The mapping of a Set onto the characteristic function is a bijection

Question

Let $X$ be a Set. For all Subsets $ A \subset X$ the characteristic function of A is defined as:

\begin{align} \chi_A(x)= \begin{cases} 1 \iff x \in A \\ 0 \iff x \notin A \end{cases} \end{align} Let $\lbrace0,1\rbrace^X$ be the Set of functions $ X \longrightarrow \lbrace 0,1\rbrace $. Further let $P(X)$ be the power set of $X$.

Show that the following function is a bijection: \begin{align} P(X) & \longrightarrow \lbrace 0,1\rbrace ^X \\ A & \longmapsto \chi_A \end{align} I have struggled with this problem for a few days now and I believe one of my biggest issues is to create the desired set $\lbrace0,1\rbrace^X$.

http://math.stackexchange.com/questions/496429/showing-mathcalpx-is-isomorphic-to-2x — njguliyev, Sep 26 '13 at 15:21

score 1 · Accepted Answer · answered Sep 26 '13 at 15:26

$P$ is injective: Let $A,B$ be two sets with $\chi_A=\chi_B$. Then $\chi_{A\backslash B}=\chi_A-\chi_B = 0$ and thus $A\subseteq B$. In the same way $B\subseteq A$.

$P$ is surjective: Let $f\in\{0,1\}^X$, i.e. $f:X\rightarrow\{0,1\}$. Define the set $C =\{x\in X|f(x)=1\}$. Then $f=\chi_C$.

The mapping of a Set onto the characteristic function is a bijection

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