Union of Horizontal Sections is Horizontal Section of Union

Theorem
Let $X$ and $Y$ be sets.

Let $\set {E_\alpha : \alpha \in A}$ be a set of subsets of $X \times Y$.

Let $y \in Y$.

Then:


 * $\ds \paren {\bigcup_{\alpha \in A} E_\alpha}^y = \bigcup_{\alpha \in A} \paren {E_\alpha}^y$

Proof
Note that:


 * $\ds x \in \bigcup_{\alpha \in A} \paren {E_\alpha}^y$




 * $x \in \paren {E_\alpha}^y$ for some $\alpha \in A$.

This is equivalent to:


 * $\tuple {x, y} \in E_\alpha$ for some $\alpha \in A$.

This in turn is equivalent to:


 * $\ds \tuple {x, y} \in \bigcup_{\alpha \in A} E_\alpha$

This is the case :


 * $\ds x \in \paren {\bigcup_{\alpha \in A} E_\alpha}^y$