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Math Help - Set manipulation

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

    Prove that:

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

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

    x \subset y \Rightarrow \bigcup x \subset \bigcup y

    x \subset y \Leftrightarrow Px \subset Py

    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 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 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 \bigcup {\mathcal{P}\mathcal{A}=\mathcal{A}}~?

    Have you shown that  \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, [LaTeX ERROR: Convert failed] .

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

    so my \mathcal{P} \mathcal{A} contains only elements from \mathcal{A} since X \subset \mathcal{A}. We know we have all the elements since \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 \mathcal{A} giving that \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: \mathcal{A}\subseteq\mathcal{P}\bigcup {\mathcal{A}}.
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  4. #4
    MHF Contributor

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

    Does that work?
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  5. #5
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    Aha! I think by \subset you mean "implies". Is that correct?
    Last edited by DrSteve; January 25th 2011 at 03:08 PM.
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  6. #6
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  7. #7
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Plato View Post
    If t\in \mathcal{A} then \{t\}\in \mathcal{P}\mathcal{A} which implies that t\in \bigcup {\mathcal{P}\mathcal{A}}.

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