# P. Halmos Naive Set Theory Exercise 1 section 3

Consider the sets $∅,\lbrace ∅ \rbrace,\lbrace\lbrace ∅ \rbrace\rbrace$ etc.; consider the pairs such as $\lbrace ∅ ,\lbrace ∅ \rbrace \rbrace$ formed by any two of them; Consider the pairs formed any two such pairs, or else the mixed pairs formed by any singleton and any pair; and proceed so on ad infinitum. Are all sets obtained in this way unique from each other?

The question was never answered, so I am asking it again, but in a slightly different way.

I am pretty stuck on this question because I don't quite get what it is asking.

I was wondering, what does it mean by all sets; does this include the singletons $∅,\lbrace ∅ \rbrace,\lbrace\lbrace ∅ \rbrace\rbrace$ ? And also, do sets being "distinct" mean that they are not equal? And finally, if someone could answer the question, but give hints, with proof, I would appreciate it.

We have only looked at the Axiom of Specification, Extension and Pairing.

EDIT: Here is the best proof that I could think of, please let me know where I went wrong, or could improve:

It is obvious that by the remarks of Section 3, there exists sets $A = \lbrace a_1, a_2 \rbrace$ and $B = \lbrace b_1, b_2 \rbrace$ created in the specified way. And, by the equivalent formulation of the Axiom of Pairing and remarks of Section 3, $A$ and $B$ are the only such sets which contain $a_1$, $a_2$ and $b_1$, $b_2$, respectively. Now suppose, to the contrary, that $A$ and $B$ are not distinct; i.e. $A = B$. However, by the Axiom of Extension, $A = B$ implies $a_1, a_2 \in B$, and $b_1, b_2 \in A$; this is a contradiction, since by the previous remarks, $A$ and $B$ were the only sets which contained $a_1$, $a_2$ and $b_1$, $b_2$, respectively.