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Given a transitive set$$a,$$ I can prove that $$\bigcup a$$ is also transitive, but I don't quite like my method because I must first prove $$a \subseteq \mathcal{P}(\bigcup a)$$ to get at $$\bigcup a\subseteq \mathcal{P}(\bigcup a).$$ Could you please show me a smarter proof?

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Also, are you asking about a subset of a transitive set, or about the union over a transitive set? – Asaf Karagila Jul 3 '11 at 15:33
Asaf, I am interested in proving $$\bigcup a$$ is transitive, then $$\bigcup \bigcup a$$, etc. – user11750 Jul 3 '11 at 16:57
Asaf, I can't find the transparent check mark by the vote count. Sorry, I feel inadequate in navigating this site. – user11750 Jul 3 '11 at 17:00
When you have an answer which you want to accept, just below the arrow to downvote it there is a transparent check mark. Click it and it will be accepted. – Asaf Karagila Jul 3 '11 at 17:05
As for the union of a union, if $\bigcup a$ is transitive, by the proof below $\bigcup\bigcup a$ is also transitive, and so on. Also, you might want to edit the question to incorporate your exact question, and not just "approximations" to it and complements in the comments. – Asaf Karagila Jul 3 '11 at 17:06
up vote 3 down vote accepted

Suppose that $a$ is transitive. That is $b\in c\in a\implies b\in a$.

Now take $d\in b\in\bigcup a$, then $b\in c\in a$, therefore $b\in a$, therefore $b\subseteq\bigcup a$ therefore $d\in\bigcup a$.

To add on the "A subset of a transitive set is transitive" is clearly false:
Take the transitive set $4=\{0,1,2,3\}$ (where $0=\varnothing$ and $n+1=n\cup\{n\}$) then $\{3\}$ is a subset of $4$ but not transitive itself.

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