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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 $$A \cup B=B \cup A?$$

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Maybe something along the lines of (possibly infinite) permutation groups. –  user31373 Jul 14 '12 at 22:35

2 Answers 2

up vote 3 down vote accepted

Let $K$ be an index set, and $S(K)$ be group of bijections from $K$ to $K$. Then one can say that for $\sigma\in S(K)$ and arbitrary family $\{A_k:k\in K\}$ we have $$ \bigcup\limits_{k\in K} A_k=\bigcup\limits_{k\in K} A_{\sigma(k)} $$

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@PeterTamaroff, even if you intriduce new notation it still will be $O(K)=S(K)$. As for generalization it is possible but not necessary. –  no identity Jul 14 '12 at 22:44
    
Oh, right. Silly me. Thank you. –  Pedro Tamaroff Jul 14 '12 at 22:45

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:

If $\sigma$ is any permutation of the index set $K$ then $$\bigcup_{k\in K}A_k=\bigcup_{k\in K}A_{\sigma(k)}$$

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.

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Peter, if you prove that $\cal A_1=A_2$, then $\bigcup\cal A_1=\bigcup A_2$. If $S$ is any permutation of $K$ then $\{A_k\mid k\in K\}=\{A_k\mid k\in S(K)\}$. –  Asaf Karagila Jul 14 '12 at 22:50
    
@Peter: However Norbert describes a set of permutations of $K$, so the index set is different, and thus the family itself may have changed. –  Asaf Karagila Jul 14 '12 at 22:52
    
Because now your index set is different. Recall that writing $\{A_k\mid k\in K\}$ is to say "I have a function from $K$ into the universe" but now you want to say "I have a function from $S(K)$ into the universe" and there is no guarantee that the range of this function is the same. –  Asaf Karagila Jul 14 '12 at 22:58
    
No, I cannot drop by the chat for a second. Let me clarify. The notation $\mathcal A=\{A_k\mid k\in K\}$ is to say $\mathcal A$ is the range of a function from $K$ into some power set, and $A_k$ is the image of $k$. Look at the comment you wrote, $k\in S(K)$. However $S(K)$ is a group of permutations of $K$, which is not $K$ at all. In fact it is usually of a much larger cardinality. So now you have a different function, which means that the sets may be different. All this, even though you use $A_k$ to denote your sets. Hooray for quantified variables! –  Asaf Karagila Jul 14 '12 at 23:02
    
Yes, you misunderstood what $S(K)$ meant. Read Norbert's answer again. It says $S(K)$ a set of permutation of $K$, then for $\sigma\in S(K)$ ... the set $S(K)$ is a collection of functions, not the image of $K$ under some permutation. Note that my first reply assumed that you meant $S$ to be a specific permutation of $K$, but I figured that since you accepted his answer you read it too. :-) –  Asaf Karagila Jul 14 '12 at 23:08

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