Talk:Transfinite Recursion/Uniqueness of Transfinite Recursion

Instead of supposing $y \subseteq z$, can't we just use $y \cap z$? --Dfeuer (talk) 22:11, 22 December 2012 (UTC)


 * I don't follow. $\subseteq$ is a relation, $\cap$ is an operation. So the first is a statement, the second is a term which does not stand alone and needs to be endowed with a predicate in order to be a statement. --prime mover (talk) 22:52, 22 December 2012 (UTC)


 * I mean changing the conclusion to read $\forall x\in y \cap z$, etc. --Dfeuer (talk) 23:32, 22 December 2012 (UTC)


 * I don't see it would make a difference, myself. Given that ordinals are defined as being related to each other as a bunch of nested subsets, it seems more natural to me to present the proof in that form: $y \subseteq z$ means $y$ is less than or equal to $z$. Granted that $y \subseteq z \iff y \cap z = y$ the two presentations would mean exactly the same - but then you'd have to bring in that fact as a link. I'm not sure yet what advantage it brings.


 * A worse criticism of this proof is that it does not specify what $G$ is. It appears (from looking at other pages) to be a mapping, but no indication is given as to what its domain and codomain are, which IMO is a serious omission. Unfortunately, as I don't have the source work cited (which if it's anything like the presentation of its contents is a substandard work) I can't directly interpret exactly how this proof is to work. If you can make sense of it, you'd be doing the world a favour. --prime mover (talk) 06:04, 23 December 2012 (UTC)


 * I think I understand the general intention, but I'll have to ponder a bit to give $G$ a proper domain. Its codomain isn't really an issue, I don't think. --Dfeuer (talk) 06:37, 23 December 2012 (UTC)