# Can we have $x\in A$ and $x\in A\times B$?

Is it possible, for sets $A$ and $B$, to have $x\in A$ and $x\in A\times B$?

It seems unlikely to me, but maybe some degenerate case? $x=\emptyset$?

Consider $A = \{x, (x, y)\}$ and $B = \{y\}$.

Then, $(x, y) \in A$ and $(x, y) \in A \times B$.

Let $A=\{0,(0,1)\}$. So $A$ has two elements; one element is a number and the other element is an ordered pair of numbers. Let $B=\{1\}$ and $x=(0,1)$. Then $x\in A$ and also $x\in A\times B$.

I will take the given answers, and 1-up them.

Let $X$ be some non-empty set. For example $X=\{\varnothing\}$. Define by induction, $A_0=X$ and $A_{n+1}=A_n\cup(A_n\times A_n)$. Now define $A=B=\bigcup A_n$.

Then $A\times B=A\times A\subseteq A$. If $(x,y)\in A\times A$, then for some large enough $n$ both $x$ and $y$ are elements of $A_n$, and therefore $(x,y)\in A_{n+1}\subseteq A$. And since $X$ is non-empty, $A$ is non-empty as well.