# Formal definition of definite condition of membership $\in$ in set theory?

We have sets $$\{ x \}$$ and $$\{ \{ x \} \}$$. Then it holds that $$x \in \{ x \}$$ but $$x \notin \{ \{ x \} \}$$. It seems that the condition of membership ($$\in$$) presupposes that only those things in a set $$A$$ which are only in $$A$$ and in no set in $$A$$, are the members of $$A$$. More simply, only those things that are in $$A$$ in its "first layer" are the members of $$A$$. But apart from using natural language, how can one define $$\in$$? Is this even possible in set theory or do we have to use something outside of it (such as first-order logic) to formally define $$\in$$?

• In what I would call "usual" set-theory there are two primitive notions. This in the sense that they are not defined. It is not defined what a set is and secondly the relation $\in$ is not defined. There are other concepts of set-theory that concern e.g. "urelements" but I am not familiar with that. Commented Mar 18, 2020 at 9:29
• Wow, thanks for the insight. So we just take for granted that $\in$ concerns only those things in sets which are on their "first layers"? Commented Mar 18, 2020 at 9:31
• We don't "take it for granted"; it's what the symbols mean. Commented Mar 18, 2020 at 9:34
• @MaliceVidrine OK, my clumsy wording is at fault here. My concern is that what $\in$ means is determined solely by our natural-language expression of this concept. In contrast, logical systems usually include their own precise semantics which do not fall prey to vagueness or reference to our (imprecise) natural languages. Commented Mar 18, 2020 at 9:37
• It may help to realize that the $\{x\}$ is not actually a term in the language of set theory. The expression $y\in\{x_1,x_2,\ldots,x_n\}$ is a shorthand for something equivalent to $\exists z(y\in z \wedge \forall w(w\in z \Leftrightarrow w=x_1\vee\ldots\vee w=x_n))$. There are no natural language ambiguities involved in deciding whether or not $x\in\{\{x\}\}$ if you know what the actual set theoretic statement is (and the axioms involved). Commented Mar 18, 2020 at 18:56

Regarding your statement " It seems that the condition of membership (∈) presupposes that only those things in a set A which are only in A and in no set in A, are the members of A.": I don't know why it seems that way, but it's not so. Say for example $$S=\{1,\{1\}\}.$$Then $$1$$ is an element of $$S$$, even though it's also an element of an element of $$S$$.

The fact that $$x\notin\{\{x\}\}$$ has nothing to do with that. By definition $$S=\{\{x\}\}$$ has exactly one element, namely $$\{x\}$$; since $$x\ne\{x\}$$ this says $$x$$ is not an element of $$S$$.

Since the OP tagged his question with philosophy, they should accept/contemplate on the following:

• The universe of objects that can be examined are sets and the $$\in$$ relation is used
to determine when two sets are equal.

• There exists a unique object in set theory defined by

$$\tag 1 (\exists X) \, (\forall x) \; [x \notin X]$$

Their 'marching orders' (allowing them to enter the paradise of set theory) is to study and analyze 'logically coherent frameworks' allowing them to 'expand off' of the above 'ground-floor philosophical foundation' containing at least this one object, that is named, in natural language, the empty set; it is denoted by $$\emptyset$$.

One path of study that has been intensely scrutinized can be found in a wikipedia outline article:

$$\quad$$ Zermelo–Fraenkel set theory

Since the OP asked about first-order logic, they should closely examine the leading/introductory paragraph in the Axioms section of that article.

...how can one define $$\in$$?
In formal abstract set theory both sets and the membership relation $$\in$$ are primitive concepts. So no benefits can be accrued by attempting to describe a set as, say, consisting of all the objects in its "first layer". Rather, the framework/rules allow one to 'play the game' in the sense of David Hilbert,
The notion of membership $$\in$$ is not defined in set theory, it is assumed. It is also an axioms, that whenever two sets satisfy $$\forall x, x\in A \Leftrightarrow x\in B$$, then $$A = B$$. So it makes sense to define sets by only specifying their elements.
Then the set $$\{ z \}$$ is defined to be the set so that $$z\in\{z\}$$, and $$\forall y,y\neq z \Rightarrow y\notin\{z\}$$. So you see in your example, it is true that $$x\notin\{\{x\}\}$$, but that is not a property of the membership relation, it is the definition of the set $$\{\{ x\}\}$$. You can show this because $$x\neq \{x \}$$.