Let $A$ be a function s.t. $\mbox{dom}(A)=I\times J$.
Prove that: $\bigcup_{i \in I,j \in J}A_{ij}=\bigcup_{i \in I}\left ( \bigcup_{j \in J}A_{ij}\right )$.
The problem is that I cannot interpret the right-hand side of the equation.
By the book, if $A$ is a function with $\mbox{dom}A=I$,
$\bigcup_{i \in I}A_{i}:=\bigcup\mbox{rng}(A)$.
On the left-hand side of the equality I have to prove, we clearly have a function.
On the right-hand side, no matter how I rearrange it, I cannot interpret the nested union by definition.
Let me explain.
By definition, on the left-hand side I have
$\bigcup_{i \in I,j \in J}A_{ij}\\
:=\bigcup_{(i,j)\in I\times J}A(i,j)\\
:=\bigcup\mbox{rng}A.$
Let's see an example with:
$I\times J=\{(0,0),(0,1),(1,0),(1,1)\},\\
A(0,0)=\{0\},\\
A(0,1)=\{1\},\\
A(1,0)=\{2\},\\
A(1,1)=\{3\}.\\$
Then
$A=\{\\
((0,0),\{0\}),\\
((0,1),\{1\}),\\
((1,0),\{2\}),\\
((1,1),\{3\})\}.$
And in the following calculations we can see what we get left-hand side.
$\mbox{rng}A\\
=\{y:\exists x[(x,y)\in A]\}\\
=\{\{0\},\{1\},\{2\},\{3\}\}$.
$\bigcup_{(i,j)\in I\times J}A(i,j)\\
=\bigcup \mbox{rng}A\\
=\{x:\exists y \in \mbox{rng}A(x \in y)\}\\
=\{x:\exists y \in \{\{0\},\{1\},\{2\},\{3\}\}(x \in y)\}\\
=\{0,1,2,3\}$.
For the right hand side we define the following functions.
Let $i \in \{0,1\}$.
$A\upharpoonright (\{i\}\times \{0,1\})=A\cap (\{(i,0),(i,1)\}\times \mbox{rng}A),\\
A\upharpoonright (\{0\}\times \{0,1\})=A\cap (\{(0,0),(0,1)\}\times \mbox{rng}A),\\
A\upharpoonright (\{1\}\times \{0,1\})=A\cap (\{(1,0),(1,1)\}\times \mbox{rng}A)$.
We have then:
$\mbox{rng}A\upharpoonright (\{0\}\times \{0,1\}),\\
=\{y:\exists x[(x,y)\in A\upharpoonright (\{0\}\times \{0,1\})]\},\\
=\{\{0\},\{1\}\}$.
$\mbox{rng}A\upharpoonright (\{1\}\times \{0,1\}),\\
=\{y:\exists x[(x,y)\in A\upharpoonright (\{1\}\times \{0,1\})]\},\\
=\{\{2\},\{3\}\}$.
$\bigcup_{j \in J}A\upharpoonright (\{0\}\times \{0,1\})_{j}\\
:=\bigcup \mbox{rng}A\upharpoonright (\{0\}\times \{0,1\})\\
=\{x:\exists y \in \mbox{rng}A\upharpoonright (\{0\}\times \{0,1\})(x \in y)\}\\
=\{x:\exists y \in \{\{0\},\{1\}\}(x \in y)\}\\
=\{0,1\}$
$\bigcup_{j \in J}A\upharpoonright (\{1\}\times \{0,1\})_{j}\\
=\{1,2\}$.
So it is clear by this example that taking the union for $i=1,2$ we get $\{1,2,3,4\}$.
Still, I don't have the definition for the nested union so I cannot really prove the theorem not knowing what am dealing with.
Is there a more general definition of infinite union I am not aware of?
Notation from: Introduction to Set Theory, Donald Monk.
Best Answer
Forget that theorem about rng. What you want to prove is obvious/trivial. Show that $x$ is an element of the RHS if and only if $x$ is an element of the LHS.
Probably starting like so: $x\in\bigcup_i\bigcup_j A_{i,j}$ if and only if there exists $i$ such that $x\in\bigcup_jA_{i,j}$.