Infinite-limit temperaments are horribly screwed up

I was reading Gene's Zeta tuning page on the Wiki, and thought that it
might be useful if I studied the structure of infinite-limit
temperaments to gain a better understanding of what's going on.
Unfortunately, it looks like a complete mess.

The group of infinite-limit vals is isomorphic to the "Baer-Specker
group" Z^N, the direct product of countably many Z's, so I spent the
last few days reading up on it. Naively, we might expect that
subgroups of this group can be taken as representing infinite-limit
temperaments, such as an infinite-limit 7&12 extension of meantone,
and an infinite-rank temperament eliminating nothing but 81/80, for
instance.

Indeed, finitely-generated subgroups of Z^N are straightforward
enough; they represent infinite-limit finite-rank temperaments (which
might be contorted). Some cracks begin to show when we consider
infinite-limit temperaments: the subgroup of vals which tempers out
nothing but 81/80 is of uncountably infinite rank, despite the
quotient group of Q+* by 81/80 being of countably infinite rank.
Indeed, the entire group Z^N of vals itself, representing
infinite-limit JI, is of uncountably infinite rank, while Q+* is
countable, somehow also representing infinite-limit JI.

Things take a turn for the worse once you start considering the
subgroup of Z^N of vals with only finitely many nonzero entries. This
is the direct sum of countably many Z's and is sometimes written as
Z^inf. For instance, the vals <1 2 3 0 0 0 ...| and <6 0 9 4 0 0 0
...| are in this subgroup, but <1 1 1 1 1 1 ...| is not.

Treated as an infinite-limit infinite-rank temperament, this is a
total monstrosity. Despite there being countably many monzos, the
group Hom(Z^inf, Z) of "tmonzos" dual to this "temperament" is now
uncountable, and contains things like |1 1 1 1 ...>, the product of
all primes. Furthermore, it doesn't look like this subgroup actually
tempers anything out; if you take the intersection of kernels of every
val in this subgroup, you get nothing.

Since we're now clearly observing glitches in the Matrix, you might
then wonder if there are subgroups of Z^N which aren't even "dual" to
anything at all, meaning that they can't be expressed as Hom(G, Z) for
some group G. There is no hope for you if you take this route, as it's
undecidable in ZFC.

In short, the whole thing is fantastically screwy, and this poses some
challenges for anyone looking to tie Gene's ideas about zeta in with
regular temperament theory in its current form.

-Mike