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/
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Received on Wed Sep 02 2009 - 18:21:39 PDT