# 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. – drhab Mar 18 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"? – Gregor Perčič Mar 18 at 9:31
• We don't "take it for granted"; it's what the symbols mean. – Malice Vidrine Mar 18 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. – Gregor Perčič Mar 18 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). – Malice Vidrine Mar 18 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 $$x\neq \{x \}$$.