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):



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