Is the “set” of all sets which contain $1$ a set?

Is the "set" of all sets which contain $1$ a set under ZF?

It does not contain itself, as it contains sets which contain $1$, but does not contain $1$ itself (as $1$ is not a set which contains $1$), and this has been the cause of a "failure to be a set" in all the cases that I've considered.

What axiom(s) of ZF is violated, if any, and why?

• If there were such a set you can apply the axiom of union to it, you would get a set that contains all sets. – blueInk Mar 18 '18 at 20:47
• There are many interesting classes that are not sets in ZF, like the class of all groups, or rings or partial orders or ... Self-membership is not the issue for these classes. – Rob Arthan Mar 18 '18 at 21:44
• Incidentally, while it doesn't contain itself we can still extract a failure of regularity (the axiom disallowing sets which contain themselves) from it; see the second bulletpoint of my answer. – Noah Schweber Mar 18 '18 at 22:11
• All of the "big ones", really. Union, separation, replacement, power set, regularity... You can get away with infinity, extensionality, and to some extent choice. This is witnessed by the fact NFU+AC is consistent, and there you can have the set of all sets which contain $1$. – Asaf Karagila Mar 21 '18 at 15:37

Such a class is as large as the class of all sets (and hence is not a set), because for each set $S$ it contains $S\cup\{ 1\}$, and hence it can be bijected with the sets $\not\owns 1$. If they lived in a set, and our original class were a set too, we would have identified two sets whose union is the class of all sets.

This class (call it "$C$") is not a set. There are a couple different ways to see this. For example:

• Consider the subclass $O$ of ordinals in $C$. If $C$ is a set, so is $O$ (by Separation), but $O\cup\{0, 1\}$ is the class of all ordinals, which is not a set.

• Let $D=\{C,1\}$. We have $C\in D\in C$, and this contradicts Regularity. (Specifically, the set $\{C, D\}$ has no $\in$-minimal element.)

• The transitive closure of $C$ is the class of all sets, which of course isn't a set.

Note that there are really two reasons why $C$ is not a set: it's "too big," and it leads to a failure of regularity (even though as you observe it doesn't contain itself - the failure of regularity here is not as direct, but it's still present).

• Also, in the spirit of a big list, $\{1\} \cup \{ (2, x) : x \in W\}$ has a surjection onto $W$, for any particular set $W$, so if the collection of sets containing $1$ was a set, it would have a surjection onto its powerset. Or, put another way, if we let $W = V$ and assume the result is a set then we get a failure of the axiom of replacement. – Carl Mummert Mar 21 '18 at 15:45