[FOM] generalizing Schanuel

[Andreas Blass]

Joe Shipman wrote (18 July 2003) that Schanuel's construction of the
Eudoxus reals can be generalized by replacing the additive group of
integers with any infinite abelian group.  For the integers, "bounded" and
"finite" are the same; in general, "bounded" may be undefined, so Joe
suggested using "finite" in the definitions of "almost homomorphisms
modulo almost equality".  He asked what happens when the group of integers
is replaced with some other specific groups: Z x (Z/2), Z x Z, Z x Z x Z,
(Q,+), (Q-{0},*), Q/Z, or R.  I don't know the answers to most of these
questions, but let me describe what I know and what I believe, in the hope
that someone with more time to devote t this will figure out the full
story.
	First, if you start Schanuel's construction with the reals R in
place of the integers, you get a huge mess.  The reason is that, as an
additive group, R is completely described by the fact that it's a
continuum-dimensional vector space over the rationals.  A homomorphism
from R to R can be specified by giving arbitrary values on any basis of R
as a vector space over Q (called a Hamel basis).  So there are 2^continuum
homomorphisms, all nonequivalent modulo almost equality, and that's
without even looking at *almost* homomorphisms.
	Something similar, but not quite so bad, happens with (Q-{0},*),
which is the product of a cyclic group of order 2 and a
countable-dimensional vector space over the rationals.  You do get
something of cardinality only the continuum, but it doesn't look much like
the reals.
	At the other end of the scale is Z x (Z/2).  Here, I think the
construction just produces the reals.  Any almost homomorphism f from this
group into itself induces an almost homomorphism g from Z to Z by taking
g(n) to be the first component of f(n,0); conversely, any g induces an f
by taking f(n,p) to be (g(n),0).  These transformations are inverse to
each other up to almost equality. The same argument applies with any
finite group in place of Z/2.
	For Z x Z, I think you get the ring of 2 by 2 matrices over R, but
I haven't checked this carefully.  Similarly, Z x Z x Z should give 3 by 3
matrices, etc.
	The case of (Q,+) looks quite interesting.  Any almost
homomorphism f of Q into itself should give a real number as the limit of
f(n)/n as n tends to infinity, and all real numbers arise in this way.  So
the ring we get includes the field of reals and has a projection to the
reals, but it is not just the reals, i.e., that projection isn't
one-to-one.  For example, if we let [x] denote the largest integer not
exceeding x, then f(x)=x-[x] defines an almost homomorphism, not almost
equal to 0, but projecting to the real number 0.  This "fractional part"
function represents an idempotent element in the ring, as fof=f.  The
complementary idempotent is represented not only by the almost
homomorphism that sends x to [x] but by any "reasonable" rounding map.
What I mean by that is that you can define an almost homomorphism g by
sending each rational x to an arbitrarily chosen one of the 100 integers
closest to x, and all the g's you can build this way are almost equal.
And if you allow, say, half-integers instead of integers as the values of
g, nothing changes.
	By the way, if you used "bounded" rather than "finite" in defining
"almost", then the construction starting with Q appears to give R.
	No comment on Q/Z, because I haven't thought about it.
	This doesn't look very foundational, so let me add one comment a
little closer to foundations.  I said that starting with R you get a huge
mess, but the argument there depended on the axiom of choice, to get a
Hamel basis.  If you drop the axiom of choice and work instead in ZF +
dependent choice + "all sets of reals have the Baire property" (which is
consistent relative to an inaccessible cardinal by work of Solovay, and
without the inaccessible cardinal by work of Shelah), then things look
much nicer.  Every homomorphism from R to R is multiplication by a
constant.  I wouldn't be surprised if the almost homomorphisms are also
pretty decent; maybe the situation looks somewhat like that for (Q,+).

Andreas Blass