Is there a non empty set X such that $X \subseteq X\times X$? I came up with this idea but I can't seem to prove it. Maybe I have to prove that X is the set of every set (such that X doesn't exist), or that X is empty, but I am not getting anything done and I would like a bit of help. Just for reference, I am just starting first year of undergrad, so I would prefer an answer that doesn't require much knowledge beyond set theory. Thanks a lot.

Assuming the axiom of regularity (a standard ZFC axiom) the existence of such a set is not possible.
Assume otherwise that $X$ is a non-empty set satisfying $X \subseteq X \times X$ and take $Y = X \cup \bigcup X$. By the mentioned axiom there is an $\in$-minimal element $y \in Y$.

*

*If $y \in X$, then $y = \left< a, b \right> = \{ \{ a \}, \{ a, b \} \}$ for some $a, b \in X$. But then $\{ a \} \in \bigcup X$ and $\{ a \} \in y$, so $y$ is not $\in$-minimal.


*Else $y \in \bigcup X$, so there is some $x \in X$ with $y \in x$. But then again we can write $x = \{ \{ a \}, \{ a, b \} \}$ where $a, b \in X$. Then $y = \{ a \}$ or $y \in \{ a, b \}$. But $a \in y$ and $a \in X$ so again $y$ is not $\in$-minimal. $\blacksquare$
