Re: The seven step series

From: Mirek Dobsicek <m.dobsicek.domain.name.hidden>
Date: Tue, 01 Sep 2009 14:30:40 +0200

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 Tue Sep 01 2009 - 14:30:40 PDT

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