Theorem. Let $I,J$ be sets and let $\alpha \in I$ and $\beta \in J$ be labels such that $A_{\alpha}$ and $B_{\beta}$ are indexed sets. Then $\left( \bigcup\limits_{\alpha \in I} A_{\alpha} \right) \cap \left( \bigcup\limits_{\beta \in J} B_{\beta} \right) = \bigcup\limits_{(\alpha, \beta) \in I \times J} A_{\alpha} \cap B_{\beta}$.

Proof.

Two sets are equal iff every element of one set is element of the other and vice versa. From the definition of a union we have $$y \in \bigcup\limits_{(\alpha, \beta) \in I \times J} A_{\alpha} \cap B_{\beta} \leftrightarrow y \in (\exists (\alpha, \beta) \in I \times J) (y \in A_{\alpha} \wedge y \in B_{\beta}).$$ Whereas from the definition of the Cartesian product it follows that $$(\alpha, \beta) \in I \times J \leftrightarrow (\exists x \in I)(\exists y \in J) (\alpha, \beta) = (x, y),$$ which then serves as a new quantifier in the identity $$((\exists \alpha \in I) \: y \in A_{\alpha} \wedge (\exists \beta \in J) \: y \in B_{\beta}) \leftrightarrow (\exists \alpha \in I)(\exists \beta \in J) (y \in A_{\alpha} \wedge y \in B_{\beta}).$$ The rhs translates to $$(\exists \alpha \in I)(\exists \beta \in J) (y \in A_{\alpha} \wedge y \in B_{\beta}) \leftrightarrow (y \in A_{\alpha_1} \vee y \in A_{\alpha_2} \vee ...) \wedge (y \in B_{\beta_1} \vee y \in B_{\beta_2} \vee ...) \leftrightarrow (\exists \alpha \in I) \: y \in A_{\alpha} \wedge (\exists \beta \in J) \: y \in B_{\beta} \leftrightarrow y \in \left( \bigcup\limits_{\alpha \in I} A_{\alpha} \right) \cap \left( \bigcup\limits_{\beta \in J} B_{\beta} \right).$$ $\Box$

2015 FEB 15 (v1.0)
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