Date: Nov 10, 2013 6:40 AM
Author: Paul
Subject: Re: Possible major blunder in Rado's version of Canonical Ramsey<br> Theorem that goes far beyond omitting proof steps

On Sunday, November 10, 2013 11:02:11 AM UTC, David Hartley wrote:
> In message <45d5b781-2ea8-4455-b6ff-c7ead6d7bd7b@googlegroups.com>, Paul
>
> <pepstein5@gmail.com> writes
>

> >Agreed that, (when amended using the contributions on this thread), the
>
> >newer approach is simpler. But it also proves _far less_. The
>
> >original Erdos-Rado version doesn't assume the axiom of choice
>
> >(referred to as "Zermelo's axiom" in that paper). However, the newer
>
> >paper does assume the axiom of countable choice by asserting the
>
> >existence of B' = {b0, b1...}
>
> >
>
> >I'm sure that Erdos could have given a particularly simple presentation
>
> >if he didn't care about avoiding choice.
>
>
>
> I haven't studied the Erdos-Rado paper in detail, but they state that
>
> their proof of the canonical theorem from Ramsey's theorem does not use
>
> AC. It is the original proof of Ramsey's theorem which did, but they
>
> give a version which avoids it.
>
>
>
> Anyway, in the later paper B' is a subset of the natural numbers so no
>
> choice axiom is needed to order it.
>


No, B' is a subset of A which is not countable, in general.

Your initial point about Erdos-Rado stating that they avoid AC is a repetition of my point. In summary, the Erdos-Rado result proves Canonical Ramsey without AC whereas Rado deduces Canonical Ramsey from AC.

It is the second paper that errs in not mentioning choice, while comparing itself to the first paper.

Paul Epstein