On the concept of a set containing itself I know that the axiom of regularity and axiom of pairing together imply that a set cannot contain itself. Here is my attempt at proving that a set cannot contain itself using these axioms.
Assume $A$ contains itself. By pairing, there exists a set $C$ satisfying that
$$
C = \{A, A\} = \{A\} 
$$
By regularity,
$$A\cap C = \emptyset$$
But, $A\in A$ and thus $A\in A\cap C$.
contradiction!
Also, what if I took the set
$$
\ldots\{\{\{\}\}\}\ldots
$$
with countably infinite brackets. That is, the singleton of the singleton of the singleton (and so on infinitely many times) of the empty set.
Wouldn’t this set contain itself? The singleton of the singleton of the singleton (countably many times) of the empty set, contains the singleton of the singleton of the singleton (countably many times minus 1) of the emptyset.
And, countably many minus 1 is still the same as countably many, so the number of brackets are still in bijection?
Is this technically not a set because of the axiom of regularity (or some other axiom), or does this set not contain itself for some other reason I’m missing?
 A: Writing down something does not make it real. I can write the sentence "My unicorn is the King of Alabama", but the reality is that unicorns do not exist and Alabama is not a monarchy.
Writing down $\dots\{\dots\{\}\dots\}\dots$ does not make it a meaningful mathematical statement or object. If you're saying that the expression is infinite, you need to make sure that it is also grammatical, that is to say, that it fits the syntax. But the syntax of first-order logic is only using finite expressions.
Now you can say that really what you mean is that we define by recursion the sets $a_0=\{\}$ and $a_{n+1}=\{a_n\}$, and this set is "the limit". But why does that limit exist? What is that limit? Why does it exist? Even from a meta-theoretic point of view. If it's somehow the union, then that union would be $\{a_n\mid n<\omega\}$, which is definitely not a singleton.
There is no reason for that limit to exist anyway. The expression is just without meaning. Even if you omit the axiom of regularity, even if you do postulate that there is $x$ such that $x=\{x\}$, the expression doesn't make sense. Do note, by the way, that if it would imply that $x=\{x\}$ means that $x$ "is that set", then such $x$ would have to be unique. But it is just as consistent that there is a proper class of distinct sets satisfying $x=\{x\}$. So, again, the problem is that the expression is too wishy-washy to have any actual meaning to it.
