TITLE: Is $\{\{\emptyset\}\}$ actually a set?
QUESTION [5 upvotes]: I think this question is not asked here. I apologize in advance if I am wrong.
I have the following two definitions (Joaquín Olivert, Estructuras de álgebra multilineal, 1996):
Class.- A class is an abstract object $C$ which permits to decide if their elements belong to it or not.
Set.- A class $C$ is said to be a set if there exists a class $D$ such that $C\in D$. A class which is not a set is called a proper class.
With this in mind, I'm trying to understand if $A=\{\{\emptyset\}\}$ is a set or not and why. It seems to me the answer should be not but with these definitions, I think $\{ A \}$ is a class and then $A$ is a set. On the other hand, with this reasoning, every class would be a set and that is false (e.g. Russell's set).
Any help please?
Thanks

REPLY [8 votes]: In ZF(C) set theory, the Axiom of Pairing states that if $x$ and $y$ are sets, then $\{x,y\}$ is a set too.  Applying this with $x=y$ gives that whenever $x$ is a set, $\{x\}$ is a set too (because it has the same elements as $\{x,x\}$ and is therefore equal to it).
Thus,

$\varnothing$ is a set because of the Axiom of the Empty Set. (Sometimes this is not considered an axiom, in which case $\varnothing$ is still a set because the Axiom of Infinity states that there exists some set, and then the Axiom of Separation can be used to remove all of its elements).
$\{\varnothing\}$ is a set because of Pairing applied to $\varnothing$.
$\{\{\varnothing\}\}$ is a set because of Pairing applied to $\{\varnothing\}$.

Your definition of "set" sounds like it doesn't belong to pure ZF(C), but more likely to NBG or MK set theory. However, these theories have similar axioms of Empty Set and Pairing, so the argument (at least at the non-formal level I'm writing at) is the same.