Intersection of Vertical Sections is Vertical Section of Intersection

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:


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




 * $y \in \paren {E_\alpha}_x$ 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:


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

giving:


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