In axiomatic set theory every set is a "collection" of "empty sets"?

Question

Based on the answers of this question: How elements are defined in axiomatic set theory

and this part of this book: (page 9)

I will examine this reasoning in depth:

Let's take a random example:

Suppose we have the set $A=\{X,Y\}$, $X$ and $Y$ are also sets, so suppose $X=\{X_1,X_2\}$ and $Y=\{Y_1,Y_2,Y_3\}$ so again $X_i$ and $Y_j$ are also sets, suppose again $X_1=\emptyset$ $X_2=\{X_2'\}$ and $X_2'=\emptyset$ and $Y_1=\{Y_1'\}$, $Y_2=\emptyset$ and $Y_3=\emptyset$ assume $Y_1'=\{K_1,K_2\}$, where $K_1=\emptyset$ and $K_2=\emptyset$. Then

$A=\{X,Y\}=\{\{X_1,X_2\},\{Y_1,Y_2,Y_3\}\}=\{\{\emptyset,\{X_2'\}\},\{\{Y_1'\},\emptyset,\emptyset\}\}=\{\{\emptyset,\{\emptyset\}\},\{\{\{K_1,K_2\}\},\emptyset,\emptyset\}\}=\{\{\emptyset,\{\emptyset\}\},\{\{\{\emptyset,\emptyset\}\},\emptyset,\emptyset\}\}=\{\{\emptyset,\{\emptyset\}\},\{\{\{\emptyset\}\},\emptyset,\emptyset\}\}$

Then in our example $A=\{\{\emptyset,\{\emptyset\}\},\{\{\{\emptyset\}\},\emptyset,\emptyset\}\}$

So my question is: every set "looks like" our set $A$?

Thanks in advance

Answer 1

Not necessarily, it may depend on the precise axioms. For example there might be an axiom $\exists x\colon x=\{x\}$. But I assume your book will (justifiedly) avoid that. Then it is hard to formalize "looks like". The most important difference is that $A$ might be infinite, and so might be its elements. But any path you take down into the jungle of braces, you will end up at nothingness sooner or later indeed. There is one axiom specifically responsible for this:

Axiom of Foundation: If $A$ is a nonempty set, then there exists $B\in A$ such that there is no $C$ with $C\in B$ and $C\in A$.

From a set that would allow an infinite descent or something not ultimately ending in empty sets, one could construct (using the other axioms) a set that would fail foundation

Answer 2

In $\sf ZFC$ we can prove the following thing.

Consider the following hierarchy,

$V_0=\varnothing$, if $V_\alpha$ was defined then $V_{\alpha+1}=\mathcal P(V_\alpha)$, and if $\{V_\alpha\mid\alpha<\delta\}$ were all defined, then $V_\delta=\bigcup_{\alpha<\delta}V_\alpha$ for a limit ordinal $\delta$.

Then $V=\bigcup_{\alpha\in\mathsf{Ord}}V_\alpha$. That is, for every set $x$ there exists some ordinal $\alpha$ such that $x\in V_\alpha$.

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What does that mean? Well, if we consider the construction, then if $x_0$ is a set, and $x_1\in x_0$ and $x_2\in x_1$ and so on, after finitely many steps you'll have some $x_n=\varnothing$. Of course the value of $n$ may depend on what is $x_1$ and what is $x_2$ and so on, but we are guaranteed that it will be finitely many steps.

Why is that? This is because of the axiom of foundation, or axiom of regularity, which asserts that $\in$ is well-founded, as Hagen wrote, if $A$ is a non-empty set, then $A$ is disjoint from one of its elements.

The most immediate consequence of this axiom is that there are no infinite decreasing sequences in $\in$, namely, $\ldots\in x_{n+1}\in x_n\in\ldots\in x_0$. So when we begin descending, we have to stop at some point, why? Because we have reached the empty set.

If this axiom is not assumed, then it is consistent that there are counterexample to that, namely sets like $x=\{x\}$. But having $\in$ well-founded is a good thing, as it allows to use $\in$-induction and have things like the Mostowski collapse lemma (whatever it means, and whatever its importance is). And the things that we can do with sets which contradict the axiom of foundation end up as being not-as-important, and without too much hassle we can often do them in $\sf ZFC$ as well.