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Math Help - Set theory. Is the converse true?

  1. #1
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    Set theory. Is the converse true?

    1. The problem statement, all variables and given/known data

    Prove that \cup_{x \in C} 2^{x} \subseteq 2^{\cup C}

    2. Relevant equations

    \cup_{x \in C} 2^{x} = \{ A | \exists x \in C, A \subseteq 2^{x} \}

    2^{x} is the powerset of x. i.e. 2^{x} = \{ y | y \subseteq x \}

    3. The attempt at a solution

    Suppose A \in \cup_{x \in C} 2^{x} . Then,

    \exists x \in C, A \in 2^{x}

    \exists x \in C, A \subseteq x

    A \subseteq ( \cup C )

    A \in 2^{\cup C}

    Therefore, A \in \cup_{x \in C} 2^{x} \Rightarrow A \in 2^{\cup C}

    Therefore, \cup_{x \in C} 2^{x} \subseteq 2^{\cup C}

    But I think there might be something wrong with my proof. Because why can't I start assuming A \in 2^{\cup C} and go to A \in \cup_{x \in C} 2^{x} . That means A \in 2^{\cup C} \Rightarrow A \in \cup_{x \in C}  2^{x} and therefore 2^{\cup C} \subseteq \cup_{x \in C}  2^{x} also. That means \cup_{x \in C} 2^{x} = 2^{\cup C}

    Is there something wrong with this proof?
    Last edited by omoplata; June 10th 2011 at 09:46 AM. Reason: wrong notation
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  2. #2
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    To me that seems like odd use of the '{ }' notation.

    Ordinarily, \cup_{x \in C} \{ 2^{x} \} = {A | Ex(x in C & A in {Px})} = {A | Ex(x in C & A = Px)}.
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  3. #3
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    But I glean that the actual problem is this:

    Show {A | Ex(x in C & A in Px)} subset PUC

    So, let x in C and A subset x. Show A subset UC.

    Suppose z in A. So z in x in C. So z in UC.

    Done.
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  4. #4
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    Quote Originally Posted by MoeBlee View Post
    To me that seems like odd use of the '{ }' notation.

    Ordinarily, \cup_{x \in C} \{ 2^{x} \} = {A | Ex(x in C & A in {Px})} = {A | Ex(x in C & A = Px)}.
    Sorry, it should be \cup_{x \in C} 2^{x}, not \cup_{x \in C} {2^{x}}. I edited the original post and corrected it.
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  5. #5
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    Quote Originally Posted by MoeBlee View Post
    But I glean that the actual problem is this:

    Show {A | Ex(x in C & A in Px)} subset PUC

    So, let x in C and A subset x. Show A subset UC.

    Suppose z in A. So z in x in C. So z in UC.

    Done.
    Thanks. Now I understand why I can't go backwards. z in UC does not imply z in x.
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  6. #6
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    So I take it you also wanted to check whether

    PUC subset {A | Ex(x in C & A in Px)} is a theorem.

    But counterexample:

    Let C = {{0} {1}}. Let A = {0 1}.

    There, A in PUC but there is no x in C such that A in Px.
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