[FOM] Are (C,+) and (R,+) isomorphic?

[Robert M. Solovay]

In my model for "All sets Lebesgue measurable" the two groups are not 
isomorphic. As I recall the proof goes along the following lines: 1) 
(Shown in my paper) every set in R or C has the property of Baire; 2) 
(classical as I recall) this implies every map from C to R which is linear 
is continuous.

Caution: I haven't stopped to reconstruct the proof of 2) just now.

 	--Bob Solovay



On Mon, 20 Feb 2006, Miguel A. Lerma wrote:

> I am conducting an elementary math problem solving group and,
> unexpectedly, the solution to one of the (supposedly "elementary")
> problems has led to a question of foundations.
>
> The solution to the problem involves a group-isomorphism between (C,+)
> and (R,+), i.e., between the additive groups of complex and real
> numbers.  But are they really isomorphic?  The only proof I have in
> mind resorts to the fact that they are Q-vector spaces of the same
> dimension (the cardinality of the continuum), so they are isomorphic
> as vector spaces over Q (rational numbers), and consequently they are
> isomorphic as additive groups.
>
> However that is a highly non-constructive proof, and am not sure
> whether it would work without resorting to the Axiom of Choice.  So,
> this is the question: is there any model of ZF (without AC) in which
> (C,+) and (R,+) are not isomorphic?
>
>
> Miguel A. Lerma
>
> --
>  Miguel A. Lerma
>  Math Comp Sys Admin (former Math Lecturer)
>  NU Math Problem Solving Group Coordinator
>  Department of Mathematics     <mlerma at math.northwestern.edu>
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