Talk:Correspondence Theorem (Set Theory)

Before setting out to prove this, let me remark that there is no such thing as a "set of partitionings". For, given a partitioning $g: I \to \mathcal P \left({S}\right)$ and any bijection $f: J \to I$, we can precompose with $f$ to obtain a new partitioning. Since $J$ is arbitrary, it cannot be that the collection of all partitionings is a set. Rather, I think the theorem statement must use the images (i.e., partitions) instead. --Lord_Farin (talk) 14:37, 7 November 2012 (UTC)


 * That's what Blyth calls it, but then he does over-complicate things at times. I never got to grips with why. --prime mover (talk) 18:26, 7 November 2012 (UTC)


 * I have no sources on the subject, so I'll leave it to you to decide an appropriate formulation. In the mean time, I'll add a questionable tag. --Lord_Farin (talk) 20:40, 7 November 2012 (UTC)


 * In that case it will just have to swing, then, because I have no time to look at it. Unless you want to delete it and everything on the subject. --prime mover (talk) 22:47, 7 November 2012 (UTC)