[FOM] A question about uncountable torsion-free divisible groups
Thomas Forster
T.Forster at dpmms.cam.ac.uk
Fri Mar 23 05:44:36 EDT 2012
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; DPMMS ph: +44-1223-337981;
> mobile +44-7887-701-562.
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>
>
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