How does the Axiom of Regularity apply to $A=\{1,2,3\}$?

Question

I cannot understand the axiom of regularity. It says that any generic non-empty set $A$ contains an element $X$ such that $ X\cap A = \emptyset $. How can this be true? If, for example I have a set $A = \{1,2,3\}$, according to this axiom there must be an element inside of this set such that its intersection with $A$ is the empty set, but $1,2,3$ are part of $A$, how can the intersection be an empty set?

Answer 1

Well, what is $1$ and what is $2$ and what is $3$?

The Axiom of Regularity talks about sets in the language of set theory. So it presupposes that everything is a set. But $1,2$ and $3$ are not part of the language, so you need to specify how you interpret these objects as sets.

The standard way is by the von Neumann ordinal assignments: $0=\varnothing$ and $n+1=n\cup\{n\}$.

So $1=\{0\}=\{\varnothing\}$ and $2=\{0,1\}=\{\varnothing,\{\varnothing\}\}$, and $3=\{0,1,2\}=\{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}\}$.

So $A=\{1,2,3\}$ is a particular set, and you can check now that $1\cap A=\varnothing$.

Answer 2

" whenever I try to chase down a chain of members , it must stop at some finite stage. You can think of it in this way. We have a string of sets $x_1, x_2, x_3....$ where each is a member of the preceding one ; that is : $..... x_3\in x_2\in x_1$. This we shall call a descending membership chain. Then the axiom ( Axiom of Foundation/ Axiom of Regularity ) is this : any descending membership chain is finite "

in Crossley et alii , What is mathematical logic? OUP, 1972 (Chapter 6 " Set Theory", pp. 62-63)

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So the axiom is aimed at ruling out the possibility of having such a descending chain.