0}} \geq - \epsilon [/tex]
    [tex] \frac{1}{n_0{0}} \leq \epsilon [/tex]
    [tex] n_{0} \geq \frac{1}{ \epsilon } [/tex]


    Therefore, [tex] sup(A) = 1 [/tex]
     
  16. Sep 21, 2010 #15
    What if I rewrote the last part as:

    Since [tex] A \subseteq B [/tex], [tex]\forall x \in B [/tex] and [tex] \forall y \in A [/tex] then
    [tex] y \leq x [/tex] so [tex] a \leq x \leq b [/tex],
    then [tex] a \leq b [/tex].

    Therefore [tex] sup(A) \leq sup(B) [/tex].
     
  17. Sep 21, 2010 #16

    vela

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    I might have misread your intent. I assumed the above is your definition for sup(B). What is your definition of the supremum? The definition is the key to this problem.
    Note that 1 isn't an element of A. The supremum of a subset isn't necessarily in the subset. In many of your attempts at the proof so far, you seem to assume sup(A) is in A and sup(B) is in B.
     
    Last edited: Sep 21, 2010
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