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Math Help - Set equality

  1. #1
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    Set equality

    I need to prove this plz help

    <br />
A,B,C \ and\ D\ are\ non\ empty\ sets\ with\ \left|A\right|=\left|B\right| and\ \left|C\right|=\left|D\right|
    <br />
1. Prove\ that\ \left|A\times C\right|=\left|B \times D\right|
    <br />
2. \left|A^{n}\right|= \left|B^{n}\right|<br />
    <br /> <br />
3.\left|P\left(A\right)\right|=\left|P\left(B\righ  t)\right|<br /> <br />
    With P i mean the power set

    thx guys
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  2. #2
    Senior Member JaneBennet's Avatar
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    |A|=|B|\ \mbox{and}\ |C|=|D|\ \Rightarrow\ \exists\ \mbox{bijections}\ \mathrm{f}:A\to B\ \mbox{and}\ \mathrm{g}:C\to D. Now show that the following functions are bijections.

    (1) \mathrm{h}:A\times C\to B\times D,\ \mathrm{h}(a,c)=(\mathrm{f}(a),\mathrm{g}(c))

    (2) \mathrm{h}:A^n\to B^n,\ \mathrm{h}(a_1,\dots,a_n)=(\mathrm{f}(a_1),\ldots\  ,\mathrm{f}(a_n))

    (2) \mathrm{h}:\mathcal{P}(A)\to\mathcal{P}(B),\ \mathrm{h}(X)=\{\mathrm{f}(a):a\in X\}
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  3. #3
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    From the given, there are bijections: f:A \leftrightarrow B\,\& \,g:C \leftrightarrow D.
    Now define a function \Phi :\left( {A \times C} \right) \leftrightarrow \left( {B \times D} \right),\,\,\Phi (a,c) = \left( {f(a),g(c)} \right).
    Prove that \Phi is a bijection.

    I have no idea what the notation in part b means.

    For part c, use the \chi function:  \chi _E (x) = \left\{ {\begin{array}{lr}    1 & {x \in E}  \\   0 & {x \notin {\rm E}}  \\ \end{array}} \right.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kuntah View Post
    I need to prove this plz help

    <br />
A,B,C \ and\ D\ are\ non\ empty\ sets\ with\ \left|A\right|=\left|B\right| and\ \left|C\right|=\left|D\right|
    <br />
1. Prove\ that\ \left|A\times C\right|=\left|B \times D\right|
    Hint: |A \times C| = |A||C|

    2. \left|A^{n}\right|= \left|B^{n}\right|<br />
    Hint: \left| A^n \right| = | \underbrace{A \times A \times A \cdots A}_{\mbox{n times}}| = \underbrace {|A||A||A| \cdots |A|}_{\mbox{n times}}

    3. \left|P\left(A\right)\right|=\left|P\left(B\right)  \right|<br /> <br />
    With P i mean the power set
    Hint: |P(A)| = 2^{|A|}

    EDIT: ...ok, so both JaneBennet and Plato have different methods from mine, and theirs are similar, so chances are what I did was foolishness . you may ignore this post. i'll leave it up for anyone to comment on it.
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    EDIT: ...ok, so both JaneBennet and Plato have different methods from mine, and theirs are similar, so chances are what I did was foolishness . you may ignore this post. i'll leave it up for anyone to comment on it.
    The posters proved this for any sets in general. You assumed it were finite sets and applied finite numbers.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    The posters proved this for any sets in general. You assumed it were finite sets and applied finite numbers.
    i thought so, thanks
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