On 02 Sep 2009, at 17:16, Mirek Dobsicek wrote:
> > Bruno Marchal wrote: >> Ouh la la ... Mirek, >> >> You may be right, but I am not sure. You may verify if this was not >> in >> a intuitionist context. Without the excluded middle principle, you >> may >> have to use countable choice in some situation where classical logic >> does not, but I am not sure. > > Please see > http://en.wikipedia.org/wiki/Countable_set > the sketch of proof that the union of countably many countable sets is > countable is in the second half of the article. I don't think it has > anything to do with the law of excluded middle. I was thinking about the equivalence of the definitions of infinite set (self-injection, versus injection of omega), which, I think are inequivalent without excluded middle, but perhaps non equivalent with absence of choice, I don't know) > > Similar reasoning is described here > http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2008;task=show_msg;msg=1545.0001 I am not sure ... I may think about this later ... > > >> My opinion on choice axioms is that there are obviously true, and >> this >> despite I am not a set realist. > > OK, thanks. > >> I am glad, nevertheless that ZF and ZFC have exactly the same >> arithmetical provability power, so all proof in ZFC of an >> arithmetical >> theorem can be done without C, in ZF. This can be seen through the >> use >> of Gödel's constructible models. > > I am sorry, but I have no idea what might an "arithmetical provability > power" refer to. Just give me a hint ... By arithmetical provability power, I mean the set of first order arithmetical sentences provable in the theory, or by a machine. I will say, for example, that the power of Robinson Arithmetic is smaller than the power of Peano Aritmetic, *because* the set of arithmetical theorems of Robinson Ar. is included in the set of theorems of Peano Ar. Let us write this by RA < PA. OK? Typically, RA < PA < ZF = ZFC < ZF + k (k = "there exists a inaccessible cardinal"). The amazing thing is ZF = ZFC (in this sense!). Any proof of a theorem of arithmetic using the axiom of choice, can be done without it. > >> I use set theory informally at the metalevel, and I will not address >> such questions. As I said, I use Cantor theorem for minimal purpose, >> and as a simple example of diagonalization. > > OK. Fair enough. > >> I am far more puzzled by indeterminacy axioms, and even a bit >> frightened by infinite game theory .... I have no intuitive clues in >> such fields. > > Do you have some links please? Just to check it and write down few new > key words. This is not too bad, imo, (I should have use "determinacy", it is a better key word): http://en.wikipedia.org/wiki/Determinacy Bruno http://iridia.ulb.ac.be/~marchal/ --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~----------~----~----~----~------~----~------~--~---
