Talk:Ω-consistent theory

message: Samuel Wantman left an edit summary asking why "Omega" shouldn't be capitalized. I'm not claiming it's completely clear-cut, but here's my argument:  Ideally the article should be at ω-consistent theory, with the lower-case ω. If we use the Greek letter, then the capitalization is very important; the phrase "Ω-consistent theory", though (as far as I know) having no standard meaning, could imaginably be taken to mean "consistent in Woodin's Ω-logic", an immensely stronger notion.
The part that's a little less than clear-cut is the association "Omega/Ω :: omega/ω". This is standard, though, to anyone who uses TeX. --Trovatore 17:05, 31 August 2006 (UTC)
replies: text: I've changed the note to reflect that ω-consistent theory is the ideal name. -- Samuel Wantman 18:49, 31 August 2006 (UTC)
replies: text: No, it isn't.  Gene Nygaard 02:05, 3 December 2006 (UTC)
thread_title: lowercase

message: The notation \Sigma^m_n is used without definition. Can someone please provide one? -- Hairy Dude (talk) 18:02, 16 November 2007 (UTC)
replies: text: There is a prominent link to the arithmetical hierarchy in the article. Just click it. -- EJ (talk) 10:05, 20 November 2007 (UTC)
replies: text: The symbolism always be defined when first used, not later -- otherwise there's no associative relationship (as happened in my reading of this. So I came here to the talk page and found Hairy Dude's complaint). So I added a footnote. Maybe someone can word-smith it; put in the link right there, for example. Bill Wvbailey (talk) 13:28, 25 September 2008 (UTC)
thread_title:  Notation 

message: It looks to me that the ω-rule makes the theorems of T no longer recursively enumerable.  E.g. suppose H is a Turing machine that does not halt, and P_H(x) is the predicate "H does not halt at step x".  Since PH is a theorem at every x (provable by running H for that many steps), "H does not halt" is a theorem of T under the ω-rule.  The ω-rule gives T a built-in halting oracle.  Is this worth remarking on in the article? —Preceding unsigned comment added by 67.117.147.133 (talk • contribs) 19:38, 25 October 2008 
:Well, your argument doesn't seem to completely close the deal, since the true statements of the form "H halts" are in fact recursively enumerable. I think your conclusion is true; probably you just need to work a tiny bit harder to get it.
::Oh, whoops, sorry, you're talking about "H does not halt". Sure, that sounds right. I think you can actually do quite a lot better. --Trovatore (talk) 20:49, 25 October 2008 (UTC)
replies: text: I doubt this is the right place for detailed discussion of the ω-rule. Fundamentally ω-consistency is a lot weaker than consistency with the ω-rule or the property of having an ω-model (not sure of the relationship between the latter two — may not be very hard to figure out; I just don't know off the top of my head). --Trovatore (talk) 20:45, 25 October 2008 (UTC)
thread_title:  ω-rule 

message: The page says, "Let T be PA together with the axioms c ≠ n for each natural number n, where c is a new constant added to the language. Then T is arithmetically sound (as any nonstandard model of PA can be expanded to a model of T), but ω-inconsistent (as it proves \exists x\,c=x, and c ≠ n for every number n)."  Does this mean that Skolem's proof of existence of non-standard models of arithmetic provides an alternative proof of Goedel's theorem?  Tkuvho (talk) 15:11, 15 August 2010 (UTC)
replies: text: It doesn't. A theory can be complete and still have nonstandard models. In fact, the Lowenheim-Skolem theorem shows that every first-order theory with an infinite model has nonstandard models, whether the theory is complete or incomplete. So just showing that PA has nonstandard models doesn't show PA is incomplete. You could replace PA in the example above with the (complete) theory of the standard model \mathbb{N}. — Carl (CBM · talk) 19:40, 15 August 2010 (UTC)
thread_title:  Skolem