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\title{Measuring the classification difficulty of 
countable torsion free abelian groups}         % Enter your title between curly braces
\author{Greg Hjorth}        % Enter your name between curly braces
\date{\today}          % Enter your date or \today between curly braces
\maketitle


\section{THE PROBLEM}       % Enter section title between curly braces

\bigskip 

\underline{Question:} Can we hope to classify countable torsion free abelian groups? 

\bigskip 

Already a few remarks should be made about this question. 

First of all the word ``classify" is somewhat plastic in its meaning. Someone 
might for instance take the question to mean whether there is any sense at 
all in which we can understand countable torsion free abelian groups, and I 
am sure ``classification" takes on different hues across different 
guilds and mathematical specialties. 

I will take the word ``classify" to mean "completely classify by some class of 
invariants." Here I have in mind something like the Ulm invariants for countable 
abelian $p$-groups or Baer's classification for the rank one case. 

Secondly one might wonder about the restriction to this particular class of 
groups. Here I would respond by saying that we cannot hope to classify everything, 
and some restrictions probably are inevitable. Abelian groups represent the topic 
of this conference, and should be easier and more hopeful than general groups; 
and the choice of torsion free further removes potentially distracting details. 
As for confining ourselves to the countable case, cardinality $\aleph_0$, I 
would mention the kinds of set theoretical complexities which can arise when 
one considers uncountable discrete structures. Frequently one is led into 
independence results, and considering subtle combinatorial properties, such as the 
behavior of the non-stationary ideal, 
and even classification schemes which would be virtually 
perfect in the countable case, such as the Ulm invariants, may begin to fail 
when we pass to $\aleph_1$. 

Even granting these restrictions, we may want to take a sceptical stance. 
After all, if a classification scheme was going to be found then 
surely it would have made itself known already. 

So perhaps it would be better to ask: 

\bigskip 

\underline{Question:} What would establish that there is no way to completely classify 
countable torsion free abelian groups? 

\bigskip 

Some possible answers: 

\bigskip 

\underline{Answer 1:} (An appeal to empirical evidence.) No classification scheme has been 
forthcoming. We have waited long enough. It is safe to assume that nothing here 
is possible. 

\bigskip 

For some people this will already this will already be enough. And if this is your 
position, then the remainder of the talk is unlikely to hold much interest. 

Personally I am inclined to at least look for a deeper explanation of this empirical 
fact, so let me push on: 

\bigskip 

\underline{Answer 2:} We may be able to reduce some other truly horrendous classification 
problem to that of countable torsion free abelian groups. For instance, perhaps 
there is a way we can show them to be as hard to classify as general countable 
groups. 

\bigskip 

Still this is a little bit question begging, since we might then want some 
confirmation of the intuition that general countable groups are unclassifiable. 

\bigskip 

\underline{Answer 3:} Perhaps we can develop an abstract theory of invariants, and show that 
there is no {\it reasonable} way to assign certain classes of objects, such as 
the Ulm invariants or Baer's invariants for rank one, as complete invariants. 

\bigskip 

In fact, it turns out, following work of Friedman, Kechris, Louveau, Stanley, and 
others, that such a theory has been developed. In this talk I want to discuss how 
that theory bears on the classification problem for countable torsion free abelian 
groups. 

The meaning of {\it reasonable} is subject to some negotiation. I will begin 
by considering the explication which takes a {\it reasonable reduction} to be one 
that is Borel in some appropriate Borel structure. 

\def\R{\mathbb R}

Other explications are possible. For instance, absolutely $\Ubf{\Sigma}^{\rm HC}_1$, 
as discussed several sections below. 
Or using reductions in $L(\R)$. And there are various other 
exotic classes that logicians find natural to consider. 
The Borel category has the advantage of being 
one which is widely used mathematically and does indeed include most commonly 
accepted classification schemes. 

However the reader would not go too far wrong to simply think of a {\it reasonable 
classification} as being one which does not make an egregious appeal to the 
existence of a well ordering of $\R$ in assigning its invariants. 

\bigskip 

\def\N{\mathbb N}


\section{SPACES OF ABELIAN GROUPS} 

\bigskip 

\underline{Definition:} Let AbGrp $=\{H=(+^H, -(\cdot)^H)\in \N^{\N\times\N}\times 
\N^\N| H$ defines an abelian group structure on $\N\}$. 

\bigskip 

In the discrete topology, $\N$ is a separable completely metrizable space. 
Thus $\N^{\N\times\N}$ and $\N^\N$ are separable completely metrizable spaces 
in the product spaces, as is $\N^{\N\times\N}\times 
\N^\N$. We call this kind of space a {\it Polish space}. 

AbGrp is a closed subset of a Polish space, and hence again Polish in the 
subspace topology. 

\bigskip 

\underline{Definition:} TFA$=\{H\in{\rm AbGrp}: H$ is torsion free$\}$ and TFA$_n=\{H 
\in {\rm TFA}: H $  has rank $\leq n\}$. 

\bigskip 

\def\Q{\mathbb Q}

Again TFA is a closed subset of AbGrp, and hence again Polish. TFA$_n$ is not 
closed, but instead $G_\delta$ -- that is to say, defined by a countable intersection 
of open sets -- and hence Polish.\footnote{For a proof of the general fact that 
a $G_\delta$ subset of a Polish space is again Polish one can see Kechris' 
{\bf Classical descriptive set theory}. This is also a good reference for 
other general facts about Polish spaces and Borel sets.} 

There are other ways in which we can model these objects. For instance, in 
Simon Thomas' papers\footnote{See for instance 
{\it On the complexity of the classification problem for torsion-free abelian groups of rank two},  to appear in {\bf Acta Mathematica} or 
{\it On the complexity of the classification problem 
for torsion-free abelian groups of finite rank,} to appear in the 
{\bf Bulletin of Symbolic Logic}} 
he takes the space of subgroups of $\Q^n$ to provide a 
Borel structure on the torsion free abelian groups of rank $\leq n$. It turns 
out that from the point of view of the kinds of questions we will be considering, 
such choices are immaterial. All the known ways of providing a Borel structure 
give the same results. 

\bigskip 

\underline{Definition:} For $X, Y$ Polish, 
a function $f: X\rightarrow Y$ is Borel if for any 
open set $O$ we have 
$f^{-1}[O]$ is Borel -- that is to say, 
in the $\sigma$-algebra generated by the open sets. 

\bigskip 

\section{A FIRST APPROXIMATION: SMOOTHNESS} 

\bigskip 

\underline{Definition:} An equivalence relation $E$ on a Polish space $X$ is {\it smooth} 
(Mackey\footnote{See {\it Infinite dimensional group representations}, 
{\bf Bulletin of the American Mathematical Society}, vol. 69(1965)}) 
or {\it tame} or {\it concretely classifiable} (Kechris) if there is a 
Borel function 
\[f: X\rightarrow Y,\] 
for some Polish space $Y$, such that $\forall x_1, x_2\in X$
\[x_1 E x_2 {\: \rm iff } \: f(x_1)=f(x_2).\] 

\bigskip 

\underline{E.g.} If we take $Y=\R$, then this would correspond to assigning real numbers 
as complete invariants. 

\bigskip 

\underline{Alas:} Almost no real life equivalence relations are smooth.\footnote{One 
of the few exceptions to this lament, and something very much in the mind of George 
Mackey, is given by the irreducible unitary representations of countable finite by 
abelian groups considered up to isomorphism. They {\it are} smooth.} 

\bigskip 

\underline{E.g.} $\cong|_{{\rm TFA}_1}$ (isomorphism of rank 1 torsion free abelian 
groups) is {\it not} smooth.\footnote{An elementary proof of this fact can be 
found in the exercises at the end of $\S$3.1 of {\bf Classification and orbit 
equivalence relations,} G. Hjorth, AMS Surveys and Monograph Series, Rhode Island, 
2000. 

A sceptical reader might be concerned that the inability to assign reals or points 
in some other concrete space as complete invariants for rank 1 TFA groups is purely an 
artifact of our decision to work in the Borel category. This would be a 
very reasonable 
concern. 

It turns out not to be the case. The same obstruction reappears even considering 
much more generous class of functions, but a full discussion of this would require 
an excursion into foundational issues.}  

\bigskip 

\section{A BETTER APPROXIMATION: BOREL REDUCIBILITY} 

\bigskip 

\underline{Definition:} For $E, F$ equivalence relations on Polish 
$X, Y$, we write 
\[E\leq_B F\] 
if there is a Borel function $f:X\rightarrow Y$ such that $\forall x_1, x_2\in X$ 
\[x_1 E x_2\Leftrightarrow f(x_1) F f(x_2).\] 
In other words, for any $x\in X$, $[f(x)]_F$ (the $F$-equivalence class of $f(x)$) is a 
complete invariant for $[x]_E$. 

We can then write $E<_B F$ if $E\leq_B F$ holds but $F\leq_B E$ fails. 

\bigskip 

\underline{E.g} Consider $\{0,1\}^\N=_{\rm df} 2^\N$, the space of 
infinite binary sequences in the product topology. For $\vec x=(x_0, x_1,...), 
\vec y=(y_0, y_1,...)$, set 
\[\vec x E_0 \vec y\] 
if 
\[\exists N\forall n> N (x_n=y_n).\] 

So this is the equivalence relation of eventual agreement, and under the 
natural identification of $2^\N$ with ${\cal P}(\N)$ (the power set of $\N$), 
one has 
\[2^\N/E_0\sim {\cal P}(\N)/{\rm Finite}.\] 

Sets considered up to finite difference are not totally unreasonable 
objects to try to 
assign as complete invariants, and indeed there is the following classical 
theorem: 

\bigskip 

\underline{Theorem:} (In effect, Baer) $\cong|_{{\rm TFA}_1}\leq_B E_0$. 

\bigskip 

Indeed, this is precise. One can also show $E_0\leq_B\cong|_{{\rm TFA}_1}$. 
And indeed it was shown by Harrington, Kechris, and Louveau\footnote{L. Harrington, 
A.S. Kechris,  A. Louveau, {\it  A Glimm-Effros dichotomy for Borel equivalence relations},  
{\bf Journal American Mathematical Society,} 
vol. 3(1990), pp. 903--928. } that $E_0$ corresponds to the {\it next} level of classification 
difficulty after smoothness. 

For a few years it was open whether the same thing is true for rank 2. This was 
ultimately shown to be false. 

\bigskip 

\underline{Theorem:} (Hjorth) \footnote{See {\it Around nonclassifiability for countable torsion 
free abelian groups,} 
{\bf Abelian groups and modules (Dublin, 1998)}, 
pp. 269--292, Trends Math., Birkh�user, Basel, 1999. Here I should mention as an aside that 
Simon Thomas has recently shown that for every $n$ we have 
\[\cong|_{{\rm TFA}_n}<_B \cong|_{{\rm TFA}_{n+1}},\] 
thereby subsuming this earlier result. 

I would have been inclined to consider this the final word on the abstract question 
of the classification difficulty of the finite rank TFA groups, but after the talk 
someone pointed out a further issue which is unresolved. We do not know whether 
$\cong|_{{\rm TFA}_2}$ lies {\it directly after} $E_0$ in this hierarchy of classification 
difficulties -- that is to say, if $E<_B \cong|_{{\rm TFA}_2}$, must it be the case that 
$E\leq_B E_0$?}
 $\cong|_{{\rm TFA}_2}\not\leq E_0$. 

\bigskip 

And thus in particular we would have to look well beyond $E_0$ in 
order to find complete invariants for $\cong|_{{\rm TFA}}$, the infinite dimensional 
case. 

\bigskip 

\underline{Definition:} For $x, y\in 2^{\N\times\N}$, set $xF_2 y$ if 
\[\{x(n, \cdot)|n\in\N\}=\{y(n, \cdot)|n\in\N\};\] 
that is to say, $\forall n_2\exists n_2, n_3\forall m$ 
\[x(n_1, m)=y(n_2, m),\]
\[x(n_3, m)=y(n_1, m).\] 

\bigskip 

Thus $F_2$-equivalence classes correspond to something like countable sets of 
reals.\footnote{Here and beyond I am somewhat lazily assuming that 
the sequence $(x(n,\cdot))_{n\in\N}$ has no repetitions, and thus $x F_2 y$ if and 
only if $\{x(n, \cdot)|n\in\N\}=\{y(n, \cdot)|n\in\N\}$; in fact it can be argued that 
we lose no generality if we restrict our attention to $x$ for which the sequence 
$n\mapsto x(n, \cdot)$ is one to one.} 

\bigskip 

\underline{Fact:} At each $n$, $\cong|_{{\rm TFA}_n}\leq_B F_2$. 

\bigskip 

\def\Z{\mathbb Z}

This is proved as so: For each $H$ a rank $n$ torsion free abelian group, 
and $\vec g= g_1,..., g_n\in H$, we let $\theta_0(H, \vec g)=\{(k_1,...k_n, \ell)\in 
\Z^{n+1}: \ell$ divides $k_1\cdot g_1+k_2\cdot g_2+...k_n\cdot g_n\}$. Then 
\[\theta(H)=\{\theta_0(H, \vec g): \vec g\in H^n\}\] 
gives us an element of ${\cal P}_{\aleph_0}(\N^{n+1})$ as a complete invariant. 
It is not a big step to turn these invariants into elements of ${\cal P}_{\aleph_0}(\N)$, 
and from there to pass to a Borel reduction to the equivalence relation $F_2$. 

The general rule of thumb is that any class of countable structures which are in 
some sense {\it finitely generated} or in some sense have {\it finite rank} are 
reducible to $F_2$ by a Borel function.\footnote{For a discussion of results in 
this direction, as well as general schemes for classifying countable structures, one 
can see G. Hjorth, A.S. Kechris, {\it Borel equivalence relations and 
classifications of countable models,} 
{\bf Annals of Pure and Applied Logic,} 
vol. 82(1996), pp. 221--272.} 

So we might at least hope that something like {\it countable sets of reals} can 
stand as complete invariants for $\cong|_{\rm TFA}$. 

\bigskip 

\underline{Definition:} For $x, y\in 2^{\N^{n+1}}$, set 
\[x F_{n+1} y\] if 
\[\{[x(m, \cdot,...)]_{F_n}|m\in\N\}=\{[y(m,\cdot,...)]_{F_n}|m\in\N\}.\] 

\bigskip 

In other words, $F_2$-equivalence classes correspond to something like elements 
of ${\cal P}_{\aleph_0}(\N)$, $F_3$-equivalence classes correspond to elements of 
${\cal P}_{\aleph_0}({\cal P}_{\aleph_0}(\N))$, and so on. 

In fact, as given by Friedman and Stanley\footnote{{\it A Borel reducibility 
theory for classes of countable structures,} {\bf Journal of Symbolic Logic,} 
vol. 54(1989), 
pp. 894--914} one can iterate this definition out through the countable 
ordinals, and define $F_\alpha$ for each $\alpha < \omega_1$. They also observe: 

\bigskip 

\underline{Theorem:} (Friedman-Stanley) For each $n$, 
\[F_n <_B F_{n+1}.\] 

\bigskip 

At this point we can finally give the chief negative result regarding the isomorphism 
relation on countable torsion free abelian groups. 

\bigskip 

\underline{Theorem:} (Hjorth\footnote{{\it Torsion free abelian groups}, 
available on line at 

\centerline{\texttt{www.math.ucla.edu/\^{}greg/research.html}}}) For every $n$, 
\[F_n\leq_B \cong|_{\rm TFA}.\] 

\bigskip 

Combining this with Friedman-Stanley: 

\bigskip 

\underline{Corollary:} At every $n$, 
\[\cong|_{\rm TFA}\not\leq_B F_n.\] 

\bigskip 

(And as might be expected, this goes out through the ordinals. At every countable $\alpha$, 
$F_\alpha <_B \cong|_{\rm TFA}.$) 

\bigskip 

\section{WHAT WE WOULD ULTIMATELY WANT TO PROVE} 

\bigskip 

\underline{Definition:} For ${\cal L}$ a 
countable language, with relations $R_1, R_2,...$, having arities 
$a(1), a(2),...$, and function symbols $F_1, F_2,....$ having arities $b(1), b(2),...$, we let 
Mod$({\cal L})$ be the space 
\[\prod_i 2^{\N^{a(i)}}\times \prod_j \N^{\N^{b(j)}}.\] 
We can define the isomorphism relation, $\cong|_{{\rm Mod}({\cal L})}$, on this 
space in the obvious way. 

We then say 
an isomorphism invariant $K\subset$Mod$({\cal L})$ is said to be {\it universal} 
if given any other countable ${\cal L}'$ we have 
\[\cong|_{{\rm Mod}({\cal L}')}\leq_B \cong|_K.\] 

\bigskip 

\underline{E.g.1} Graphs on $\N$ can be viewed as a subset of $2^{\N^2}$. It is a folklore 
result that the isomorphism relation on this class of countable structures is universal. 


\underline{E.g.2} (Friedman-Stanley) 
Countable linear orderings are universal in this sense. 

\underline{E.g.3} (? folk?) Isomorphism on countable groups is universal. 

\underline{E.g.4} (Camerlo-Gao) Countable Boolean algebras are universal. 

\bigskip 

So....

\bigskip 

\underline{Question:} Are countable torsion free abelian groups universal? 

\bigskip 

Here I have to admit -- with great shame and hanging of the head -- that I had 
previously announced in print\footnote{{\it Around the classification of countable 
torsion free abelian groups}, Dublin proceedings, as cited above.} a positive solution to 
this problem. The proof was flawed, though it turns out that the result above 
regarding the $F_{\alpha}$'s is coming close, since they play a 
special role 
in the general investigation of isomorphism of countable structures. 

\bigskip 

\underline{Theorem:} (Dana Scott\footnote{See 
{\it Invariant Borel sets,} 
{\bf Fundamenta Mathematica,} vol. 56(1964) pp. 117--128.}) 
For each countable language ${\cal L}$ there are Borel sets 
$(A_\alpha)_{\alpha\in \aleph_1}$ such that: 

\leftskip 0.4in 

\noindent (a) the space of 
${\cal L}$-structures with underlying set $\N$ equals 
\[\bigcup_{\alpha\in\aleph_1} A_\alpha;\] 

\noindent (b) at each $\alpha$, 
\[\cong|_{\bigcup_{\beta\leq \alpha}A_\alpha}\leq_B F_\alpha.\] 

\leftskip 0in 

\bigskip 

There are various consequences of his result which can probably be 
considered folklore. 
For instance, a Borel set of countable structures has a  
Borel isomorphism relation if and only if it is Borel reducible to some 
$F_\alpha$, if and only if it is included in some $\bigcup_{\beta<\alpha}A_\beta$, 
some countable $\alpha$. 

\bigskip 

\underline{Question:} (Friedman-Stanley) Let ${\cal L}$ be a countable language and let 
$K\subset {\rm Mod}({\cal L})$ be an isomorphism invariant Borel subset. Suppose 
at each $\alpha< \aleph_1$ we have 
\[F_\alpha \leq_B \cong|_K.\] 

Must $\cong|_K$ be universal? 

\bigskip 

Therefore, at the very least, we can say that either countable torsion free abelian 
groups are universal, or their failure to be represents an entirely new phenomena. 

\bigskip 

\section{DETAILS} 

\underline{Definition:} A function $F$ is {\it absolutely} $\Ubf{\Sigma}_1^{\rm HC}$ if: 

\leftskip 0.4in 

\noindent (a) there is $x\in 2^\N$ and $\psi$ in the language of set theory such 
that 
\[F(a)=b\] 
if and only if there is a countable transitive structure ${\cal M}$ 
containing $a, b, x$ and satisfying 
\[{\cal M}\models \psi(a, b, x);\] 

\noindent (b) (this part is more technical) the formulation of (a) continues to 
define a total function through all generic extension. 

\leftskip 0in 

\bigskip 

Following G\"{o}del's work we know that (a) alone is not sufficient to guarantee 
a function is {\it nicely behaved}. For instance a function satisfying (a) alone 
from $\R$ to $\R$ may fail to be Lebesgue measurable. 

It is a kind of folklore result that if we add (b) in addition then we do indeed 
obtain all the nice properties we could hope for -- such as being universally 
measurable.\footnote{See for instance 
$\S$ 9.1{\bf Classification and orbit equivalence 
relations}, cited above.} For various purposes these kinds of functions actually 
give a 
kind of better fit to the notion of {\it reasonable reduction} or {\it reasonable 
schema of classification}. 

\bigskip 



\underline{E.g.1} For $p$ a prime, TA$_p=\{H\in {\rm AbGrp}| H {\rm \: is \: 
a \: } p{\rm -group}\}$, there is an 
absolutely $\Ubf{\Sigma}_1^{\rm HC}$ 
\[U: {\rm TA}_p\rightarrow 2^{<\omega_1}\] 
such that 
\[H_1\cong H_2 \Leftrightarrow U(H_1)\cong U(H_2).\] 
This comes out of the Ulm classification of $p$-groups. 

\bigskip 

\underline{E.g.2} For any countable language ${\cal L}$, there is an 
absolutely $\Ubf{\Sigma}_1^{\rm HC}$ 
\[S: {\rm Mod}({\cal L})\rightarrow {\rm HC}\:\:{\rm (the \: hereditarily \: countable \: sets)}\] 
such that 
\[{\cal M}_1\cong {\cal M}_2\Leftrightarrow S({\cal M}_1)=S({\cal M}_2).\] 
(D. Scott) Moreover we can write 
\[{\rm HC}=\bigcup_{\alpha<\aleph_1}V_{\alpha}\cap {\rm HC}\] 
and think of $V_\alpha\cap {\rm HC}$ as being a subset of the $F_\alpha$-equivalence 
classes. 

\bigskip 

Here I would be more optimistic about a limited conjecture: 

\bigskip 

\underline{Conjecture:} ${\cong}|_{\rm TFA}$ is universal with respect to 
absolutely $\Ubf{\Sigma}_1^{\rm HC}$ functions. That is to say, for any countable 
language ${\cal L}$, we may reduce $\cong|_{{\rm Mod}({\cal L})}$ to 
$\cong|_{\rm TFA}$ by the use of an absolutely $\Ubf{\Sigma}_1^{\rm HC}$. 


\bigskip 


\section{SOMETHING ABOUT THE PROOFS} 

The argument that $F_2\leq_B \cong\:|_{\rm TFA}$ at least is very simple. Indeed somewhat 
misleadingly so. $F_2$-equivalence classes can be coded up in countable structures just 
using unary predicates, and the model theory without relations or functions is 
extremely simple. For $F_3$ and beyond relations are necessary, and the proof of 
$F_3\leq_B \cong|_{\rm TFA}$ is more involved that than the sketch below. 

I will also further simplify this sketch by skipping over any argument that the 
reduction is Borel. 

\bigskip 

\def\G{\cal G}

Let $(q_n)_{n\in\N}, (p_n)_{n\in\N}$ be sequences of distinct primes. For $x\in 2^{\N\times
\N}$, for which we can assume $(x(n, \cdot))_{n\in\N}$ is one to one, 
we define an abelian group ${\G}_x$ as follows: At each $\ell\in\N$ we set 
\[g_{x,\ell}\in{\G}_x\] 
so that $g_{x,\ell}$ is divisible by all powers of $p_n$ if $x(\ell, n)=1$ and divisible 
by all powers of $q_n$ if $x(\ell, n)=0$. We then let ${\G}_x$ be the abelian group generated 
by these $\{g_{x,\ell}: \ell\in\N\}$ and all the divisors we have just insisted on. 

The isomorphism type of ${\G}_x$ encodes 
$[x]_{F_2}\sim\{x(\ell, \cdot): \ell \in\N\}$. 
We can reconstruct the latter from the 
former. 

Here goes. 

Say that $g\in{\G}_x$ is {\it good} if for all $n$ either $g$ is divisible by all powers 
of $p_n$ or it is divisible by all powers of $q_n$. Then for $g\in{\G}_x$ good we let 
\[d(g)\in 2^\N\] 
be defined by 
\[(d(g))(n)=1\] 
if and only if $g$ is divisible by all powers of $p_n$. 

And then one indeed obtains 
\[\{x(\ell, \cdot)|\ell\in\N\}=\{d(g)| g\in {\G}_x{\rm \: good}\}.\] 
The other details are routine, and thus it is shown that 
\[F_2\leq_B \cong|_{\rm TFA}\] 

\bigskip 

Here we can get some insight by recalling a general fact (see the Friedman, Stanley paper 
cited above): 

\bigskip 


\underline{Fact:} There is no absolutely $\Ubf{\Sigma}_1^{\rm HC}$ function 
\[\theta: 2^{\N\times\N}\rightarrow 2^{<\omega_1}\] 
such that 
\[x F_2 y\Leftrightarrow \theta(x) =\theta(y).\] 

\bigskip 

In particular this gives a result first obtained by Garvin Melles using very 
different means: 

\bigskip 

\underline{Corollary:} (Melles) We cannot classify countable torsion free abelian 
groups by elements of $2^{<\omega_1}$ using 
absolutely $\Ubf{\Sigma}_1^{\rm HC}$ functions.\footnote{G. Melles, 
{\it One cannot show from ZFC that there is an Ulm-type classification of the 
countable torsion-free abelian groups,} {\bf Set theory of 
the continuum (Berkeley, CA, 1989)}, pp. 293--309, 
Mathematical Sciences Research Institute Publications, 26, Springer, New York, 1992.} 



















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