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.
Best Answer
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
$$A_0\cup(A_1\cup A_2)=(A_0\cup A_1)\cup A_2\;.$$