We have

\[ \chi _{\mathcal{P}\webleft (X\webright )}\webleft (\chi _{x},\chi _{U}\webright )=\chi _{U}\webleft (x\webright ) \]

for each $x\in X$, giving an equality of functions

\[ \chi _{\mathcal{P}\webleft (X\webright )}\webleft (\chi _{\webleft (-\webright )},\chi _{U}\webright )=\chi _{U}, \]

where

\[ \chi _{\mathcal{P}\webleft (X\webright )}\webleft (U,V\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} \mathsf{true}& \text{if $U\subset V$,}\\ \mathsf{false}& \text{otherwise.} \end{cases} \]

We have

\begin{align*} \chi _{\mathcal{P}\webleft (X\webright )}\webleft (\chi _{x},\chi _{U}\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} \mathsf{true}& \text{if $\webleft\{ x\webright\} \subset U$,}\\ \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}\webleft (x\webright ). \end{align*}

This finishes the proof.


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