Does the set containing all sets contain itself?

Question

A set containing all sets by definition contains itself. Is that even possible, for a set besides the empty set?

Answer 1

In most modern set theories the collection of all sets is not a set itself. Namely, if $X$ is a set, then there is a set $Y$ such that $Y\notin X$.

Set theories like Zermelo-Fraenkel prove this by showing that if $X$ is a set, then $\{x\in X\mid x\notin x\}$ is a subset of $X$ which cannot be an element of $X$. However, they also prove this using the axiom of regularity which implies that $X\notin X$ for every set $X$, and as you notice the set of all sets must be an element of itself.

However, in the set theory New Foundations there is a universal set. And yes, it is an element of itself. In this set theory, if $X$ is a set, then $\{x\in X\mid x\notin x\}$ does not define a set, so it is not an element of $X$ (it does not exist in the universe).

And while we're clearing things up, the empty set does not contain itself. It is empty, namely $\forall x(x\notin\varnothing)$, and if $\varnothing\in\varnothing$, then $\varnothing$ wouldn't be empty.

Answer 2

In ZF or ZFC there is no such thing as the set of all sets. There is a class of all sets, but it is a "proper" class, which simply means that it isn't a set. It also means that it is not an element of itself.

Within ZF or ZFC you might additionally ask whether any set (not just the universal set) is an element of itself. In ZF or ZFC there is a separate axiom called the axiom of regularity which forbids this entirely. (Note that you should be careful about the word "contains"; sometimes it appears to mean $\supseteq$ while other times it appears to mean $\ni$. You do have $\emptyset \supseteq \emptyset$ but you don't have $\emptyset \ni \emptyset$. "Is an element of" and "is a subset of" remove this ambiguity.)

In some other set theories there is a universal set. For instance this occurs in NF and NFU. In this case of course the universal set must be an element of itself, because it is also a set.

In any set theory containing a universal set (call it $U$), one must be careful because $S=\{ x : x \not \in x \}$ is paradoxically defined: regardless of what the set theory says, the question "is $S \in S$?" cannot be resolved. In ZF and ZFC, the paradox is resolved by requiring that all comprehensions are over a domain which is already a set: so you would need $\{ x \in U : x \not \in x \}$ but there is no $U$. In NF and NFU, the workaround is forbidding comprehension using a certain class of predicates, including $x \not \in x$.

Answer 3

As David Ullrich notes, the empty set does not contain itself--it does not contain anything. I suspect, perhaps, that you are confusing $\in$ with $\subseteq$ (you may find this helpful if that is, indeed, the case). Then, note that $A\subseteq A$ for all sets $A$, even when $A=\varnothing$.

On a different note, you may enjoy the following exercise that addresses Russell's paradox, what Zubzub refers to in the first comment (this particular exercise appears in A Transition to Advanced Mathematics):

A logical difficulty arises from the idea, which at first appears natural, of calling any collection of objects a set. A set $B$ is ordinary if $B\not\in B$. For example, if $B$ is the set of all chairs, then $B\not\in B$, for $B$ is not a chair. It is only in the case of very unusual collections that we are tempted to say that a set is a member of itself. (The collection of all abstract ideas certainly is an abstract idea.) Let $X=\{x : \text{$x$ is an ordinary set}\}$. Is $X\in X$? Is $X\not\in X$? What should we say about the collection of all ordinary sets?

After reflecting on that, consider the following:

Suppose that $X$ is a set. If $X\in X$, then $X$ is not an ordinary set, so $X\not\in X$. On the other hand, if $X\not\in X$, then $X$ is an ordinary set, so $X\in X$. Both $X\in X$ and $X\not\in X$ lead to a contradiction. We conclude that the collection of ordinary sets is not a set.