Re: The seven step series

From: Mirek Dobsicek <>
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.

Please see
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;task=show_msg;msg=1545.0001

> 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.


> 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
>> had to ask :-)
>> Please note that I don't have any strong opinion about the Axiom of
>> Choice. Just trying to understand it. May I ask about your opinion?
>> 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

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Received on Wed Sep 02 2009 - 17:16:13 PDT

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