Question

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 \(A\).

Answer 1

Well, we know in the finite case \(|\mathcal P (A)|=2^{|A|}\) and \(|B^A| = |B|^{|A|} = 2^{|A|}\)

Answer 2

I think this applies when \(|A|>\aleph_i\).

Answer 3

yeah it does.

Answer 4

A can have any cardinality

Answer 5

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

Answer 6

You can build a unique \(f\) from \(S\) and vise versa.

Answer 7

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 \(f[S]=1,f[S^c]=0\)

Answer 8

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 \(S\to \{0,1\}\)