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Thread: PPPP(empty set)

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

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

    I start with $\displaystyle \phi$.

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

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

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

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

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

    $\displaystyle \phi$
    $\displaystyle \{ \phi\}$
    $\displaystyle \{ \phi, \{ \phi \} \}$
    $\displaystyle \{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}$

    $\displaystyle \}$

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

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

    However, my tutor wrote that I should have 16 elements in $\displaystyle \mathcal {P} \mathcal{P}\mathcal{P}\mathcal{P}(\phi)$. 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 $\displaystyle \mathcal{P} \mathcal{P} \mathcal{P} \mathcal{P} (\phi)$.
    I start with $\displaystyle \phi$.
    $\displaystyle \mathcal{P}(\phi)=\{ \phi\}$
    $\displaystyle \mathcal{P}\mathcal{P}(\phi)=\{ \phi, \{ \phi \} \}$
    $\displaystyle \mathcal{P}\mathcal{P}\mathcal{P}(\phi)=\{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}$
    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.
    $\displaystyle \mathcal{P}\mathcal{P}\mathcal{P}(\phi)=\{ \phi , \{ \phi \}, \{\{ \phi \}\},\{ \phi, \{ \phi \} \} \}$
    The power set of a set of $\displaystyle n$ has $\displaystyle 2^n$ members. That what your tutor means.

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

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

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

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

    $\displaystyle \phi$
    $\displaystyle \{ \phi\}$
    $\displaystyle \{ \phi, \{ \phi \} \}$
    $\displaystyle \{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}$

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

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

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

    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 $\displaystyle x$ as $\displaystyle x\cup\{x\}$, so the $\displaystyle n$th set contains $\displaystyle n$ elements.
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