Add to $ZFC$ the following axiom:
Dumb Collection. For any predicate $\phi$ such that there exists a set $y$ satisfying $\phi$, there exists a nonempty set $x$ whose members are precisely the sets $y$ satisfying $\phi$ which donβt contain $x$ as a member or subset. $$\forall\phi\Big(\exists y\big(\phi(y)\big)\implies\exists x\big(x\neq\emptyset\wedge\forall z(z\in x\iff\phi(z)\wedge x\notin z\wedge x\nsubseteq z)\big)\Big).$$ Denote the set guaranteed by this axiom and a predicate $\phi$ by $x_\phi$.
This avoids Russelβs paradox since we now have that $x_{y\notin y}\notin x_{y\notin y}$ by definition, but it isnβt clear that we donβt still have other paradoxes present. If this axiom is consistent,
What is the consistency strength of $ZFC+$dumb collection?
Best Answer
Take $\phi$ to be "$y$ is a singleton", then $x$ is now the set of all singletons, except $\{x\}$ itself.
But that's impossible, since then $\bigcup(x\cup\{\{x\}\})$ is the set of all sets.