Prove that $\mathcal{P}(A) \sim \{0,1\}^A$

Question

Prove that $\mathcal{P}(A) \sim \{0,1\}^A$

We might can find a bijection function, however I am trying to prove it in other way. I want to F, G 1-1 functions and prove that those functions are inverse each other, once I did it, we know from a another proved claim that function is inverse if and only if function is bijection.

I define $F:\mathcal{P}(A) \to {0,1}^A$ by the following rule, for all $X\in \mathcal{P}(A)$, we have function $F(X): A \to \{0,1\}$ and for all $a\in A$ we define function $F(X)(a)=0$ its easy to prove that this function is 1-1.

Now I have to define $G:\{0,1\}^A \to \mathcal{P}(A)$ by the rule: for all $f\in \{0,1\}$, $G(f)=?$

My question is how can we define $G(f)$ to be 1-1 function? Moreover, any suggestions on this strategy will be highly appreciated.

If your $F$ maps every $X$ to the map $a\mapsto 0$, then this is certainly not injective once $A$ is nontrivial — Hagen von Eitzen, Jun 01 '19 at 14:38

score 2 · Accepted Answer · answered Jun 01 '19 at 14:42

2

For each $X\subseteq A$, define the characteristic function $\chi_X:A\rightarrow\{0,1\}$ such that $\chi_X(a)=1$ iff $a\in X$. This gives the bijection $P(A)\rightarrow\{0,1\}^A:X\mapsto \chi_X$.

answered Jun 01 '19 at 14:42

Wuestenfux

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Prove that $\mathcal{P}(A) \sim \{0,1\}^A$

1 Answers1