bio | website | mathnet.ru/php/… |
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visits | member for | 2 years, 11 months |
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stats | profile views | 1,460 |
Sep 21 |
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“Axiom of global choice”
François, I don't understand you. It's not me who writes books about "conglomerates" or articles in Wikipedia about "strong form of global axiom of choice". My question is addressed to people who understands what these words mean. If they will tell me that this is explained in a paper by Ackermann or by somebody else, I would thank them and start to read this paper. How can I explain what theory they have in mind if the only things I know about it is a mentioning in a book (without references) and an article in Wikipedia (with irrelevant references)? |
Sep 21 |
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“Axiom of global choice”
Michael, according to Neumann-Goedel-Bernays, sets are defined exactly as those classes which are elements of other classes. This implies that these "conglomerates" can't be classes, because otherwise we obtain that each conglomerate of classes is actually a conglomerate of sets, and there is no difference between the "strong" and the "weak" forms of the axiom of choice. So conglomerates must be introduced axiomatically: there must be a new axiomatic system, a theory, with conglomerates as objects, and with the results explaining why this theory extends the usual set theory. |
Sep 21 |
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“Axiom of global choice”
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Sep 21 |
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“Axiom of global choice”
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Sep 20 |
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“Axiom of global choice”
@Zhen Lin: You swallow the most intriguing part of the explanation: how can a class (which is not a set) be an element of someting else? |
Sep 20 |
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“Axiom of global choice”
By justification I mean an accurate definition of the new tool together with the analysis of whether it is compatible with the other tools of the theory. For example, when I apply usual axiom of choice (in Wikipedia it is called the "weak form"), I have in mind 1) one of the textbooks, for example the one by J.Kelly "General topology", where (in appendix) the classes are defined (axiomatically) and functions (as a construction of a theory) and the axiom of choice is formulated, and 2) the results of Goedel and Cohen which show that this axiom is independent from the other axioms of the theory. |
Sep 20 |
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“Axiom of global choice”
@Zhen Lin: Do you mean what they call in Wikipedia "strong form of the axiom of global choice"? My question is where can I read about it. |
Sep 20 |
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“Axiom of global choice”
David, but if I consider a classical category, which is defined in a standard way on the base of set theory, like, for example, a category of topological vector spaces, what should I change in the set theory for obtaining the theorem that this category has a skeleton? |
Sep 20 |
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“Axiom of global choice”
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Sep 20 |
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“Axiom of global choice”
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Sep 20 |
asked | “Axiom of global choice” |
Sep 6 |
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Trying to find Russian paper from 1947
Most of Russian mathematical joournals are available in electronic form here: mathnet.ru/#, but this is not the one. You can try to contact I.V.Volovich directly, his address is here: mathnet.ru/php/person.phtml?option_lang=rus&personid=8846 I am sure he will help. |
Jun 18 |
awarded | Good Question |
May 12 |
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the Richardson theorem and the base identities problem
Thank you, now I recall that in the university our lecturer told something similar. But I suppose, it's too difficult to communicate each other in this way. If you have time and forbearance, you can contact me by e-mail, it is indicated at the website that I gave at my page here in Mathoverflow. Or you can just edit your answer and put there a name of a book about all this. |
May 12 |
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the Richardson theorem and the base identities problem
I need somebody to talk about this. When reading texts on this topic I see a lot of unclear places. For example, according to definition - en.wikipedia.org/wiki/Recursive_set - recursive set must lie in the set of positive integers. When Richardson says that the set of true identities is not recursive, this means that the set of all identities is enumerated, I suppose... So there must be a standard procedure for numeration formulas, is it? What is this procedure? Or I don't understand something? |
May 12 |
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the Richardson theorem and the base identities problem
Or, perhaps, the situation is more simple: is it possible that giving a recursive base for all true identities is equivalent to giving an algorithm which describes all true identities? If yes, then, apparently these two problems are again the same (up to obvious replacements)? |
May 12 |
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the Richardson theorem and the base identities problem
Is it possible, that Richardson and Gurevich do the same, but for different classes of identities, and with different results? Can I understand the Richardson therem as follows: he describes some class of identities (although, in contrast to what Gurevich does, I don't understand what exactly this class is) and proves that there is no a recursive base for this class? |
May 12 |
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the Richardson theorem and the base identities problem
@Gerhard: If I were a specialist, I could of course formulate a question in such a way that it would be much easier for another specialist to give an answer, since there would not be necessity to explain elementary things to non-specialists. But in this case there would not be a necessity for me to ask elementry questions, since, being a specialist, I could understand everything independently, without other specialists. |
May 12 |
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the Richardson theorem and the base identities problem
Gerhard, I don't feel the difference. The identities for Gurevich are true not only in his abstract model, but for functions from $N^k$ to $N$ as well. And, as far as I understand, for Richardson this is the same, but the difference is that he considers functions from $R$ to $R$. |
May 12 |
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the Richardson theorem and the base identities problem
@Gerhard: "The set of identities satisfied by the real numbers with exponentiation , addition and multiplication is a set which has a logically equivalent, recursive, and non finite subset. The set of terms in the Richardson theorem which are equivalent to 0 is a nonrecursive subset of the set of all terms used in the context of the theorem." - This must mean that Gurevich and Richardson use the same terminology, do they? |