Vertical 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 $x \in X$.

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:

$y \in \paren {A \times B}_x$

if and only if:

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

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:

$y \in \paren {A \times B}_x$

if and only if:

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

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 $x \not \in A$.

$\blacksquare$