Union of Horizontal Sections is Horizontal Section of Union

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Theorem

Let $X$, $Y$ and $A$ be sets.

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

Let $y \in Y$.


Then:

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

where:

$\ds \paren {\bigcup_{\alpha \mathop \in A} E_\alpha}^y$ is the $y$-horizontal section of $\ds \bigcup_{\alpha \mathop \in A} E_\alpha$
$\paren {E_\alpha}^y$ is the $y$-horizontal section of $E_\alpha$.


Proof

Note that:

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

if and only if:

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

From the definition of the $y$-horizontal section, this is equivalent to:

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

This in turn is equivalent to:

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

Again applying the definition of the $y$-horizontal section, this is the case if and only if:

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

So:

$\ds x \in \bigcup_{\alpha \mathop \in A} \paren {E_\alpha}^y$ if and only if $\ds x \in \paren {\bigcup_{\alpha \mathop \in A} E_\alpha}^y$

giving:

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

$\blacksquare$