Constructing a sequence up to infinity

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Consider the sets $x_1, x_2,x_3,\ldots$ I was wondering if we could construct a sequence of sets as follows: \begin{equation} x_1=\{x_2\},\;x_2=\{x_3\},\;\ldots \end{equation} and continue to infinity. My initial thoughts were that no, because for any $x\in x_1$, it happens that $x\in x$, given how it's constructed. And this cannot happen because of the separation axiom in ZFC. At least that's my idea.

I just don't know how to put it mathematically, or if the idea is wrong. Could anyone please help?

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The Regularity axiom prohibits sequences such as you imagine, more accurately described as running down to infinity. Such a sequence would be an infinite descending $\in$ chain, thus $\{x_n\,|\,n\in\mathbb{N}\}$ would have no $\in$-minimal element.

Notice that if $a = \{b\}$, then $b = \bigcup a$. (The converse does not hold!) Starting with any $x_0$, you can define a sequence of iterated unions:

$$ x_{n+1} = \bigcup x_n. $$ By Regularity, eventually $x_n = \emptyset$, so of course $\emptyset = x_{n+1} \notin x_n = \emptyset$.