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Math Help - PPPP(empty set)

  1. #1
    Super Member Showcase_22's Avatar
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    PPPP(empty set)

    So I need to find \mathcal{P} \mathcal{P} \mathcal{P} \mathcal{P} (\phi).

    I start with \phi.

    \mathcal{P}(\phi)=\{ \phi\}

    \mathcal{P}\mathcal{P}(\phi)=\{ \phi, \{ \phi \} \}

    \mathcal{P}\mathcal{P}\mathcal{P}(\phi)=\{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}

    (After \mathcal{P} (\phi), i'm just taking the union of the sets before it. Therefore (writing the elements as a list)):

    \mathcal{P}\mathcal{P}\mathcal{P}\mathcal{P}(\phi)  = \{

    \phi
    \{ \phi\}
    \{ \phi, \{ \phi \} \}
    \{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}

    \}

    To me this makes sense since i'm constructing the natural numbers:

    \phi=0
    \mathcal{P} (\phi)=\{ \phi \} =1
    \mathcal{P} \mathcal {P} (\phi)=\{\phi, \{ \phi \} \}=2 etc

    However, my tutor wrote that I should have 16 elements in \mathcal {P} \mathcal{P}\mathcal{P}\mathcal{P}(\phi). How is he getting all these extra elements?
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    So I need to find \mathcal{P} \mathcal{P} \mathcal{P} \mathcal{P} (\phi).
    I start with \phi.
    \mathcal{P}(\phi)=\{ \phi\}
    \mathcal{P}\mathcal{P}(\phi)=\{ \phi, \{ \phi \} \}
    \mathcal{P}\mathcal{P}\mathcal{P}(\phi)=\{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}
    To me this makes sense since i'm constructing the natural numbers:
    However, my tutor wrote that I should have 16 elements in
    You missed a set at the third level.
    \mathcal{P}\mathcal{P}\mathcal{P}(\phi)=\{ \phi , \{ \phi \}, \{\{ \phi \}\},\{ \phi, \{ \phi \} \} \}
    The power set of a set of n has 2^n members. That what your tutor means.

    That said, that is not the way the natural numbers are defined.
    This is
    0=\emptyset
    1=\{0\}=\{\emptyset\}
    2=\{0,1\}
    3=\{0,1,2\}
    .\quad   \vdots
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  3. #3
    MHF Contributor
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    Since I have already started writing...

    If |A|=n, then |\mathcal{P}(A)|=2^n. Therefore, indeed, |\mathcal{P}^4(\emptyset)|=16.

    Quote Originally Posted by Showcase_22 View Post
    \mathcal{P}\mathcal{P}\mathcal{P}(\phi)=\{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}
    You need one more element: \{\{\emptyset\}\}.

    \mathcal{P}\mathcal{P}\mathcal{P}\mathcal{P}(\phi)  = \{

    \phi
    \{ \phi\}
    \{ \phi, \{ \phi \} \}
    \{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}

    \}
    If we denote the elements of \mathcal{P}^3(\emptyset) as follows: a_1=\emptyset, a_2=\{\emptyset\}, a_3=\{\{\emptyset\}\}, a_4=\{\emptyset,\{\emptyset\}\}, then your list can be written as follows:

    \emptyset, \{a_1\}, \{a_1,a_2\}, \{a_1,a_2,a_4\}.

    It is missing 12 sets, including \{a_2\}, \{a_4\}, \{a_1,a_4\}, \{a_2,a_4\}, as well sets that contain a_3.

    To me this makes sense since i'm constructing the natural numbers
    Though you can encode natural numbers in this way, a standard construction (von Neumann ordinals) defines the successor of x as x\cup\{x\}, so the nth set contains n elements.
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