A help to understand the generalized version of the associative law of union While I was looking for a better understanding of the the concept of families (which is not yet entirely clear) in the Halmos book, I found me with this: 
Let $\left\{ I_j \right\}$ be a family of sets with domain $J$; write $K = \bigcup_j I_j$ and let $\left\{ A_k \right\}$ be a family of sets with domain $K$. It is not difficult to prove that: 
$$\bigcup_k A_k = \bigcup_{j\in J} \bigg( \, \bigcup_ {i\in I_j}A_i\, \bigg) $$  
So, I have two question about this: 
First: Is the next proof correct?
$$\bigcup_k A_k = \bigcup_{j\in J} \bigg( \, \bigcup_ {i\in I_j}A_i\, \bigg) $$  
($\Rightarrow$) Suppose $z\in \bigcup_k A_k $. Then there is some $k\in K$ such that $z\in A_k$. But since  $K = \bigcup_j I_j$, $k\in K$ means  $k\in I_j$ for at least one $j\in J$. So, $z\in A_k$ for some $k\in I_j$, i.e., $z\in \bigcup_{k\in I_j} A_k$; for at least one $j\in J$. So then, $z\in \bigcup_{k\in I_j} A_k$ for some $j\in J$, i.e., $z\in \bigcup_{j\in J} \big( \, \bigcup_ {k\in I_j}A_k\, \big).$ 
($\Leftarrow$) Now suppose $z \in \bigcup_{j\in J} \big( \, \bigcup_ {i\in I_j}A_i\, \big)$. Then there is some $j\in J$ such that $z \in \bigcup_ {i\in I_j}A_i$. For $z \in \bigcup_ {i\in I_j}A_i$ in turn there is an $i\in I_j$ such that $z\in A_i$. Let's define the set $K := \bigcup_j I_j$ so clearly $i \in K$. Then there exists an $i \in K$ such that $z\in A_i$, so $z\in \bigcup_{i\in k} A_i$. $\;\;\; \Box$       
And second: At the end of the paragraph the author says: "This is the generalized version of the associative law of union". But I cannot see how that generalized the associative law. Could somebody explain me the reason for which it is the generalized form, if it is not too much trouble, please? 
As usual thanks in advance.  
 A: Your proof is correct. I guess one way to see why this is a generalization is to note that usually the associative law of union is stated via three sets:$$A \cup (B \cup C) = (A\cup B)\cup C.$$
We can also write it in this form:
$$A_1 \cup (A_2 \cup A_3) = (A_1\cup A_2)\cup A_3.$$
The LHS corresponds to the case where $I_1 = \{1\}$, $I_2 = \{2,3\}$, and $J=\{1,2\}$, and the RHS corresponds to the case where $I_1 = \{1,2\}$, $I_2 = \{3\}$ and $J=\{1,2\}$. They are both equal to $\bigcup_{k\in K} A_k$ with $K = \{1,2,3\}$
A: $\cup_k A_k$ is defined in your equations to be a union of unions, with potentially infinitely many sets being unioned together in each union and potentially infinitely many unions being joined together into the final union.  So the fact that the "union of unions" is equal to the union of all the constituents is a generalization of the basic associative law fact that $(A \cup B) \cup C$ = $A \cup (B \cup C)$, because you could define the union-of-unions any way you want, distributing the sets among the individual unions being unioned together, just as long as the final collection of constituent sets is the same.  Hope that helps.  And yes, your proof looks correct.  :) 
A: Your proof is fine.
Consider $I_0=\{0\}$, $I_1=\{1,2\}$, and $J=\{0,1\}$. Then $K=\{0,1,2\}$, so the result essentially says that 
$$\bigcup_{k\in K}A_k=A_0\cup(A_1\cup A_2)\;.$$
If instead you set $I_0=\{0,1\}$ and $I_1=\{2\}$, it essentially says that
$$\bigcup_{k\in K}A_k=(A_0\cup A_1)\cup A_2\;.$$
Indirectly, therefore, it says that
$$A_0\cup(A_1\cup A_2)=(A_0\cup A_1)\cup A_2\;.$$
A: Another form of generalized associated law is as follows:
\begin{align*}
\bigcap_{i\in I}\bigcap_{j\in J}A_{i.j}&=\bigcap_{j\in J}\bigcap_{i\in I}A_{i.j}\\
\bigcup_{i\in I}\bigcup_{j\in J}A_{i.j}&=\bigcup_{j\in J}\bigcup_{i\in I}A_{i.j}.
\end{align*}
The essence of "associated" is: no matter which part we intersect or unit first the result are the same. This form is more intuitive than the form you give, but I think the essence of the both are the same.
