Vertical Section of Characteristic Function is Characteristic Function of Vertical Section
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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:
- $\map {\chi_E} {x, y} = 1$ if and only if $\tuple {x, y} \in E$.
From the definition of the $x$-vertical section, we then have:
- $\map {\chi_E} {x, y} = 1$ if and only if $y \in E_x$.
So:
- $\map {\paren {\chi_E}_x} y = 1$ if and only if $y \in E_x$.
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 $x$-vertical section, we then have:
- $\map {\chi_E} {x, y} = 0$ if and only if $y \not \in E_x$.
So:
- $\map {\paren {\chi_E}_x} y = 0$ if and only if $y \not \in E_x$
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.
$\blacksquare$