How can we define the set $\{\{\}, \{\{\}\}, \{\{\{\}\}\}, ...\}$? How do you define the set of elements which can be made by repeatedly wrapping the null set in another set: $\{\{\}, \{\{\}\}, \{\{\{\}\}\}, ...\}$?
I've tried $A = \{x\ |\ x=\emptyset \lor \exists y\in A : x = \{y\}\}$, but that would give us a recursive formula, because $y\in A$ is replaced by that same formula.
Is this set definable, and if yes, how?
 A: How about:
$$\{x\mid\forall A,(\{\}\in A\land(\forall a\in A,\\\{a\}\in A))\Rightarrow x\in A\}$$
To make that easier to read, let me define the following. A set $A$ is upwards if $a\in A$ implies $\{a\}\in A$. (I made up this word just now.) A set $A$ is called totally upwards if it's upwards and it contains $\{\}$. (I also just made this up.) Then the above reads:
$$\{x\mid\forall A,\\A\text{ is totally upwards }\Rightarrow x\in A\}$$
A: What axioms of set theory are you using?  The existence of an infinite set is independent of the other axioms.
ZFC usually takes the axiom of infinity to be something like 
$$
\exists Z ( \emptyset \in Z \wedge \forall x \in Z (x \cup \{x\} \in Z) )
$$
but your axiom would be just as good.
Do you see how to use my axiom to construct the set you are looking for?
A: Yes, because there's an axiom that says you can do exactly that, because you can't derive it from the intuitively more "basic" axioms. It is called the axiom of infinity, and states that there is a non-empty set $S$ such that whenever $X \in S$, then $(\{X\} \cup X) \in S$ as well.
