# [FOM] A question about uncountable torsion-free divisible groups

Christian Espindola christian.espindola at gmail.com
Thu Mar 22 06:04:41 EDT 2012

One little observation regarding the general question. To conclude that
those two vector spaces have basis of the same cardinality one would also
need to use that the cardinality of a basis (if it exists) is the same as
that of the vector space. This uses that m times aleph_0 is equal to m for
all infinite cardinals, which, unless the uncountable cardinal is an aleph,
relies on some choice principle (while it is strictly less than AC, it
implies that every infinite set has a countable subset, for instance).

On Wed, Mar 21, 2012 at 12:32 PM, Thomas Forster
<T.Forster at dpmms.cam.ac.uk>wrote:

>
> ...from a colleague of mine here.
>
> It's an uncountably categorical theory: given any two of these
> things of the same uncountable power, think of them as vector
> spaces over Q. They are then vector spaces of the same dimension
> over the one vector space, and so are isomorphic - once was has
> a basis! This is where AC comes in. Can one do it with anything
> strictly weaker than full AC? I know that if every vector space
> has a basis then one gets back AC... but we aren't going to
> assume that *every* vector space has a basis.
>
> (Actually my colleague's specific question was about how much
> choice one needs to prove that \$\Re\$ and \$\Re^2\$ are iso as
> abelian groups, but the general question seems to be of interest)
>
>  Can anyone shed any light..?
>
>
>
> URL:  www.dpmms.cam.ac.uk/~tf <http://www.dpmms.cam.ac.uk/%7Etf>; DPMMS
> ph: +44-1223-337981;
> mobile +44-7887-701-562.
>
>
>
>
>
>
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