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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 \(A\).
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, we know in the finite case \(\mathcal P (A)=2^{A}\) and \(B^A = B^{A} = 2^{A}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think this applies when \(A>\aleph_i\).

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0A can have any cardinality

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You could create a homomorphism as well. \[ \forall S\subseteq A: \exists f\in B^A: f(x)= \begin{cases}1 &x\in S \\ 0 &x\notin S\end{cases} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can build a unique \(f\) from \(S\) and vise versa.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, 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 \(f[S]=1,f[S^c]=0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is a sideeffect 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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