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 $\in$?

  • 1
    $\begingroup$ 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. $\endgroup$
    – drhab
    Commented Mar 18, 2020 at 9:29
  • $\begingroup$ 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"? $\endgroup$
    – God bless
    Commented Mar 18, 2020 at 9:31
  • 3
    $\begingroup$ We don't "take it for granted"; it's what the symbols mean. $\endgroup$ Commented Mar 18, 2020 at 9:34
  • $\begingroup$ @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. $\endgroup$
    – God bless
    Commented Mar 18, 2020 at 9:37
  • 2
    $\begingroup$ 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). $\endgroup$ Commented Mar 18, 2020 at 18:56

3 Answers 3


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.

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.


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 $x\neq \{x \}$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .