Let $X$ be a set and let $U\subset X$ be a subset of $X$.
We have
\[ \chi _{\mathcal{P}(X)}(\chi _{x},\chi _{U})=\chi _{U}(x) \]
for each $x\in X$, giving an equality of functions
\[ \chi _{\mathcal{P}(X)}(\chi _{(-)},\chi _{U})=\chi _{U}, \]
where
\[ \chi _{\mathcal{P}(X)}(U,V)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} \mathsf{true}& \text{if $U\subset V$,}\\ \mathsf{false}& \text{otherwise.} \end{cases} \]
We have
\begin{align*} \chi _{\mathcal{P}(X)}(\chi _{x},\chi _{U}) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} \mathsf{true}& \text{if $\left\{ x\right\} \subset U$,}\\ \mathsf{false}& \text{otherwise} \end{cases}\\ & = \begin{cases} \mathsf{true}& \text{if $x\in U$}\\ \mathsf{false}& \text{otherwise} \end{cases}\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\chi _{U}(x). \end{align*}
This finishes the proof.
