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:

$y \in Y \setminus E_x$

if and only if:

$y \in Y$ 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:

$\paren {\paren {X \times Y} \setminus E}_x = Y \setminus E_x$

$\blacksquare$