Union of Vertical Sections is Vertical Section of Union
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Theorem
Let $X$ and $Y$ be sets.
Let $\set {E_\alpha : \alpha \in A}$ be a set of subsets of $X \times Y$.
Let $x \in X$.
Then:
- $\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x = \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
where:
- $\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x$ is the $x$-vertical section of $\ds \bigcup_{\alpha \in A} E_\alpha$
- $\paren {E_\alpha}_x$ is the $x$-vertical section of $E_\alpha$.
Proof
Note that:
- $\ds y \in \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
- $y \in \paren {E_\alpha}_x$ for some $\alpha \in A$.
From the definition of the $x$-vertical section, this is equivalent to:
- $\tuple {x, y} \in E_\alpha$ for some $\alpha \in A$.
This in turn is equivalent to:
- $\ds \tuple {x, y} \in \bigcup_{\alpha \in A} E_\alpha$
Again applying the definition of the $x$-vertical section:
- $\ds y \in \paren {\bigcup_{\alpha \in A} E_\alpha}_x$
So:
- $\ds y \in \bigcup_{\alpha \in A} \paren {E_\alpha}_x$ if and only if $\ds y \in \paren {\bigcup_{\alpha \in A} E_\alpha}_x$
giving:
- $\ds \paren {\bigcup_{\alpha \in A} E_\alpha}_x = \bigcup_{\alpha \in A} \paren {E_\alpha}_x$
$\blacksquare$