Interpretations of the Axiom of Regularity in ZFC

Question

I am currently studying the Zermelo-Fraenekl Set Theory and I have some problems with understanding the Axiom of Regularity:

Firstly, I found this version of the axiom on Wiki

I understand that x and y are all referring to sets. Then there is a problem: Consider the set {1,2,3}. This set is non-empty but there not does exist a "y" in x such that "y intersects x" is the empty set because none of the members of x are sets (In this case, it does not make sense to talk about "y intersect x" since intersection is a binary connective between two sets).

Secondly, here is another version of this axiom:

I have some troubles with understanding the second part of this experssion (after the implication arrow). Why is this version equivalent to the version above? Thanks so much if someone could give a hand. I really appreciate!

Answer 1

In set theory, everything is a set. Specifically, the most common interpretation of $1$ is $\{\{\}\}$, and $2$ becomes $\{\{\}, \{\{\}\}\}$. In fact, your set $\{0,1, 2, 3\}$ (with a small addition) is actually what we would denote as $4$.

As for why they are equivalent, one says "if $x$ is non-empty, there is an element in $x$ that doesn't intersect $x$", while the other says "if there is an element in $x$ (i.e., if $x$ is non-empty), then there is an element $y$ such that $y$ is an element of $x$ and at the same time, there is no set that is an element of both $x$ and $y$ at the same time (i.e. $x$ and $y$ do not intersect)".

Answer 2

So in ZF set theory every object is a set, in particular $1,2,3$ denote sets ($1= \{\emptyset\}$ usually, for instance), so it does make sense to talk about their intersection.

The second version is equivalent because of the following : the right part of the implication states that $x$ is not empty and so corresponds with the $x\neq \emptyset$ part of the first version. The second part precisely states that their exists a $y$ such as in the first version : indeed $y\cap x=\emptyset$ is equivalent to $\neg \exists z, z\in y\cap x$, which is equivalent to $\neg \exists z, z\in x \land z\in y$.