Paul
Posts:
676
Registered:
7/12/10


Re: Possible major blunder in Rado's version of Canonical Ramsey Theorem that goes far beyond omitting proof steps
Posted:
Nov 10, 2013 12:23 PM


On Sunday, November 10, 2013 4:30:56 PM UTC, David Hartley wrote: > In message <BWpz$uCS15fSFw6Q@212648.invalid>, David Hartley > > <me9@privacy.net> writes > > >As far as I know (which isn't far) Ramsey's theorem only applies to > > >countable sets so Rado's proof will only work with countable A. > > > > That was a bit silly of me. For any A, if Ramsey's theorem applies to an > > infinite subset B of A then it applies to A. So, if you're worrying > > about choice, it is certainly true for at least any set with a > > denumerable subset. > > > > The statement and proof of Ramsey's theorem in the ErdosRado paper > > apply only to sets of natural numbers. They can easily be extended to > > any countably infinite set but not obviously to an infinite set with no > > denumerable subset. So if you want the Canonical theorem to apply to > > all infinite sets, neither proof works without at least a weak form of > > choice. > > > > I suspect that Rado's proof can be modified to show that if Ramsey's > > theorem extended to all infinite sets holds in ZF, then so does the > > extended canonical theorem, but I haven't checked it through carefully. >
Agreed with everything that you've said in this subthread. However, I have discovered a problem among these papers (aside from problems in my own understanding). Ramsey's original paper is at the URL: www.cs.umd.edu/~gasarch/BLOGPAPERS/ramseyorig.pdf
The underlying set for that paper is "an infinite class" with no suggestion whatsoever that this set is countable. ErdosRado then make at least two errors in their exposition. 1) They say that they are proving Ramsey's original theorem but they aren't because theirs is weaker, in assuming a countable underlying set. 2) This is much more serious. They praise their own proof of Ramsey's theorem as being "choicefree" and contrast this with Ramsey's proof. This is misleading because the presentation can only be choicefree if the underlying set is countable.
I don't think a contemporary referee would be happy with the ErdosRado statements about how their proof is choicefree.
Paul Epstein

