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

[Harvey Friedman]

On 2/20/06 10:26 AM, "Miguel A. Lerma" <mlerma at math.northwestern.edu> wrote:

> 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?

Let h be such an isomorphism. Look at h on the reals inside C. Then we get
h:R into R which is an additive isomorphism onto a proper additive subgroup
of R.  

THEOREM. (ZFCD). Every Borel measurable h:R into R such that for all x,y in
R, h(x + y) = h(x) + h(y), is multiplication by a constant. It suffices to
assume that h has the Baire property.

Since there is a model of ZFDC such that every set of reals has the Biare
property, we see that there is a model of ZFDC such that every h:R into R
such that for all x,y in R, h(x, + y) = h(x) + h(y), is multiplication by a
constant.

I think this is all well known, especially to Solovay.

Harvey Friedman