Does the Set of All Self-Membered Sets Contain Itself?

In ordinary set theory, the answer is no: no set is a member of itself, so the collection of self-membered sets is empty. The empty set is not a member of itself.

There is a small but important qualification. In standard ZF set theory, “the set of all sets” is not available as a set. Here the phrase can be understood as the collection of sets satisfying the property $x\in x$. The Axiom of Foundation says that no set satisfies it.

Why no set contains itself in ZF

The Axiom of Foundation (also called Regularity) says that every nonempty set $A$ has an element $y$ with no elements in common with $A$:

$$\exists y\in A\quad y\cap A=\varnothing.$$

Suppose, for contradiction, that some set $x$ were a member of itself. Apply Foundation to the singleton $A=\{x\}$. Its only possible element is $x$, but $x\in x$ means that $x$ and $\{x\}$ have an element in common: namely, $x$. That contradicts the requirement that the chosen element have no elements in common with $A$.

Thus $x\notin x$ for every set $x$. The collection described by “all sets $x$ such that $x\in x$” has no members. Its set representation is $\varnothing$, and $\varnothing\notin\varnothing$.

What “all sets” means here

ZF does not contain a universal set whose members are every set. If it did, familiar set-forming rules would lead to Russell’s paradox. So it is imprecise to treat “the set of all sets” as an ordinary set in standard set theory.

For this particular property, however, Foundation settles the issue: there are no self-membered sets. In ZF, you can form the subset of any given set $A$ consisting of its self-membered elements,

$$\{x\in A:x\in x\},$$

and that subset is always empty. The phrase “the collection of all self-membered sets” is therefore an informal description of an empty class, represented by the empty set—not a universal set.

Why this is not Russell’s paradox

Russell’s paradox considers the opposite property: being a set that does not contain itself. If unrestricted comprehension allowed a set $R=\{x:x\notin x\}$, then asking whether $R\in R$ would give

$$R\in R\quad\Longleftrightarrow\quad R\notin R,$$

which is a contradiction. The condition $x\in x$ does not produce that contradiction. If you write $S=\{x:x\in x\}$ and substitute $S$ for $x$, you get $S\in S$ if and only if $S\in S$—a tautology that, by itself, tells you nothing. In ZF, Foundation supplies the additional fact that makes the answer definite: no set is self-membered.

A related way to see why there is no universal set is to start with any set $A$ and form the relative Russell set $R_A=\{x\in A:x\notin x\}$. This is a legitimate subset of $A$, but it cannot itself be an element of $A$: if it were, the defining property would imply $R_A\in R_A$ exactly when $R_A\notin R_A$. So $R_A$ lies outside $A$. No set can contain every set.

What changes if Foundation is dropped?

Some alternative set theories do not assume Foundation and allow sets that are members of themselves. In such a theory, the collection of self-membered sets need not be empty. Whether that entire collection forms a set, and whether it contains itself, depends on the theory’s other axioms and rules for forming sets.

The definition alone still does not decide the self-membership question: substituting the collection for $x$ again yields a tautology. You need the axioms of a specific set theory to determine what exists. Under the standard ZF convention, Foundation gives the clear answer: the collection is empty and does not contain itself.

Common mistakes

• Confusing the two properties: Russell’s paradox uses $x\notin x$; this question uses $x\in x$.
• Assuming a universal set exists: Standard ZF has no set of all sets. “All sets” must be handled as a class-like description or through subsets of a given set.
• Treating a tautology as a solution: $S\in S\leftrightarrow S\in S$ is consistent but uninformative. The answer in ZF comes from Foundation, not from that substitution.