# Notation for the union of a system of sets

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?

• The function you are looking for in (1) is $C\mapsto \bigcup_{a \in C}F_{a}$. Is that not formal enough for you? – Carsten S Dec 25 '13 at 22:30
• 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)? – Carsten S Dec 25 '13 at 22:31
• @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). – user36546 Dec 25 '13 at 23:26
• Great, problem solved :) – Carsten S Dec 26 '13 at 0:17
• @CarstenSchultz Not for the 3rd question yet... – user36546 Dec 26 '13 at 2:10