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If $x$ is a set, and $\mathcal{P}x$ and $\cup x$ are both transitive sets, does $x$ necessarily have to be transitive?

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No. To find a counterexample, consider $x'$ which is transitive, and take $x=x'\setminus\{\varnothing\}$.

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    $\begingroup$ Say, $x=\{1\}$. Then $\mathcal P(x)=\{0,x\}$. $\endgroup$ – Andrés E. Caicedo Mar 1 '14 at 17:33
  • $\begingroup$ Yeah, that's what I had in mind, but it works pretty much with any larger transitive set (not with smaller ones, though). $\endgroup$ – Asaf Karagila Mar 1 '14 at 17:37

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