# What's the meaning of an element that belongs to the same element?

In classical set theory, if I consider that $x$ is an element, which means it is not a set, can I write $x \in x$ ? If yes, what this would mean? Correct me if I am wrong, but I don't need to have some hierarchy here to apply the belonging operator, it means if I consider that exist a basic element, like the numbers, the operator is not only defined to the sets?

• That $x$ is an element does not mean $x$ is not also a set. Sets are elements of other sets. – Michael Hardy Jul 6 '15 at 22:41
• Such a set is called a Quine atom and its existence is negated in ZF(C) set theory by the Axiom of Regularity. – Hagen von Eitzen Jul 6 '15 at 22:43
• What's the sense if I write 1 $\in$ 1 or 1 $\in$ 2 for example? – Mike Jul 6 '15 at 22:45
• @ExampleMo If we define $1=\{\emptyset\}$ and $2=\{\emptyset,\{\emptyset\}\}$ as usual, then $1\notin 1$, $1\in 2$, and $2\notin 2$. Also $1\subseteq 2$. – Hagen von Eitzen Jul 6 '15 at 22:46

You wrote $x\in x$. This means that you can in fact write that. What does it mean? It means that you claim that $x$ is an element of itself. But $x$ is just a placeholder, a free variable.
First-order logic is not typed. You can impose stratification which disallows $x\in x$ or $x\notin x$ as legit formulas. This is what Quine suggested as a restriction for the comprehension axiom schema. This is what type theory is all about, having different types of objects and syntactic restriction about what you can and cannot write about these objects.
But in first-order logic, this is not the case. The formula $x\in x$ is a well-formed formula in the language of set theory. And then you should probably ask, is it even possible that such $x$ exists? The answer is that assuming the standard axioms of set theory (read: $\sf ZFC$) the answer is negative, the axiom of regularity implies that $\forall x(x\notin x)$. But without the axiom of regularity it is consistent that there are such objects in the universe.
You could argue that $1\in 1$ is just a false sentence. And you'd be right, assuming $\sf ZFC$ anyway. But you should also remember that in modern set theory everything is a set, including the elements of other sets. Again, you can change the settings of your universe and have objects which are not sets, but foundational research has shown that you gain nothing and lose nothing in terms of what you can do and prove about "everyday mathematics", other than perhaps your peace of mind knowing that $1\in 2$ and $\ln(2)\subseteq 1$ are true statements under some interpretations.
That $x$ is an element does not mean $x$ is not also a set. Sets are elements of other sets. But if you write $x\in x$, that means that $x$ is an element, since $x\in x$ means $x$ is an element of $x$. It also means $x$ is a set, since something is an element of it.
However, in the most usual forms of set theory, nothing is an element of itself; the statement $x\in x$ is always false. And $x\subset x$ is always true.