Reading Roitman's revised edition of her 'Introduction to Modern Set Theory', I got a bit confused about her proof of the claim that under regularity $V$, the universe of sets is equal to $\bigcup_{\alpha \in ON} V_\alpha$, the von Neumann universe, especially about the role of choice principles in that proof.

Here's her proof. Let a set $x$ be big, if $x \not \in \bigcup_{\alpha \in ON} V_\alpha$. Every big $x$ has some big element. Otherwise we would have $x \subseteq \bigcup_{\alpha \in ON} V_\alpha$, which is impossible. So, if there's some big $x$, $x$ is the beginning of some infinite descending $\in$-chain, contradicting regularity.

Here's my question: Doesn't inferring the existence of some infinite descending $\in$-chain require the axiom of Dependent Choice (DC)? For under DC from every big $x$ containg some big $y$ we can conclude that there's some countably infinite sequence of big sets $f$ with $f(0) = x$ and $f(n +1) \in f(n)$. Is there a way of avoiding choice principles in Roitman's proof?


1 Answer 1


You're right that as stated, this proof requires DC. You can adjust it like this.

Fix $x$ big, and let $X_0=\{y \in x: y \text{ is big}\}.$ Then for each $n,$ let $X_{n+1}=X_n \cup \{y \in \cup X_n: y \text{ is big}\}.$ Notice that $\cup_{n < \omega} X_n$ is nonempty and has no $\in$-minimal element, contradicting regularity.


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