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 $\in$?
 A: 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 $S$.
A: 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.

The OP asks

...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,

Mathematics is a game played according to certain simple rules with
  meaningless marks on paper.

A: I think you are doing things the wrong way :
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 $x\neq \{x \}$. 
