# Re: The seven step series

From: Mirek Dobsicek <m.dobsicek.domain.name.hidden>
Date: Wed, 02 Sep 2009 17:16:13 +0200

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.

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.

Similar reasoning is described here

> 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 ...

> 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.

Cheers,
Mirek

> On 01 Sep 2009, at 14:30, Mirek Dobsicek wrote:
>
>> The reason why I am puzzled is that I was recently told that in
>> order to
>> prove that
>>
>> * the union of countably many countable sets is countable
>>
>> one needs to use at least the Axiom of Countable Choice (+ ZF, of
>> course). The same is true in order to show that
>>
>> * a set A is infinite if and only if there is a bijection between A
>> and
>> a proper subset of A
>>
>> (or in another words,
>>
>> * if the set A is infinite, then there exists an injection from the
>> natural numbers N to A)
>>
>> Reading the proofs, I find it rather subtle that some (weaker)
>> axioms of
>> choices are needed. The subtlety comes from the fact that many
>> textbook
>> do not mention it.
>>
>> In order to understand a little bit more to the axiom of choice, I am
>> thinkig if it has already been used in the material you covered or
>> whether it was not really needed at all. Not being able to answer
>> it, I
>>
>> Please note that I don't have any strong opinion about the Axiom of
>>
>> Mirek
>>
>>
>>
>>
>>
>> Bruno Marchal wrote:
>>> Hi Mirek,
>>>
>>>
>>> On 01 Sep 2009, at 12:25, Mirek Dobsicek wrote:
>>>
>>>
>>>> I am puzzled by one thing. Is the Axiom of dependent choice (DC)
>>>> assumed
>>>> implicitly somewhere here or is it obvious that there is no need for
>>>> it
>>>> (so far)?
>>> I don't see where I would have use it, and I don't think I will use
>>> it. Cantor's theorem can be done in ZF without any form of choice
>>> axioms. I think.
>>>
>>> Well, I may use the (full) axiom of choice by assuming that all
>>> cardinals are comparable, but I don't think I will use this above
>>> some
>>> illustrations.
>>>
>>> If you suspect I am using it, don't hesitate to tell me. But so far I
>>> don't think I have use it.
>>>
>>> Bruno
>>>

--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups "Everything List" group.
To post to this group, send email to everything-list.domain.name.hidden
To unsubscribe from this group, send email to everything-list+unsubscribe.domain.name.hidden
For more options, visit this group at http://groups.google.com/group/everything-list?hl=en
-~----------~----~----~----~------~----~------~--~---
Received on Wed Sep 02 2009 - 17:16:13 PDT

This archive was generated by hypermail 2.3.0 : Fri Feb 16 2018 - 13:20:16 PST