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:
- $\ds x \in \bigcup_{\alpha \mathop \in A} \paren {E_\alpha}^y$
- $x \in \paren {E_\alpha}^y$ 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:
- $\ds \paren {\bigcup_{\alpha \mathop \in A} E_\alpha}^y = \bigcup_{\alpha \mathop \in A} \paren {E_\alpha}^y$
$\blacksquare$