# Is true that $\bigcup\Big(\bigcup( A\times B)\Big)=A\cup B$ for any pair of sets $A$ and $B$

elementary-set-theoryexamples-counterexamplessolution-verification

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?

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:

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