Intersection of Vertical Sections is Vertical Section of Intersection

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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:

$\ds \paren {\bigcap_{\alpha \in A} E_\alpha}_x = \bigcap_{\alpha \in A} \paren {E_\alpha}_x$

where:

$\ds \paren {\bigcap_{\alpha \in A} E_\alpha}_x$ is the $x$-vertical section of $\ds \bigcap_{\alpha \in A} E_\alpha$
$\paren {E_\alpha}_x$ is the $x$-vertical section of $E_\alpha$.


Proof

Note that:

$\ds y \in \bigcap_{\alpha \in A} \paren {E_\alpha}_x$

if and only if:

$y \in \paren {E_\alpha}_x$ for all $\alpha \in A$.

From the definition of the $x$-vertical section, this is equivalent to:

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

This in turn is equivalent to:

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

Again applying the definition of the $x$-vertical section, this is equivalent to:

$\ds y \in \paren {\bigcap_{\alpha \in A} E_\alpha}_x$

So:

$\ds y \in \bigcap_{\alpha \in A} \paren {E_\alpha}_x$ if and only if $\ds y \in \paren {\bigcap_{\alpha \in A} E_\alpha}_x$

giving:

$\ds \paren {\bigcap_{\alpha \in A} E_\alpha}_x = \bigcap_{\alpha \in A} \paren {E_\alpha}_x$

$\blacksquare$