Talk:Tarski's undefinability theorem

message: Not really an English word, even though for some odd reason it does get a fair number of Google hits. Should be moved to Tarski's undefinability theorem, or better, Tarski's theorem on the undefinability of truth, which I think is the standard name. --Trovatore 05:42, 9 March 2006 (UTC)
replies: text: Looking it up in www.m-w.com, I find "indefinable" is an English word, which I suppose I knew. I still maintain it's not a mathematical English word. It's used for qualities like someone's indefinable charm, not for sets that have no definition of a specified form. --Trovatore 05:49, 9 March 2006 (UTC)
thread_title: Indefinability?

message: The proof does assume that every L-formula has a Gödel number,  Does this mean that the result does not hold for languages that are uncountable sets?  Taemyr (talk) 18:48, 10 January 2008 (UTC)
replies: text: If you could define a language that consists of an uncountable infinite set of formulas, this proof probably wouldn't hold anymore. However, proofs in such a language are not machine-checkable, and such a language must allow infinitely long formulas, i.e. formulas with infinitely many symbols. So such a language would certainly not be usefull in mathematical logic, and it is arguable if it can be considered a formal language at all. Somejan (talk) 09:53, 1 July 2008 (UTC)
replies: text: Such a language needn't involve infinitely long formulas; it could just have an uncountable number of constants, for example. Anyway, the proof would still go through, for a suitably expanded notion of Gödel number. The point is, there can be no definable surjection from T to the definable predicates on T, whatever the type T is (be it integers, strings, real numbers, pizza toppings, or anything else). (Of course, this is just Cantor's theorem in the category of definable maps, as was essentially pointed out by Lawvere). -Chinju (talk) 08:57, 21 November 2008 (UTC)
thread_title:  Gödel numbers 

message: If I understand the hyperarithmetical theory article, the truth predicate for PA can be written as a formula in second-order arithmetic.  Is that correct?  Should this article mention it?  66.127.55.192 (talk) 14:37, 7 February 2010 (UTC)
replies: text: It would be an example that the truth predicate for one language can be expressed in a stronger language.  It's not really surprising, and does not contradict the theorem.  Taemyr (talk) 17:05, 7 February 2010 (UTC)
replies: text: Of course it doesn't contradict the theorem, but it's not obvious to a non-expert.  I only realized it after looking at that other article earlier. 66.127.55.192 (talk) 20:31, 7 February 2010 (UTC)
replies: text: I added a paragraph; is that what you meant? — Carl (CBM · talk) 23:43, 7 February 2010 (UTC)
replies: text: Yes, thanks.   66.127.55.192 (talk) 00:10, 8 February 2010 (UTC)
thread_title:  second order arithmetic 

message: I'm familiar with "complete" as used in complexity (Complete_(complexity)).  For that usage, "complete" is wrong here (because the arithmetical hierarchy does not collapse, as stated in the proof containing the referenced sentence). I would expect the term "hard" here instead.  —Preceding unsigned comment added by Enoksrd (talk • contribs) 20:37, 22 June 2010 (UTC)
replies: text: You're completely correct. Thanks for pointing this out. — Carl (CBM · talk) 23:12, 22 June 2010 (UTC)
thread_title:  Error?: "However, T* is $\Sigma^0_k$  complete for all $k$." Should "complete" be "hard" instead? 

message: I have trouble believing the theorem statement is actually correct. I think that the formula True(x) cannot be a function of another formula, but must be a function of the code number of x, as usual practice when the diagonalization lemma is invoked. For instance, the hypothetic True is assumed to define the set of code numbers of true sentences, so it should be a predicate on code numbers. Moreover, in the proof, how do you otherwise define a formula S referring to itself?
Hence, I would guess that the correct statement of the theorem is the following.
Let L be the language of first-order arithmetic, and let N be the standard structure for L. Thus (L, N) is the "interpreted first-order language of arithmetic." Let T denote the set of  L-sentences true in N, T* the set of code numbers of the sentences in T, and g the function taking L-sentences to their code numbers. The following theorem answers the question: Can T* be defined by a formula of first-order arithmetic?
Tarski's undefinability theorem:  There is no L-formula True(x) which defines T*.
That is, there is no L-formula True(x) such that for every L-formula x, True(g(x)) ↔ x is true.
A similar change needs to be done in the subsequent proof sketch.
--Blaisorblade (talk) 15:09, 3 February 2012 (UTC)
replies: text: Thank you for the correction, you're quite right. — Carl (CBM · talk) 12:24, 4 February 2012 (UTC)
thread_title:  Theorem formulation