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Given a nonempty set $A$, define

$$F=\{f:A \to \{0,1\}\mid f \text{ is a function} \}.$$

Prove that $\mathcal{P(A)}$ is equipotent to $F$ ($\mathcal{P(A)}$~$F$), where $\mathcal{P}$ is the power set of $A$.

How would one prove this?

I try

Definate a function,, with $B \subseteq \mathcal{P}(a) $

\begin{array}{rcl} \phi: \mathcal{P}(A) & \to & F \\ B & \to & X_b:A \to \{0,1\} \\&& x \to X_B(x) = && \end{array}

Where $X_B(x)=1 $ if $x\in B$ and $X_B(x)=0 $ if $x \notin B$

How proof that $\phi$ is bijetive ??

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    Your definition looks strange. – 2017-01-17

3 Answers 3

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Suppose $G \ne H$ that means there is either a $x \in G$ so that $x \not \in H$ or there is an $x \in H$ so that $x \not \in G$. Wolog assume $x \in G; x\not \in H$. Then $\phi(G)(x) = 1$ but $\phi(H)(x) = 0$ so $\phi(G) \ne \phi(H)$ so $\phi$ is injective.

Let $f: X \rightarrow \{0,1\}$. Let $B = \{x\in X| f(x) = 1\}$. Then $B \subset X$. $\phi(B)(y) = 1$ if $y \in B$ which would mean $f(y) = 1$. And $\phi(B)(y) = 0$ if $y \not \in B$ which would mean $f(y) \ne 1$ which would mean $f(y) = 0$. So $\phi(B)(y) = f(y); \forall y \in X$. So $\phi(B) = f$ and $\phi$ is surjective.

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$$\phi(B_1)=\phi(B_2)\iff \chi_{B_1}=\chi_{B_2}\iff B_1=B_2$$

Then take $f:A\rightarrow\{0,1\}$ and $B=\{x\in A:f(x)=1\}$. Then prove $F(B)=f$

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  • Given a subset $S$ of $A$, define the function $f_S(x)=\mathbb{I}(x\in S)$.

  • Given a function $g:A\to\{0,1\}$, define the set $S_g=\{x\in A:g(x)=1\}$

Then, the mappings $S\mapsto f_S$ and $g\mapsto S_g$ are mutually inverse, and provide a bijection between the sets $F$ and $\mathcal{P(A)}$.