Let $A$ be a set such that each member of $A$ is a non-empty set. Then $A \times X \neq x$ for all $X \in A, x \in \bigcup A$.

Question

I tried so hard but to no avail while proving the statement:

Let $A$ be a set such that each member of $A$ is a non-empty set. Then $A \times X \neq x$ for all $X \in A, x \in \bigcup A$. Here $A \times X$ is the Cartesian product of $A$ and $X$.

Please shed some light! Thank you for your help!

Answer 1

$A$ is a collection of sets, $X$ is a set in $A$ and $x$ is an
element. $x = A\times X$ has to violate regularity. Proof.

There is some $K \in A$ with $x \in K$ and some $y \in X$.
Thus $x \in K \in \{K,y\} \in \{\{K\}, \{K,y\}\} = (K,y) \in A \times X = x$.

Answer 2

$A$ is not empty and from $X\in A$ it follows that $X$ is not empty, so let $y$ denote some element of $X$.

Now let $x\in a\in A$.

Then $\{\{a\},\{a,y\}\}=\langle a,y\rangle\in A\times X$.

Assumption $A\times X=x$ then leads to $\langle a,y\rangle\in x$.

But then $x\in a\in\cup\cup x$ so that $x\overline{\in}x$.

However, with induction on $\overline{\in}$ it can easily be shown that:$$\forall x[\neg x\overline{\in}x]$$

So if the axiom of regularity is accepted (which makes induction on $\overline{\in}$ legitimate) then the assumption $A\times X=x$ must be rejected.

Here $\overline{\in}$ denotes the transitive closure of $\in$.