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Math Help - Trouble with unique existance proof

  1. #1
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    Trouble with unique existance proof

    Exercise 3.7.1 of "How To Prove It" (Vellemen 2006)

    Suppose F is a family of sets. Prove that there is a unique set A that has the following two properties:

    (a) F \subseteq \mathcal{P}(A)
    (b) \forall B (F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)

    So what I want to show is:

    \exists !A[F \subseteq \mathcal{P}(A) \land \forall B(F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)]

    I have the existence part down:

    Suppose A = \cup F. Then F \subseteq \mathcal{P}(A).

    Suppose y \in F and x \in y and B such that F \subseteq B.

    A = \cup F, so clearly x \in A. Since F \subseteq \mathcal{P}(B) then it
    follows that x \in B, therefore A \subseteq B. Since x, y and
    B were arbitrary, \forall B(F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B).

    Where I'm having trouble is with the uniqueness proof. I believe that in order to prove uniqueness,
    I have to show that

    \forall C \{ [F \subseteq \mathcal{P}(C) \land \forall B(F \subseteq \mathcal{P}(B) \rightarrow C \subseteq B)] \rightarrow C = \cup F \}

    The first half of proving the equality is trivial, that is F \subseteq \mathcal{P}(C) so clearly \cup F \subseteq C. It's the
    other half that is giving me trouble, that is how do I show that C \subseteq \cup F? I can see that it is true but for some reason I can't come up with well reasoned argument as to why.
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  2. #2
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    I'm tired and completely didn't think about A = \cup F. I also haven't done set theory in awhile now. Good insight! You are right. The first half of the uniqueness proof is trivial since it follows from (b) and that A exists with the given properties. Given the definition you provided for uniqueness, wouldn't A be included in the \forall B? If so, does it now follow from (a) then that:

    F \subseteq P(A) \Rightarrow C\subseteq A

    Then the proof is complete, assuming everything else is correct.
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  3. #3
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    Quote Originally Posted by bryangoodrich View Post
    I'm tired and completely didn't think about A = \cup F. I also haven't done set theory in awhile now. Good insight! You are right. The first half of the uniqueness proof is trivial since it follows from (b) and that A exists with the given properties. Given the definition you provided for uniqueness, wouldn't A be included in the \forall B? If so, does it now follow from (a) then that:

    F \subseteq P(A) \Rightarrow C\subseteq A

    Then the proof is complete, assuming everything else is correct.
    Hey that's great Yeah, that completes the proof. I don't know why, but sometimes if I don't get how to do a certain proof right away, I'll get stuck in this awful whirlpool of circular reasoning that I just can't seem to get out of.
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