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"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)


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

Just before this passage, the author introduces the question by considering the fact that when I have a family of sets I can take the union of this family... that is I can chase the members of the members of this family ...

Is the axiom equivalent to saying that we want the operation of taking the union of the union of the union ... of a family of sets to be stationary at some stage or that this operation finally results in the empty set? Or neither perhaps?

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If a set $X$ is well-founded (essentially, if it contains no infinite $\in$-descending chains), then indeed $\emptyset$ belongs to its transitive closure, that is, either $X=\emptyset$ or $\emptyset\in\bigcup X$ or $\emptyset\in\bigcup\bigcup X$ or... However, this does not mean that there is some $n$ such that the result of iterating the union operation $n$ times on $X$ coincides with iterating it more than $n$ times, as Hagen showed in his example in the other answer.

The axiom of regularity is equivalent to the assertion that all sets are well-founded so, indeed, under regularity, iterating union eventually produces the empty set among the elements of the resulting set.

It is consistent with the other axioms of set theory that regularity fails and there is a set $\Omega$ such that $\Omega=\{\Omega\}$. Note that $\emptyset$ does not appear in the process starting with $\Omega$ because, in fact, $\Omega=\bigcup\Omega$.

(I actually do not know whether regularity may fail and still $\emptyset$ belongs to the transitive closure of any set, though I suspect this is not possible. Recall that the transitive closure of $X$ is $X\cup\bigcup X\cup\bigcup\bigcup X\cup\dots$)

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Not really.

If some set $x$ is $n$ levels "down" in $A$, then in $\bigcup A$ is is only $n-1$ levels down and hence in $\underbrace{\bigcup\cdots \bigcup}_nA$, it reaches the "surface". But this doesn't mean that everything reaches the "surface" together.

For example, let $A_{n,0}=n$, $A_{n,k+1}=\{A_{n,k}\}$, and $X=\{\,A_{n,n}\mid n\in\omega\,\}$. Then we have

$$\underbrace{\bigcup\cdots\bigcup }_nX=\{\,A_{m,m-n}\mid m\ge n\}\cup (n-1)$$ so this sequence is not stationary (but of course $X$ is regular).

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  • $\begingroup$ Hagen, what do you mean by regular? Well-founded? $\endgroup$ – Andrés E. Caicedo Apr 20 at 14:10

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