## 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 $$A$$.

1. anonymous

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

2. anonymous

I think this applies when $$|A|>\aleph_i$$.

3. zzr0ck3r

yeah it does.

4. zzr0ck3r

A can have any cardinality

5. anonymous

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

You can build a unique $$f$$ from $$S$$ and vise versa.

7. anonymous

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

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 $$S\to \{0,1\}$$