Complement of Vertical Section of Set is Vertical Section of Complement
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Theorem
Let $X$ and $Y$ be sets.
Let $E \subseteq X \times Y$.
Let $x \in X$.
Then:
- $\paren {\paren {X \times Y} \setminus E}_x = Y \setminus E_x$
where:
- $\paren {\paren {X \times Y} \setminus E}_x$ is the $x$-vertical section of the set difference $\paren {X \times Y} \setminus E$
- $E_x$ is the $x$-vertical section of $E$.
Proof
Note that from the definition of set difference, we have that:
- $y \in Y \setminus E_x$
- $y \in Y$ and $y \not \in E_x$.
That is, from the definition of the $x$-vertical section:
- $y \in Y$ and $\tuple {x, y} \not \in E$.
This is equivalent to:
- $\tuple {x, y} \in \paren {X \times Y} \setminus E$
From the definition of the $x$-vertical section, this is then equivalent to:
- $y \in \paren {\paren {X \times Y} \setminus E}_x$
So we have:
- $y \in Y \setminus E_x$ if and only if $y \in \paren {\paren {X \times Y} \setminus E}_x$.
So:
- $\paren {\paren {X \times Y} \setminus E}_x = Y \setminus E_x$
$\blacksquare$