Let $\{I_j\}$ be a family of sets indexed by $J$ and let $$K=\bigcup_{j\in J}I_j$$
Then let $\{A_k\}$ be a family of sets indexed by $K$. The generalization of the associative law for unions is that
$$\bigcup_{k\in K}A_k=\bigcup_{j\in J}\left(\bigcup_{i\in I_j}A_i \right)$$
What I interpret this as is: "To take the union over $K$, pick an $I_j \in K$, perform the union of all $A_i$ such that $i\in I_j$, and for each $j\in J$ unite all this possible unions to get $\bigcup_{k\in K}A_k$. What this is saying is that the order in which the $j$ and thus the $I_j$ are picked is of no importance in the ultimate union. The above is a generalization of $$(A\cup B)\cup C=A\cup (B\cup C)$$
How can I find the analogous generalization for $$A \cup B=B \cup A?$$
Best Answer
Commutativity simply means that the order of operations is not important. One can see the obviousness of this claim for unions since: $$x\in\bigcup_{\lambda\in\Lambda}A_\lambda\iff\exists\lambda\in\Lambda:x\in A_\lambda$$
Now to say that the order does not matter is to say that if we shift the indices around (but do not remove any index!) then the result would be the same. Mathematically this translates to saying:
One last remark is that one should note that the axioms of ZF do not speak of $A\cup B$ directly, but rather $\bigcup\{A,B\}$, the set which is the union over the pair $\{A,B\}$.
In that aspect $\bigcup_{k\in K}A_k$ should be written as $\bigcup\{A_k\mid k\in K\}$, in which case it is obvious that any permutation of $K$ keeps the union the same, as the set over which we take the union remains the same.