Is true that $\bigcup\Big(\bigcup( A\times B)\Big)=A\cup B$ for any pair of sets $A$ and $B$? So I'd like to know if the identity
$$
\bigcup\Big(\bigcup (A\times B)\Big)=A\cup B
$$
is true. So I tried to prove it by the following arguments. First of all I observed that any element of $A\times B$ is a set such as
$$
\big\{\{a\},\{a,b\}\big\}
$$
for any $a\in A$ and for any $b\in B$. Now if $x$ is an element of $\bigcup(A\times B)$ then there exists $X\in A\times B$ such that $x\in X$ so that
$$
\Big(x=\{a\}\,\,\,\text{for any}\,a\in A\Big)\vee\Big(x=\{a,b\}\,\,\,\text{for any}\,a\in A\,\,\,\text{and for any}\,b\in B\Big)
$$
Finally if $x$ is an element of $\bigcup\Big(\bigcup(A\times B)\Big)$ then there exists $X\in\bigcup A\times B$ such that $x\in X$ so that
$$
\Big(x\in\{a\}\,\,\,\text{for any}\,a\in A\Rightarrow x=a\in A\Big)\vee\Big(x\in\{a,b\}\,\,\,\text{for any}\,a\in A\,\text{and for any}\,b\in B\Rightarrow x=a\in A\vee x=b\in B\Big)
$$
which means that
$$
\bigcup\Big(\bigcup(A\times B)\Big)\subseteq(A\cup B)
$$
So unfortunately I am not able to prove the conversely: it seems to me that the preceding arguments can be applied conversely but I am hesitant about now so that I deliberated to post this question. Anyway it is well know that
$$
A\times B\subseteq\mathcal P\Big(\mathcal P(A\cup B)\Big)
$$
and so I think that the above argumentation can be avoided but I am yet hesitant about. So could anyone help me, please?
 A: Aside from the usefulness of the statement, I took it as an intriguing quiz.
$\newcommand{\ps}{\mathcal{P}}$
My conclusion is
$$\cup (\cup (A \times B)) \subseteq A \cup B$$
Let $x \in \cup (\cup (A \times B))$. Then there is $C \in \cup (A \times B)$ such that $x \in C$. Again, we have $D \in A \times B$ such that $C \in D$. In a single line
$$
x \in C \in D \in A \times B
$$
From Kuratowski's definition,
$$A \times B \subseteq \ps (\ps (A \cup B))$$
Hence, $D \in \ps (\ps (A \cup B))$, $D \subseteq \ps(A \cup B)$, and $C \in \ps(A \cup B)$. Again, $C \subseteq A \cup B$, which leads to $x \in A \cup B$.
(Turned out this is repetition of your proof but in a bit simpler form...)
However, as GEdgar pointed out in the comments, $ A \cup B \subseteq \cup (\cup (A \times B))$ has an obvious counterexample.
Remark: If both $A$ and $B$ are nonempty, for any $x \in A \cup B$ there are two possible cases:

*

*$x \in A \cap B$.
This leads to $(x, x) = \{\{x\}\} \in A \times B$. Hence, $\{x\} \in \cup(A \times B)$, and again, $x \in \cup(\cup(A \times B))$

*$x$ is in only one of $A$ and $B$, and the other set has some element $y$ other than $x$.
If $x \in A$ and $ y \in B $, $ (x,y) =\{\{x\},\{x,y\}\} \in A \times B$. Hence, $\{x\} \in \cup(A \times B)$. This give the same conclusion as the case $1$.
On the other hand, if $y \in A$ and $ x \in B $, $ (y,x) =\{\{y\},\{x,y\}\} \in A \times B$. Hence, $\{x,y\} \in \cup (A \times B) $ and $x \in \cup (\cup (A \times B))$.

Therefore, $\cup (\cup (A \times B)) = A \cup B$.
