So, was recently wondering wether any set could be constructed from the empty set with the axioms of ZFC set theory. I read some things like "there exists no infinite descending sequence of sets due to the axiom of regularity". So, my first question is: does this imply that ANY set can be constructed from the empty set? Because, if $x_0\ni x_1\ni\dots x_k$ is the "maximal" sequence (I hope it is clear what I mean by this), then it must be that $x_k=\varnothing$, right?
Also, while investigating, I came a cross the following question (which I don't know how to link here) in the platform:
In ZFC theory, the axiom of regularity guarantees there's no infinite descending sequence. But it seems I found one. Define $x_n=\{n,x_{n+1}\}$, then it is obvious that $x_0\ni x_1\ni x_2\ni\dots$ and to see why's no contradiction with the axiom, take any set $x_i$, there are only two elements $i$ and $x_i+1$. Neither of them intersects with $x_i$. So what's wrong with this construction?
Now, I too find myself with this question. An answer to this question was given which proves that no infinite descending sequence of sets exists (more or less the same that appears in a Wikipedia article called "Axiom of regularity". The problem, to me, is that these proofs assume that this sequence is given by a function and that the function must have a range. But, to build that function (using ZFC) you first need to present it as a subset of the cross product of two sets, one of which must contain all the $x_i's$. Namely, $S=\{x_0,x_1,…\}$, but such set does not exists as it fails the axioms of regularity! So, the function does not exists and the proof is obsolete.
Of course, bear in mind that I'm not referring to the mathematical concept when saying "sequence". Rather, I'm calling the "chain" (or however else you want to call it) of descending sets $x_0\ni x_1\ni\dots$ the sequence.
Sorry for such a long question. I hope I make myself clear… my second question is (more specifically) why is it not possible for $x_0\ni x_1\ni\dots$ to be. I understand that the axiom of regularity prohibits the existence of $S=\{x_0,x_1,\dots\}$, but why does it prohibit the existecne of the $x_i's$?
Best Answer
Let $\omega=\mathbb{N}$ be the natural numbers in a (hypothetical) model of ZFC.
First claim: There does not exist a set $A$ and a set funtion $f:\omega\longrightarrow A$ such that $f(n+1)\in f(n)$ for all $n\in\omega$.
Proof: Let $B=\operatorname{Image}(f)\subseteq A$. By the axiom of foundation there is a $c\in B$ such that $c\cap B=\emptyset$. By definition of $B$, there is some $n_c\in\omega$ such that $f(n_c)=c$. Then $f(n_c+1)\in c\cap B$ contradicting that $c\cap B$ is empty.
Second claim: There does not exist a class function $F$ whose domain contains $\omega$ as a subclass which is a set such that $F(n+1)\in F(n)$ for all $n\in \omega$.
Proof: If a class function like that existed, by the axiom schema of replacement there is a set $A$ containing as members all $F(n)$ such that $n\in\omega$. Then by the axiom schema of separation from $\omega\times A$ there is a set function $f:\omega\longrightarrow A$ as in Claim 1.
Caveat: It is possible for there be an infinite descreasing external sequence in the model $$\dots \in x_{n+1}\in x_n\in\dots \in x_0$$ Such a sequence exists only as an external subset of the domain of discourse, but it is not a set and it is not a proper class. For example, if the model has a non-standard model of $\omega$. Every non-zero element of $\omega$ has a predecessor by the $\in$-relation. So if $n$ is a non-standard member of $\omega$, it leads to an infinite descending $\in$-chain.