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zzr0ck3r

  • one year ago

Fun set theory question for budding set theorist. Let \(B^A\) be the set of all functions from \(A\) to \(B:=\{0,1\}\). Show that \(\mathcal{P}(A)\) has the same cardinality as \(B^A\) for any set \(A\).

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  1. anonymous
    • one year ago
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    Well, we know in the finite case \(|\mathcal P (A)|=2^{|A|}\) and \(|B^A| = |B|^{|A|} = 2^{|A|}\)

  2. anonymous
    • one year ago
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    I think this applies when \(|A|>\aleph_i\).

  3. zzr0ck3r
    • one year ago
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    yeah it does.

  4. zzr0ck3r
    • one year ago
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    A can have any cardinality

  5. anonymous
    • one year ago
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    You could create a homomorphism as well. \[ \forall S\subseteq A: \exists f\in B^A: f(x)= \begin{cases}1 &x\in S \\ 0 &x\notin S\end{cases} \]

  6. anonymous
    • one year ago
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    You can build a unique \(f\) from \(S\) and vise versa.

  7. anonymous
    • one year ago
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    yeah, the bijection is easy -- for \(f\in B^A\) we associate the inverse image \(f^{-1}(1)\in 2^A\), and for \(S\in2^A\) we associate \(f\) such that \(f[S]=1,f[S^c]=0\)

  8. anonymous
    • one year ago
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    this is a side-effect of the fact that sets of things in \(A\) are wholly characterized by their members (i.e. for what \(a\in A\) we have that \(a\in S\in 2^A\)), and membership is a binary map \(S\to \{0,1\}\)

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