Anatoly Vorobey (avva) wrote,
Anatoly Vorobey
avva

чудесатее и чудесатее

Вдогонку к записи о предложенном доказательстве непротиворечивости теории множеств ZF. Robert Solovay (очень известный специалист в теории множеств) из Беркли нашёл нетривиальную ошибку в доказательстве, которую автор пока не может исправить. Подробности здесь.

Тем временем Рэндалл Холмс, профессор математики из Айдахо, объявил, что придумал доказательство противоречивости PA (арифметики Пеано), причём второго порядка; это значит по сути дела, что натуральных чисел в том виде, в каком мы их обычно понимаем, нет модели; или, иными словами, он нашёл противоречие внутри стандартной модели натуральных чисел N.
Это ещё несравнимо менее вероятно, чем противоречие в теории множеств. Доказательство Холмса можно сгрузить в формате PDF с его домашней страницы, в самом начале её.

Это меня совсем уж возмутило, так что я распечатал доказательство Холмса (небольшое, 6 страниц), прочитал его и нашёл ошибку. В рассылку отправлять не стал, написал прямо ему. Копию помещаю внизу под lj-cut'ом (но её бесполезно читать, если вы не прочитали док-во Холмса, так как я использую введённую им терминологию; само же доказательство я рекомендую читать только особо любопытным или желающим самим найти ошибку. Математической пользы в нём нет).


Dear Prof. Holmes,

Let I be an interpretation for (U,n) and I' an interpretation for
(U',n). I' is said to extend I if U(i) and I(i) are subsets of
U'(i) and I'(i) for all i<=n. What does it mean for a subset of I(i) to
be a subset of I'(i)? I(i) is a pre-interpretation for (U(i),n): a
mapping which sends every assignment of variables <=n to members of the
set U(i), plus a formula <=n, to {0,1}. Members of I(i) are pairs
< <assignment,formula>, truth-value >. So technically speaking, I(i)
being a subset of I'(i) means that truth values of formulas stay the
same as long as assignments remain within original U(i)'s. That
won't preserve correctness. For example, consider a true (in the
standard model) existential sentence (Ex)phi(x), which is mapped to
"false" under some I(i) because U(i-1) is not large enough to contain
the witness for phi(x). Since it's a sentence, it'll have the same truth
value under any assignment (provided I is correct, of course). You want
this sentence to be mapped to "true" when you extend U(i-1) to include
the witness for phi(x); but under your definition the truth value of
(Ex)phi(x) can't change when I is extended to I'. Perhaps what you mean
by I' extending I is that each U'(i) extends U(i), and the set of those
pairs <assignment, formula> which are mapped to "true" by I to remain so
mapped by I'. That won't help you though, because correctness demands
not only conversion of some "false" formulas to "true" formulas under
same assignments as I is extended to I', but also conversion of some
"true" formulas to "false" formulas - for instance, negations of
existential formulas (negation being suitably defined with the help of
your neither/nor operator).
                                                                                
Basically, as you extend U(i) to U'(i), some formulas on which I(i+1)
acts change from "false" to "true", and others from "true" to "false",
under their original assignments within U(i+1). There is no relation
between I' and I that you can formalise and that will allow you to
preserve truth of formulas under extension (assuming correctness of
interpretations). You can still formalise "correct", and you can still
define a veridical interpretation to be such that it can be gradually
extended, preserving correctness, to encompass any given sets, but since
the interpretations aren't conservative extensions of each other, it
doesn't follow that a formula true in such a veridical interpretation
is true in the standard model (and in fact, "veridical" simply collapses
to "correct").
                                                                                
The only thing that can be salvaged from the proof is definability of
truth for purely existential formulas (meaning those that only have
existential quantifiers in the prenex form), which is trivial.
Subscribe
  • Post a new comment

    Error

    default userpic

    Your IP address will be recorded 

    When you submit the form an invisible reCAPTCHA check will be performed.
    You must follow the Privacy Policy and Google Terms of use.
  • 25 comments