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Thread: 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) $\displaystyle F \subseteq \mathcal{P}(A)$
    (b) $\displaystyle \forall B (F \subseteq \mathcal{P}(B) \rightarrow A \subseteq B)$

    So what I want to show is:

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

    I have the existence part down:

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

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

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

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

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

    The first half of proving the equality is trivial, that is $\displaystyle F \subseteq \mathcal{P}(C)$ so clearly $\displaystyle \cup F \subseteq C$. It's the
    other half that is giving me trouble, that is how do I show that $\displaystyle C \subseteq \cup F$? 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 $\displaystyle A = \cup F$. 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 $\displaystyle \forall B$? If so, does it now follow from (a) then that:

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

    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 $\displaystyle A = \cup F$. 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 $\displaystyle \forall B$? If so, does it now follow from (a) then that:

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

    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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