Is it consistent with $\sf NF$ or $\sf NFU$ to have a set of all transitive sets? Formally:
$\exists t \forall x (x \in t \leftrightarrow x \text { is transitive})$
Where "$x$ is transitive" means that every element of $x$ is a subset of $x$.
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asked Oct 24 at 15:24
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1 Answer
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No. See Proposition 2.1.16 in Forster's book "Set Theory with a Universal Set: Exploring an Untyped Universe".
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4$\begingroup$ Could you perhaps add a sketch of the argument? Or is it too complicated? $\endgroup$ Commented Oct 24 at 18:26
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1$\begingroup$ @EmilJeřábek, the idea of the proof is that if we have that set then we can easily have a transitive closure for each set (since that would be stratified), then the set $\{x: x \notin TC(\{ \{y\} \mid y \in x\})\}$ would be paradoxical. Where $TC$ stand for the transitive closure operator. An unpublished proof of Boffa. $\endgroup$ Commented Oct 25 at 8:15