Decide whether $\{ x | x=x\times x\}$ is a set This is an exercise my set theory teacher suggested we try and I feel like the answer is yes (because it seems to me that no set other than $\emptyset$ satisfies $x\times x=x$, then the collection would be exactly $\{\emptyset\}$, which is a set), but I am stuck trying to prove that if $x\neq\emptyset$ then $x\neq x\times x$. I have tried using the axiom of regularity and cardinality supposing $x=x\times x$ but I reach no contradiction... could you give me any clues? (the theory is ZFC)
 A: You are absolutely correct that it suffices to show that, for any $X$, if $X\neq\varnothing$ then $X\times X\neq X$. To see this, suppose for contradiction that $X$ is non-empty and equal to its Cartesian square. The idea here is that we want to construct an infinite $\in$-descending chain, which will contradict the axiom of regularity. How might we do this? Well, since $X$ is non-empty, there exists $x_0\in X$, and since $X=X\times X$ there exists $x_1,y_1,\in X$ such that $x_0=\langle x_1,y_1\rangle$. Now we have $x_1\in\{x_1\}\in x_0$, and since $x_1\in X=X\times X$ there exist $x_2,y_2\in X$ such that $x_1=\langle x_2,y_2\rangle$, whence $$x_2\in\{x_2\}\in x_1\in\{x_1\}\in x_0.$$ Hopefully it is intuitively clear that continuing in this way will give us the desired $\in$-chain. So, how do we capture the idea of this argument rigorously? Well, note that all of the elements in our chain are either elements of $X$ or of form $\{x\}$ for some $x\in X$. So, with this in mind, define $Y\subseteq\mathcal{P}(X)$ to be the set of all subsets of $X$ of size $1$. ($Y$ exists by restricted comprehension.) Now, let $Z=X\cup Y$. Since $X$ is non-empty, $Z$ is non-empty too, so by the axiom of regularity there exists $z\in Z$ such that $z\cap Z=\varnothing$. By definition of $Z$, we have two cases.

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*If $z\in Y$, then by definition of $Y$ we have $z=\{x\}$ for some $x\in X$, and then $x\in z\cap X\subseteq z\cap Z$.

*If $z\in X$, then since $X=X\times X$ we have $z=\langle x,y\rangle$ for some $x,y\in X$. But now $\{x\}\in z\cap Y\subseteq z\cap Z$.

In either case, we see that $z\cap Z$ is non-empty, which gives the desired contradiction.
