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Thread: Set manipulation

  1. #1
    Super Member Showcase_22's Avatar
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    Set manipulation

    Prove that:

    $\displaystyle \bigcup Px=x \subset P \bigcup x= P \bigcup P \bigcup x$
    where x is a set and $\displaystyle Px$ is the power set of x.

    This is the last part of the question, the former parts required me to prove:

    $\displaystyle x \subset y \Rightarrow \bigcup x \subset \bigcup y$

    $\displaystyle x \subset y \Leftrightarrow Px \subset Py$

    $\displaystyle x \subset Py \Leftrightarrow \bigcup x \subset y $

    where x and y are sets.

    I can't get anywhere with this! I've tried using $\displaystyle x \subset Px$ but I can't get it to work out. My main problem is that the first three parts of the questions gave me something to work with (ie. condition 1 allows me to use $\displaystyle x \subset y$). The last part doesn't give me any properties to use so i'm having some trouble trying to use the things I already know.

    Does anyone have any ideas?
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  2. #2
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    I have a hard time with your notation.
    Have you shown that $\displaystyle \bigcup {\mathcal{P}\mathcal{A}=\mathcal{A}}~? $

    Have you shown that $\displaystyle \mathcal{A}\subseteq\mathcal{P}\bigcup {\mathcal{A}}~? $.
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  3. #3
    Super Member Showcase_22's Avatar
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    Ohh, $\displaystyle \bigcup {\mathcal{P}\mathcal{A}=\{ t~|~(\exists X \in \mathcal{P} \mathcal{A})[t \in X] \}$.

    and we also know that $\displaystyle \mathcal{P} \mathcal{A}=\{X~|~Set(X) ~\&~ X \subset \mathcal{A} \}$

    so my $\displaystyle \mathcal{P} \mathcal{A}$ contains only elements from $\displaystyle \mathcal{A}$ since $\displaystyle X \subset \mathcal{A}$. We know we have all the elements since $\displaystyle \mathcal{A} \in \mathcal{P} \mathcal{A}$ and all the other subsets contain only some of these elements.

    Hence the unionset axiom gives all the elements in $\displaystyle \mathcal{A}$ giving that $\displaystyle \bigcup \mathcal{P} \mathcal {A}=\mathcal{A}$.

    Is that right? The only way I could see of doing it was with analysing the definitions.

    My main problem is the subset part. I can't seem to introduce anything that helps.

    So yes, this is the part i'm having trouble showing: $\displaystyle \mathcal{A}\subseteq\mathcal{P}\bigcup {\mathcal{A}}$.
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    So yes, this is the part i'm having trouble showing: $\displaystyle \mathcal{A}\subseteq\mathcal{P}\bigcup {\mathcal{A}}$.
    If $\displaystyle t\in \mathcal{A}$ then $\displaystyle \{t\}\in \mathcal{P}\mathcal{A}$ which implies that $\displaystyle t\in \bigcup {\mathcal{P}\mathcal{A}}$.

    Does that work?
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  5. #5
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    Aha! I think by $\displaystyle \subset$ you mean "implies". Is that correct?
    Last edited by DrSteve; Jan 25th 2011 at 03:08 PM.
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  7. #7
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Plato View Post
    If $\displaystyle t\in \mathcal{A}$ then $\displaystyle \{t\}\in \mathcal{P}\mathcal{A}$ which implies that $\displaystyle t\in \bigcup {\mathcal{P}\mathcal{A}}$.

    Does that work?
    But doesn't that mean $\displaystyle t\in \bigcup {\mathcal{P}\mathcal{A}}=\mathcal{A}$ so we're back to $\displaystyle t \in \mathcal{A} $?

    I'll ask my lecturer today about the subset being an implies. What i've written is exactly what's written on the example sheet.
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