Let $A$, $B$ be two sets and $B^A$ be the set of all functions that maps $A$ into $B$. How to prove that $\{0,1\}^A$ is equivalent to $P(A)$ (the power set of $A$)?
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This. Also it is more common to call such sets equinumerous. – Git Gud Feb 26 '13 at 07:25
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There is a natural bijection between the two. For any subset $K$ of $A$, define $f_K$ by $f_K(x)=1$ if $x\in K$, and $f_K(x)=0$ otherwise.
The mapping $\varphi$ that takes $K$ to $f_K$ is a bijection from the power set of $A$ to $\{0,1\}^A$.
André Nicolas
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