Which definition should I use to prove $\bigcup_i A_i \times \bigcup_j B_j = \bigcup_{i, j}(A_i \times B_j)$

Question

I'm studying some set theory from Halmos' book. He asks to prove $$\bigcup_i A_i \times \bigcup_j B_j = \bigcup_{(i, j) \in I \times J}(A_i \times B_j)$$

I want to prove it using the definition of generalized version of Cartesian product, not just taking $(a, b)$ as elements. But I struggle with the idea that elements of the set from the left side of the equality differ from the ones on the right side. Namely, on the left side we will have a set of families with an unordered pair as the index set (because it is a cartesian product of just two sets in the end) but on the right side we have a set of families with also unordered pairs as index sets but different for each family.

So, what is the proper way to prove it? Is it possible to prove it with the generalized version of the Cartesian product? And if yes, then should I introduce somehow a choice function or what other steps should I take?

P.S.: I am interested in more rigorous proof like taking an element from both sides and showing that it is also in another side. The problem I described appears on that level.

Answer 1

Equivalent are the statements:

• $(a,b)\in\bigcup_iA_i\times\bigcup_jB_j$
• $a\in\bigcup_iA_i\wedge b\in\bigcup_jB_j$
• $\exists i[a\in A_i]\wedge\exists j[b\in B_j]$
• $\exists (i,j) [a\in A_i\wedge b\in B_j]$
• $\exists (i,j) [(a,b)\in A_i\times B_j]$
• $(a,b)\in\bigcup_{(i,j)}(A_i\times B_j)$

This allows us to conclude that:$$\bigcup_iA_i\times\bigcup_jB_j=\bigcup_{(i,j)}(A_i\times B_j)$$

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edit

LHS denotes the set $F$ of functions $\{0,1\}\to(\bigcup_iA_i)\cup(\bigcup_jB_j)$ that satisy $f(0)\in\bigcup_iA_i$ and $f(1)\in\bigcup_jB_j$.

RHS denotes a collection $\bigcup_{i,j}F_{i,j}$ where $F_{i,j}$ is the set of functions $\{0,1\}\to A_i\cup B_j$ that satisfy $f(0)\in A_i$ and $f(1)\in B_j$.

Now it must be proved that $F=\bigcup_{i,j}F_{i,j}$

It is evident that $F_{i,j}\subseteq F$ for every pair $(i,j)$, showing that set on RHS is a subset of set on LHS.

If $f\in F$ then for some pair $(i,j)$ we have $f(0)\in A_i$ and $f(1)\in B_j$, so that $f\in F_{i,j}$. This shows that set on LHS is a subset of set on RHS.

Proved is now that the sets are equal.

Answer 2

(x,y) in $(\cup_i A_i) × \cup_j B_j$ iff
exists i,j with x in $A_i,$ y in $B_j$ iff
exists i,j with (x,y) in $(A_i × B_j)$ iff
(x,y) in $\cup_{i,j} (A_i × B_j)$

$(\cup_i A_i) × B_j = \cup_i (A_i×B_j)$
Let K = $\cup_i A_i$
K × $\cup_j B_j = \cup_j (K×B_i) = \cup_{i,j} (A_i×B_j)$