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Topic: Basis of module and basis of vector space qs.
Replies: 9   Last Post: Nov 21, 2013 10:08 PM

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magidin@math.berkeley.edu

Posts: 11,094
Registered: 12/4/04
Re: Basis of module and basis of vector space qs.
Posted: Nov 19, 2013 2:32 PM
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On Tuesday, November 19, 2013 12:00:10 PM UTC-6, Sandy wrote:
> Arturo Magidin wrote:
>

> > On Tuesday, November 19, 2013 7:59:57 AM UTC-6, Sandy wrote:
>
> >> Arturo Magidin wrote:
>
> >>
>
> >>> On Monday, November 18, 2013 2:17:05 PM UTC-6, Sandy wrote:
>
>
>

> >>>> Is it the case that
>
> >>>> (ii) for every vector space V,
>
> >>>> (if B and C are bases of V then, there is a bijection: B -> C)
>
>
>

> >> " then," was probably meant to be ", then" :-(.
>
>
>

> >>>> iff the axiom of choice holds
>
>
>

> >>> I don't know for sure, but my dim memory is that this is strictly weaker than the full axiom of choice. You might want to check out the website/book on equivalents to the Axiom of Choice by the Rubins.
>
>
>

> >> Thank you. I know of the book but I don't have ready access to it, so I
>
> >> shall firtle around the Internet.
>
> >
>
> > Try math.stackexchange.com; or perhaps mathoverlow.net
>
>
>
> Asked here:
>
> http://math.stackexchange.com/questions/573398/axiom-of-choice-and-vector-space-bases,
>
> where, in a reply, it is suggested that Choice may be replaced by the
>
> Boolean prime ideal theorem (or something weaker still). A link is
>
> given to
>
> http://mathoverflow.net/questions/93242/sizes-of-bases-of-vector-spaces-without-the-axiom-of-choice,
>
> much of which is over my head.


Right; so what the MO link says is that you don't need all of AC to prove that dimension, when it exists, is well defined (that is, that when a vector space has bases, then all bases have the same cardinality); you can "get away" with just the Booliean Prime Ideal Theorem. The MO link also makes the argument showing that BPIT does suffice.

However, the MO link explicitly says that the participants are not sure if "any two bases for the space are bijectable" *implies* the BPIT or some other weaker statement. That is, the precise amount of choice that is needed to conclude that any two bases must be equipollent is unclear from that post, though the participants guess that this has been settled in the literature. (Note, however, that the equivalence of "Every vector space has a basis" with the full Axiom of Choice came fairly late, I think in the 70s, so I would not be surprised if it is not completely settled how much Choice you need for (ii). But certainly, you don't need all of it, as I dimly remembered.

--
Arturo Magidin




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