Mirimanoff Paradox

Question

The class $B$ is well-founded, if every descending chain of subclasses is finite. The class $B$ is not well-founded if there exists an infinite chain of classes $B_n $ such that $ \quad \dots \in B_n \in \dots \in B_2 \in B_1 \in B. $

Paradox: The collection $\mathbf{WF}$ of well-founded sets (on a given set of atoms) is not itself a set.

Let me try to elaborate on the formulation of the paradox. Let $\mathbf{WF}$ be the set of well-founded sets; then $\mathbf{WF}$ itself is well-founded, for were $\mathbf{WF} \ni x_0 \ni x_1 \ni x_2 \ni\dots\, $, then $x_0$ would be a not well-founded member of $\mathbf{WF}$, which is preposterous. Therefore $\mathbf{WF} \in \mathbf{WF}$, so $\mathbf{WF}$ is not well-founded, i.e. contradiction.

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Is this paradox only present in the naive set theory? If the answer is no, then how is it resolved?

I think this paradox is similar to Burali-Forti antinomy, but honestly I do not understand (the resolution of) the latter one as well.

Answer 1

Paradoxes of naive set theory are examples of contradictions apparent in the naive approach that "everything is a set", as formalized by the comprehension axiom[s].

Indeed if $\bf WF$ is a set then it is well-founded, then $\bf WF\in WF$, which is a contradiction to the well-foundedness.

The resolution of this paradox, and indeed other paradoxes (like Burali-Forti that you mentioned), is that not every "property" defines a set.

Some modern set theories (e.g. $\sf ZFC$) resolve this by requiring that properties establish subsets of previously constructed sets; other set theories (e.g. $\sf NF$) resolve this by requiring that the formulas themselves will have certain syntactic properties which prevent them from defining these classes.

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And when all this has been said and done, I'd like to point out that the axiom of choice is necessary for the proof that well-foundedness is equivalent to the lack of infinite decreasing chains. The correct formulation should be that every non-empty set has an $\in$-minimal element (namely, if $x$ is not empty, then for some $y\in x$, $x\cap y=\varnothing$).

On the other hand, if you prove (without appealing to choice) that if $x$ is a well-founded set, then $\{x\}$ is a well-founded set, and that for every $x\in\bf WF$ satisfies $x\notin x$, from which we can conclude that $\bf WF\notin WF$.