axiom of regularity and power sets

Question

So axiom of regularity works for any non-empty set. Is that means that a set like this {aab, +, 50, ), (, **} and all of it's subsets are actually made of empty sets? And not just sets, but power sets? If yes, then does it means that all sets are just sets with empty set elements? But then, how to represent a set like this {aab} with power set? Like we are doing with natural numbers.

Answer 1

Generally, in set theory, one works with "pure" sets - that is sets whose every element is itself a set. So in your example, we would have to choose set-theoretic definitions of $aab$, $+$, and $50$. For example, natural numbers can be defined set-theoretically by the following recursive definition: $0 = \emptyset, n+1 = n \cup \{n\} = \{0, \ldots, n\}$, so that $1 = \{\emptyset\}, 2 = \{\emptyset, \{\emptyset\}\} = \{0,1\}$, etc. If you're doing strict ZFC mathematics, that is, working formally within the first-order system, you define everything you do as sets that ultimately are constructed from nothing but the empty set. You can do all of math this way, but it's tedious and not very helpful when you're outside of set theory. So depending on what you're trying to do, the various symbols like $aab$ can be constructed in various ways.

It is possible to do set theory with sets whose elements are not themselves sets. These are called "urelements". See the Wikipedia discussion here.

However, the idea of the axiom of regularity is less about this sort of thing, but more to prevent various pathologies in set theory that have undesirable properties. For example, it rules out the possibility that there is some set $x$ with $x \in x$. It states that for any set $x$, there is some element $y \in x$ with the property that if $z \in x$, $z \not\in y$.