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?
A subset of a transitive set is transitive
3
$\begingroup$
elementary-set-theory
-
0As 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. – 2011-07-03
1 Answers
3
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.