Vertical Section of Cartesian Product
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Theorem
Let $X$ and $Y$ be sets.
Let $A \subseteq X$ and $B \subseteq Y$, so that $A \times B \subseteq X \times Y$.
Let $x \in X$.
Then:
- $\paren {A \times B}_x = \begin{cases}B & x \in A \\ \O & x \not \in A\end{cases}$
where $\paren {A \times B}_x$ is the $x$-vertical section of $A \times B$.
Proof
Let $x \in A$.
From the definition of the horizontal section, we have:
- $y \in \paren {A \times B}_x$
- $\tuple {x, y} \in A \times B$
Since $x \in A$, this equivalent to:
- $y \in B$
So:
- $y \in \paren {A \times B}_x$ if and only if $y \in B$
giving:
- $\paren {A \times B}_x = B$ if $x \in A$.
Now let $x \in X \setminus A$.
So, by the definition of set difference, we have $x \in X$ and $x \not \in A$.
As before, we have:
- $y \in \paren {A \times B}_x$
- $\tuple {x, y} \in A \times B$
But this is equivalent to:
- $x \in A$ and $y \in B$.
Since $x \not \in A$, there exists no $y \in \paren {A \times B}_x$.
So:
- $\paren {A \times B}_x = \O$ if $x \not \in A$.
$\blacksquare$