Horizontal Section of Characteristic Function is Characteristic Function of Horizontal Section

Theorem

Let $X$ and $Y$ be sets.

Let $E \subseteq X \times Y$.

Let $y \in Y$.

Then:

$\paren {\chi_E}^y = \chi_{E^y}$

where:

$E^y$ is the $y$-horizontal section of $E$

$\chi_E$ and $\chi_{E^y}$ are the characteristic functions of $E$ and $E^y$ as subsets of $X \times Y$ respectively

$\paren {\chi_E}^y$ is the $y$-horizontal function of $\chi_E$.

Proof

We show that:

$\map {\paren {\chi_E}^y} x = \begin{cases}1 & x \in E^y \\ 0 & x \not \in E^y\end{cases}$

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

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

$\map {\paren {\chi_E}^y} x = \map {\chi_E} {x, y}$

From the definition of a characteristic function, we have:

$\map {\chi_E} {x, y} = 1$ if and only if $\tuple {x, y} \in E$.

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

$\map {\chi_E} {x, y} = 1$ if and only if $x \in E^y$.

So:

$\map {\paren {\chi_E}^y} x = 1$ if and only if $x \in E^y$.

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

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

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

$\map {\chi_E} {x, y} = 0$ if and only if $x \not \in E^y$.

So:

$\map {\paren {\chi_E}^y} x = 0$ if and only if $x \not \in E^y$

giving:

$\map {\paren {\chi_E}^y} x = \begin{cases}1 & x \in E^y \\ 0 & x \not \in E^y\end{cases}$

which was the demand.

$\blacksquare$