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

[Christian Espindola]

Thomas, you're right, I must have made the wrong calculations, since that
strategy seems actually to need full AC: if b has the same power than the
vector space V, because there is an embedding from b^2 into | V | (and one
from b into b^2), we would have b=b^2 for every infinite b, which implies
AC.

Perhaps, as you say, it's not the only way to proceed, so one might want to
try something else, if possible.

On Fri, Mar 23, 2012 at 10:44 AM, Thomas Forster
<T.Forster at dpmms.cam.ac.uk>wrote:

>
> I'm not sure that this is entirely correct, but it's certainly something
> to think about. Given two vector spaces one has no chance of showing them
> iso unless their bases have the same power, but one might be able to do
> that without showing that the bases have the same power as the spaces.
> (Tho' i admit it doesn't seem terrible hopeful!).  And the cardinal
> principle one would need for Christian's strategy is something more like
> $(b \cdot f)^{< \omega} = b$ where $b$ is the power of the basis and
> $f < < b$ is the power of the field - which looks a bit more powerful than
> the choice principle he mentions.
>
>
>
> On Thu, 22 Mar 2012, Christian Espindola wrote:
>
>  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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>>
>>
>>
>
> 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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>
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