Talk:Set has Rank

Definition of $G$
The countable union is sufficient. Because $G$ is transitive, taking more unions won't produce anything new. Furthermore, the source I linked to claims but does not prove that each transitive set is equal to $V_i$ for some ordinal $i$, suggesting I'm likely on the right track wrt the definition of $G$ if nothing else. --Dfeuer (talk) 05:36, 27 December 2012 (UTC)


 * I'm still uneasy about it. --prime mover (talk) 06:32, 27 December 2012 (UTC)


 * I'm uneasy about the fact that I haven't justified the recursive definition from appropriate axioms/theorems, but if you're uneasy with what it actually is, I don't know what to tell you. It's just a set with properties that happen to be useful. --Dfeuer (talk) 19:05, 27 December 2012 (UTC)

Does this look about right?
I think I got a decent outline down, but I'd appreciate if folks could check for gaping holes in my reasoning. --Dfeuer (talk) 18:48, 27 December 2012 (UTC)

Note: I know that nothing on the page is up to style spec. I'm just looking for thoughts on content and I'll work on style later. ---Dfeuer (talk) 19:03, 27 December 2012 (UTC)

It would be great to find a simple proof that each transitive set equals $V_i$ for some $i$. That would eliminate the mess at the end of this proof. --Dfeuer (talk) 19:56, 27 December 2012 (UTC)