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The notation $\bigcup_{i \in I}A_{i}$ denotes the union of the range of a function, and as such, is used only if we are considering some function (an indexed family).

The generalized associative law for unions can be stated like this:

$$\bigcup_{a \in \bigcup S}F_{a}=\bigcup_{C \in S}\left(\bigcup_{a \in C}F_{a}\right)$$

  1. Does the first $\bigcup$ on the RHS gain a new meaning when used in a row as above? If not, what is the function whose union of the range it represents?

  2. Concerning the second $\bigcup$, I guess we might assume that \begin{align} x \in \bigcup_{a \in C}F_{a} &\iff x \in \bigcup \text{ran }F, \quad F:C \to \{F_{a} \mid a \in C\}\\ &\iff \exists b \in \text{ran }F(x \in b)\\ & \iff \exists b(b \in \text{ran }F\land x \in b)\\ & \iff \exists b(\exists a \in C(aFb) \land x \in b)\\ &\iff \exists b(\exists a (a \in C \land aFb) \land x \in b)\\ \end{align}

    But, also in an intuitive way \begin{align} x \in \bigcup_{a \in C}F_{a} &\iff \exists a \in C(x\in F_{a})\\ & \iff\exists a (a \in C \land x \in F_{a})\\ &\iff \exists a(a \in C \land\exists b(aFb \land x \in b)) \end{align} Are these statements really equivalent?

  3. I understand the idea of the generalized associative law for unions, but how can a precise definiton for $x \in \bigcup_{C \in S}(\bigcup_{a \in C}F_{a})$ be formulated?

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    $\begingroup$ The function you are looking for in (1) is $C\mapsto \bigcup_{a \in C}F_{a}$. Is that not formal enough for you? $\endgroup$ – Carsten S Dec 25 '13 at 22:30
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    $\begingroup$ We have $ \exists b(\exists a (a \in C \land aFb) \land x \in b)\iff \exists b(\exists a (a \in C \land aFb \land x \in b)) \iff \exists a(\exists b (a \in C \land aFb \land x \in b)) \iff \exists a(a \in C \land\exists b(aFb \land x \in b))$. Does this answer (2)? $\endgroup$ – Carsten S Dec 25 '13 at 22:31
  • $\begingroup$ @CarstenSchultz Yes, it's formal enough. I didn't realise $C$ in $\bigcup_{a \in C}F_a$ was an index as $i$ is in $\bigcup_{i \in I}A_{i}$. The other comment answers (2). $\endgroup$ – user36546 Dec 25 '13 at 23:26
  • $\begingroup$ Great, problem solved :) $\endgroup$ – Carsten S Dec 26 '13 at 0:17
  • $\begingroup$ @CarstenSchultz Not for the 3rd question yet... $\endgroup$ – user36546 Dec 26 '13 at 2:10
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\begin{align} x\in \bigcup_{C\in S}\left(\bigcup_{a\in C}F_{a}\right)&\iff \exists C\in S\left(x\in \bigcup_{a\in C}F_{a}\right)\\ \\ \\ \\ &\iff \exists C\in S(\exists a\in C(x\in F_{a}))\\ \\ &\iff \exists C\in S(\exists a(a\in C\land x\in F_{a}))\\ \\ &\iff \exists C(C\in S\land \exists a(a\in C\land x\in F_{a}))\\ \\ &\iff \exists C(\exists a(C\in S\land a\in C\land x\in F_{a}))\\ \\ &\iff \exists a(\exists C(C\in S\land a\in C\land x\in F_{a}))\\ \\ \\ &\iff \exists a(\exists C(C\in S\land a\in C)\land x\in F_{a})\\ \\ &\iff \exists a(\exists C\in S(a\in C)\land x\in F_{a})\\ \\ &\iff \exists a(a\in \bigcup S\land x\in F_{a})\\ \\ &\iff \exists a\in \bigcup S(x\in F_{a})\\ \\ &\iff x\in \bigcup_{a\in \bigcup S} F_{a}\\ \end{align}

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