Vertical Section of Characteristic Function is Characteristic Function of Vertical Section

Theorem
Let $X$ and $Y$ be sets.

Let $E \subseteq X \times Y$.

Let $x \in X$.

Then:


 * $\paren {\chi_E}_x = \chi_{E_x}$

where:


 * $E_x$ is the $x$-vertical section of $E$
 * $\chi_E$ and $\chi_{E_x}$ are the characteristic functions of $E$ and $E_x$ as subsets of $X \times Y$ respectively
 * $\paren {\chi_E}_x$ is the $x$-vertical function of $\chi_E$.

Proof
We show that:


 * $\map {\paren {\chi_E}_x} y = \begin{cases}1 & y \in E_x \\ 0 & y \not \in E_x\end{cases}$

at which point we'll be done from the definition of the characteristic function of $E_x$.

From the definition of the $x$-horizontal section, we have:


 * $\map {\paren {\chi_E}_x} y = \map {\chi_E} {x, y}$

From the definition of a characteristic function, we have:


 * $\map {\chi_E} {x, y} = 1$ $\tuple {x, y} \in E$.

From the definition of the $x$-vertical section, we then have:


 * $\map {\chi_E} {x, y} = 1$ $y \in E_x$.

So:


 * $\map {\paren {\chi_E}_x} y = 1$ $y \in E_x$.

From the definition of a characteristic function, we also have:


 * $\map {\chi_E} {x, y} = 1$ $\tuple {x, y} \not \in E$.

From the definition of the $x$-vertical section, we then have:


 * $\map {\chi_E} {x, y} = 0$ $y \not \in E_x$.

So:


 * $\map {\paren {\chi_E}_x} y = 0$ $y \not \in E_x$

giving:


 * $\map {\paren {\chi_E}_x} y = \begin{cases}1 & y \in E_x \\ 0 & y \not \in E_x\end{cases}$

which was the demand.