Horizontal Section of Cartesian Product

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 $y \in Y$.

Then:


 * $\paren {A \times B}^y = \begin{cases}A & y \in B \\ \O & y \not \in B\end{cases}$

where $\paren {A \times B}^y$ is a horizontal section of $A \times B$.

Proof
Let $y \in B$.

From the definition of the horizontal section, we have:


 * $x \in \paren {A \times B}^y$




 * $\tuple {x, y} \in A \times B$

Since $y \in B$, this equivalent to:


 * $x \in A$

So:


 * $x \in \paren {A \times B}^y$ $x \in A$

giving:


 * $\paren {A \times B}^y = A$ if $y \in B$.

Now let $y \in Y \setminus B$.

So, by the definition of set difference, we have $y \in Y$ and $y \not \in B$.

As before, we have:


 * $x \in \paren {A \times B}^y$




 * $\tuple {x, y} \in A \times B$

But this is equivalent to:


 * $x \in A$ and $y \in B$.

Since $y \not \in B$, there exists no $x \in \paren {A \times B}^y$.

So:


 * $\paren {A \times B}^y = \O$ if $y \not \in B$.