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Least Upper bound proof

  1. Aug 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose that R is a partial order on A, B1 ⊆ A, B2 ⊆ A, x1 is the least upper bound of B1, and x2 is the least upper bound of B2. Prove that if B1 ⊆ B2, then (x1,x2) ∈ R.


    2. Relevant equations



    3. The attempt at a solution

    This problem has been stumping me. After assuming B1 ⊆ B2, I have been trying to construct a proof by cases. I am not even certain that this is the best way to proceed, however. The cases I have attempted are x1 ∈ B2 or x1 ∉ B2, but i haven't found a way to complete the proof this way, and so i am beginning to doubt this proof strategy (by cases). I have also attempted a proof by contradiction, but this doesn't seem to make it too far either.

    My main questions are:

    1) If this IS a proof that should proceed via cases, are there any suggestions as to which cases should be used?
    2) If this is not a proof by cases, what method of proof should be used, and in what ways does the problem suggest this?

    I have been practicing proofs in set theory for about a year and am now reviewing for a credit by examination test I will be taking this semester by performing some of the problems which occur later in the problem sets in Velleman's text on proofs/set thoery. If you have any insight, I would also like to ask your opinion as to the degree of difficulty you believe this problem is. I'm interested in assessing my mastery of the subject. The proofs which appear in the examples within the text and earlier in the problem sets are easily achieved, but the later ones usually leave me thinking for many hours, and sometimes even a day or two (oftentimes without much progress toward the solution).

    Thanks in advance,

    Syrus
     
  2. jcsd
  3. Aug 22, 2011 #2
    Complete Proof of Original post:


    Lemma: Suppose that R is a partial order on A, B1 ⊆ A, B2 ⊆ A, x1 is the least upper bound of B1, and x2 is the least upper bound of B2. Prove that if B1 ⊆ B2, x2 is an upper bound of B1.

    Proof: Assume B1 ⊆ B2. Let b1 ∈ B1. Then b1 ∈ B2. Thus, (b1, x2) ∈ R, so x2 is an upper bound of B1 (since b1 was arbitrary).

    Q.E.D.

    _________________________________________________________________________________

    Problem: Suppose that R is a partial order on A, B1 ⊆ A, B2 ⊆ A, x1 is the least upper bound of B1, and x2 is the least upper bound of B2. Prove that if B1 ⊆ B2, (x1, x2) ∈ R.

    Proof: By the lemma above, we know x2 is an upper bound of B1. But since x1 is the least upper bound of B1, it follows that (x1, x2) ∈ R.

    Q.E.D.
     
    Last edited: Aug 22, 2011
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