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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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Your accept rate is 0% at time of posting, please accept answers to the questions that you have asked, or add what is missing in the current answers. You can accept one answer per question by clicking on the transparent check mark by the vote count of the answer. –  Asaf Karagila Jul 3 '11 at 15:32
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

1 Answer 1

up vote 1 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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