in $L_{\gamma},\alpha$ has the tree property, i.e., if whenever $T\in
L_{\gamma}$, is a tree $\subseteq\alpha$, of height $\alpha$ and each level
has $A$-cardinality $<\alpha$, and the function $\beta\mapsto$ level of
$\beta$ in $T$, is in $A$ then $T$ has a branch of length $\alpha$ in $A$.

######  Remark 2.38.

We can replace $\mathbf{L}_{\beta}\models$ ZFC by $\mathbf{L}_{\beta}\models$
KP.

(B122) On Fleissner diamond, NDJFL, 22(1981), 29-35.

We prove that $ZFC+GCH+\diamondsuit^{+}_{\lambda}$ is consistent with (where
$1<\kappa<\lambda$, $\lambda$ regular $>\aleph_{0})$: there are
$S_{\alpha,\beta}\subseteq\alpha$ (for $\alpha<\lambda,\beta<\kappa$) such
that:

  * •

$\big{\\{}\\{S_{\alpha,\beta}:\beta<\kappa\\}:\alpha<\lambda\big{\\}}$ is a
diamond sequence, i.e., for every $A\subseteq X$ for stationarily many
$\alpha<\lambda$, $A\cap\alpha\in\\{S_{\alpha,\beta}:\beta<\kappa\\}$

  * •

for no function $f:\lambda\rightarrow\kappa$ is $\langle
S_{\alpha,f(\alpha)}:\alpha<\lambda\rangle$ a diamond sequence; moreover, even
$\big{\langle}\\{S_{\alpha,\beta}:\beta<\kappa\\}\setminus\\{S_{\alpha,f(\alpha)}:\alpha<\kappa\\}\big{\rangle}$
is not a diamond sequence.

(B123) Monadic theory of order and topology in ZFC, AML 23(1982), 179-198
(with Yuri Gurevich).

True first-order arithmetic is interpreted in the monadic theory of the real
line, in the monadic theory of any non-modest short chain, in the monadic
theory of Cantor Discontinuum, in the monadic theory of any non-modest
metrizable space. It was known that existence of such interpretations is
consistent with ZFC.

(B124) $\aleph_{\omega}$ may have a strong partition relation, IJM 38(1981),
283-288.

We prove the consistency of ZFC + GCH with the following (assuming con(ZFC) +
there are $\omega$ measurable cardinals):

  1. $(*)$

for any function $f$ from finite subsets of $\aleph_{\omega}$ to $\omega$,
there are pairwise disjoint $S_{n}\subseteq\aleph_{\omega},|S_{n}|=\aleph_{n}$
such that:
if $u,v$ are finite subsets of $\aleph_{\omega}$ and $(\forall
n)\big{[}|S_{n}\cap u|=|S_{n}\cap v|\leq n\big{]}$ then $f(u)=f(v)$.

Note that $(*)$ implies: every Banach space of power $\geq\aleph_{\omega}$
contains an unconditional (infinite) basis.

(B125) The consistency of ${\rm Ext}(G,\mathbb{Z})=\mathbb{Q}$, IJM 39(1981),
82.

For abelian groups, if $\mathbf{V}=\mathbf{L}$, ${\rm Ext}(G,\mathbb{Z})$
cannot have cardinality $\aleph_{0}$. We show that GCH does not imply this
(hence some complete $\mathbb{L}_{\infty,\aleph_{1}}$ sentence has exactly
$\aleph_{0}$ non-isomorphic models of power $\aleph_{1}$, but not proved
here). (Compare [Sh:91]).

The proof continues in [She77].

(127) On Boolean Algebras with no uncountably many pairwise comparable or
uncomparable elements, NDJFL 22(1981), 301-308.

We shall prove, assuming CH, the existency of a Boolean algebra of power
$\aleph_{1}$, having neither uncountable set of pairwise comparable elements
nor uncountable set of pairwise incomparable elements.

We finish commenting on generalizations to higher cardinals.

(B128) Uncountable constructions, IJM 51(1985), 273-297.

We suggest a principle which unites many proofs from
$\diamondsuit_{\aleph_{1}}$, and is similar to forcing and deduce from it
several new results. The paper is supposed to help non-set theorists to build
counterexamples by reducing the problems to questions on finitely generated
objects: (assuming $\diamondsuit_{\aleph_{1}}$).

Boolean Algebras:

1) For pedagogical reasons we build a Rubin Boolean Algebra.

2) There is a Boolean Algebra which is not 1-Rubin but among any $\aleph_{1}$
elements there are two comparable ones (even $a,b,c$ with $a\cap b=c$).

E.C. Groups: For any countable e.c.g. and uncountable $K$, there is an
uncountable $H\equiv{}_{\infty,\omega}G$, such that $K$ cannot be embeddable
into $H$.

Banach spaces: There is a non-separable Banach space $B$ so that for any non-
separable subspace $B_{1}$, $B/B_{1}$ is separable.

(B129) The number of non-isomorphic models of cardinality $\lambda$,
$\mathbb{L}_{\infty,\lambda}$ equivalent to a fixed model, NDJFL, 22(1981),
5-10.

We show that if $\mathbf{V}=\mathbf{L}$ and $\lambda$ is regular but not
weakly compact, then if $M$ is a model of cardinality $\lambda$ then
$\\{N/{\cong}:N\equiv_{\infty,\lambda}M,\ \|N\|=\lambda\\}$ has either one
number or $2^{\lambda}$ members.

We do not rely on the full power of $\mathbf{V}=\mathbf{L}$ but use a
combinatorial consequence for rgular non-weakly compact. The use of some set
theoretical assumption stronger then GCH is necessary as in [She81d] we
construct a model of ZFC + GCH where a sentence
$\psi\in\mathbb{L}_{\infty,\omega_{1}}$ has $\aleph_{0}$ non-isomorphic models
of cardinality $\aleph_{1}$. Also the restriction on $\lambda$ to be not
weakly compact is not a coincidence; see [She82].

(B130) Stability over a predicate, NDJFL, 16(1985), 361-376 (with A. Pillay).

This is a beginning of a theory suppose to measure how much $M\restriction P$
determines $M$, for this we look at $\\{M:M\restriction P=N\\}$ and count the
number of isomorphism types (over $N$). In the cases the number is not big, we
have parallels to some of the theorems on stable theories.

(B133) On the number of non-isomorphic models in $\mathbb{L}_{\infty,\kappa}$
when $\kappa$ is weakly
compact, NDJFL, 23(1982) 21-26.

We construct for $\kappa$ weakly compact a model $M$ of cardinality $\kappa$
which has exactly $\theta$ non-isomorphic models $M\equiv_{\infty,\kappa}M$ of
cardinality $\kappa$, for any $\theta\leq\kappa$ we want. This together with
[She81c] completes the answer of this problem when $\mathbf{V}=\mathbf{L}$.

(B135) Rigid homogeneous chains, Math. Proc. of Cambridge Philo. Soc.
87(1981), 7-17 (with A.M.W. Glass, Yuri Gurevich and W.C. Holland).

A chain $C$ is _rigidly homogeneous_ if for any two points $a,b$ of $C$ there
is a unique automorphism $f$ of $C$ such that $f(a)=b$. The group of
automorphisms of a rigidly homogeneous chain $C$ is isomorphic to a subgroup
of the additive group of reals. We classify rigidly homogeneous chains by
elementary properties of their automorphism groups.

The chain of integers is the only countable rigidly homogeneous chain. The
cardinality of every rigidly homogeneous chain is $\leq 2^{\aleph_{0}}$.
Ohkuma constructed rigidly homogeneous chains of cardinality $2^{\aleph_{0}}$.
We prove consistency of existence as well as consistency of non-existence of
rigidly homogeneous chains of uncountable cardinality $<2^{\aleph_{0}}$.

(B136) Construction of many complicated uncountable structures and Boolean
Algebras, IJM, 45(1983), 100-146.

This paper has three aims:

1) To make the results of [She78, Ch.VIII] on constructing models more
available for applications, by separating the combinatorial parts: for this we
have to set up a suitable framework. We believe it is useful for proving
things like: there are many non-isomorphic structures, there are many
structures no one embeddable into the other, there are rigid structures or
undecomposable ones or structures with few endomorphisms, etc.

There are few kinds of structures (mainly trees) which served as index sets,
and we have many complicated such structures. In an application we have in
some sense to interpret those structures in the one we are interested in.

2) A second aim is to strengthen the results of [She78, Ch.VIII]. In
particular we mainly are interested there in showing that there are many non-
isomorphic models of an unsuperstable theory. We got many times results on the
number of models not elementarily embeddable in each other, being a side
benefit. Here we consider the later case in more detail, and in a somewhat
stronger combinatorial results needed later.

We also consider some more kinds of index structures (mainly a tree of pairs).

3) We apply the previous results to problems on Boolean Algebras from the list
of van Douwen, Monk, and Rubin.

In particular

  1. (a)

if $\lambda=\lambda^{\aleph_{0}}$ there is a complete Boolean Algebra of power
$\lambda$ satisfying the c.c.c. with no non-trivial one-to-one endomorphism

  2. (b)

if we wave “complete” then is such $B$ of any power $\lambda>\aleph_{0}$

  3. (c)

in any $\lambda>\aleph_{0}$ there is a Bonnet-rigid Boolean Algebra.

4) To solve some problems from that list, unconnected to 1) and 2):

  1. (a)

if a Boolean Algebra $B$ has no dense subset of power $<\lambda$ then it has a
$\lambda$ pairwise incomparable element

  2. (b)

$(\diamondsuit_{\aleph_{1}})$ there is a Boolean Algebra of power $\aleph_{1}$
satisfying many rigidity conditions: Bonnet rigid endo-rigid indecomposable.

The idea of this proof is to generalize Keisler criterion for omitting types
(specific to atomless Boolean Algebras) to the case an ultrafilter is given in
each countable stage of the construction. l Continued in [Shear].

(B137) The Singular cardinals problem: independence results, London Math. Soc.
Lecture Note Ser. 87(1983) 116-134.

Assuming the consistency of a supercompact cardinal, we prove the consistency
of:

  * •

$\aleph_{\omega}$ strong limit $2^{\aleph_{\omega}}=\aleph_{\alpha+1}$,
$\alpha<\omega_{1}$ arbitrary;

  * •

$\aleph_{\omega_{1}}$ strong limit,
$2^{\aleph_{\omega_{1}}}=\aleph_{\alpha+1}$, $\alpha<\omega_{2}$ arbitrary;

  * •

$\aleph_{\delta}$ strong limit, ${\rm cf}(\delta)=\aleph_{0}$,
$2^{\aleph_{\delta}}$ arbitrarily large before the first inaccessible
cardinal; for $\aleph_{\delta}$ “large enough”.

We start with $\kappa$ supercompact, e.g. an $\omega$-sequence
$\langle\kappa_{n}:n<\omega\rangle$ through it and collapse to make $\kappa$
to $\aleph_{\omega}$. The new point is that we change our mind on what is
$\mathbf{V}$ — it becomes larger when we determine $\kappa_{n}$ for bigger
$n$.

(B138) On the structure of ${\rm Ext}(A,\mathbb{Z})$ in ZFC, JSL, 50(1985),
302-315 (with G. Sageev).

We prove:

  * •

(GCH) for any $\aleph_{0}<\kappa<\aleph_{\omega}$, and cardinals
$\nu_{p}\leq\kappa^{+}$ (for $0<p<\omega$ prime) there is a $\kappa$-free
abelian group of power $\kappa$, $\nu_{0}(G)=\kappa^{+}$, $\nu_{p}(G)=\nu_{p}$
(see [HHS78]).

(B139) On the number of non-conjugate subgroups, Algebra Universalis,
16(1983), 131-146.

Let $G$ a given group of cardinality $\lambda\,(\lambda>\aleph_{0})$ and
$\mathrm{nc}(G)$ the number of non-conjugate subgroups of $G$, so we are
interested in lower bound of the number of equivalence relations.

We prove for many uncountable cardinals $\lambda$ that if $|G|=\lambda$ then
$\mathrm{nc}(G)\geq\lambda$, (e.g. when GCH holds).

This is proved for $\aleph_{0}<\lambda<2^{\aleph_{0}}$ (for
$\lambda=\aleph_{1}$ assuming $2^{\aleph_{0}}=\aleph_{1}$ by [She80] this is
best possible); for $\lambda$ such that $\forall\mu[\lambda\neq 2^{\mu}]$. For
$\lambda=2^{\mu}$ and $\mu$ not regular non-strong limit and
$\mathrm{Ded}(\mu)>\lambda$. It is an open problem if the theorem is true for
other cardinals. Continued in [She87b].

(B140) On endo-rigid $\aleph_{1}$-strongly free abelian groups of power
$\aleph_{1}$, IJM, 40(1981), 291-295.

We prove that if $2^{\aleph_{0}}<2^{\aleph_{1}}$ then there is an abelian
group of power $\aleph_{1}$, which is strongly $\aleph_{1}$-free but endo-
rigid where:

  * •

$G$ is endo-rigid iff every homomorphism $h:G\rightarrow G$ has the form
$h(x)=n(x)$ for every $x$; (for some integer $n$).

(B141) The monadic theory of $\omega_{2}$, JSL 48(1983), 387-398 (with Y.
Gurevich and M. Magidor).

Assume consistency of ZFC plus existence of a weakly compact cardinal. Then

  1. (i)

for every $S\subseteq\omega$, ZFC plus “$S$ and the monadic theory of
$\omega_{2}$ are recursive in each other” is consistent, and

  2. (ii)

ZFC plus “the full second-order theory of $\omega_{2}$ is interpretable in the
monadic theory of $\omega_{2}$” is consistent.

(B142) The structure of saturated free algebras, Algebra Universalis,
17(1983), 191-199 (with J. Baldwin).

Let $V$ be a variety with countable similarity type. Suppose $M$ is an
uncountable algebra in $V$ which is both free and saturated.

######  Lemma 2.39.

${\rm Th}(M)$ is $\aleph_{0}$-stable.

Theorem 142.1. There exists a finite set $q_{1},q_{2},\dotsc,q_{\kappa}$ of
types (over finite sets) each with weight one such that there exist infinite
sets of indiscernibles $Y_{1},\dotsc,Y_{n}$ each $Y_{i}$ based on some $q_{i}$
such that $M$ is generated as an algebra by $Y\cup\ldots\cup Y_{n}$.

(B143) The monadic theory and the “next world”, IJM 49(1984), 55-68 (with Yuri
Gurevich).

Let $R$ be the real line in a model $\mathbf{V}$ of ZFC and $B$ be the Boolean
Algebra of regular open subsets of $R$ in $\mathbf{V}$. Let $V^{B}$ be the
corresponding Boolean valued model of ZFC.

Theorem 143.1. There is an algorithm interpreting the full second-order
$\mathbf{V}^{B}$-theory of $\aleph_{0}$ in the monadic $\mathbf{V}$-theory of
$R$.

Theorem 143.2. There is an algorithm interpreting the full second-order
$\mathbf{V}^{B}$-theory of $2^{\aleph_{0}}$ in the monadic $\mathbf{V}$-theory
of $R$ if $\mathbf{V}$ satisfies the Continuum Hypothesis.

Theorems 1 and 2 are results of applications of a general interpretation
theorem.

Continued in [She88].

(B217) There are Noetherian domains in every cardinality with free additive
groups, Abstracts Amer. Math. Soc. 7(1986), 369 (with G. Sageev).

(Based on Notices AMS 86T-03-269)

######  Theorem 2.40.

There are Noetherian rings (in fact domains) with a free additive group, in
every infinite cardinality.

######  Remark 2.41.

1) For $\aleph_{1}$ this was proved by O’Neill.

2) The work was done in Sept., ’83.

3) We thank Fuchs for suggesting to us the problem.

###### Proof.

Sketch of Proof

Let $\mathbb{Z}$ be the ring of integers, $X$ a set of distinct variables,
$\mathbb{Z}[X]$ the ring of polynomials over $\mathbb{Z}$, $\mathbb{Z}(X)$ its
field of quotients, and $R_{X}$ the additive subgroup of $\mathbb{Z}(X)$
generated by

$\big{\\{}p/q:p,q\in\mathbb{Z}[X],\ p\text{ not divisible (non-trivially) by
any integer}\big{\\}}\subseteq\mathbb{Z}(X).$

It is known that $R_{X}$ is a Noetherian domain. Let, for a ring $R$, $R^{+}$
be its additive group. For $Y\subseteq X$ we can define $\mathbb{Z}[Y]$,
$\mathbb{Z}(Y)$, $R_{Y}$ similarly. ∎

######  Lemma 2.42.

1) $R^{+}_{X}$ is a free abelian group.

2) If $n\geq 0$, $Y\subseteq X$, $x(1),\dotsc,x(n)\in X\setminus Y$ pairwise
distinct, $W=\\{x(1),\dotsc,x(n)\\}$, $W(\ell)=W-\\{x(\ell)\\}$ then
$R^{+}_{W\cup Y}\Big{/}\sum\limits^{n}_{\ell=1}R^{+}_{W(\ell)\cup Y}$ is a
free abelian group.

###### Proof.

1) Follows by 2) for $n=0,Y=X$.

2) This is phrased because it is the natural way to prove 1) by induction on
$|Y|$, for all $n$ simultaneously (a degenerated case of [She83a]). If
$|Y|>\aleph_{0}$, let $Y=\\{y(\alpha):\alpha<\lambda\\}$ with no repetitions,
so $\lambda=|Y|$, $Y_{\alpha}=\\{y(i):i<\alpha\\}$. It suffices for each
$\alpha<\lambda$ to prove that
$G_{\alpha}=:\big{(}R^{+}_{Y_{\alpha+1}}+\sum\limits^{n}_{\ell=1}R^{+}_{Y\cup
W(\ell)}\big{)}\Big{/}\big{(}\sum\limits^{n}_{\ell=1}R^{+}_{Y_{\alpha+1}\cup
W(\ell)}+R^{+}_{Y_{\alpha}\cup W}\big{)}$ is free. We now show that
$G_{\alpha}$ is isomorphic to
$G^{\prime}_{\alpha}=R^{+}_{Y_{\alpha+1}}\Big{/}\big{(}\sum\limits^{n}_{\ell=1}R^{+}_{Y_{\alpha+1}\cup
W(\ell)}+R^{+}_{Y_{\alpha}\cup W}\big{)}$. For this it is enough to show
$\bigl{(}\sum\limits^{n}_{\ell=1}R^{+}_{Y_{\alpha}\cup W(\ell)}\bigr{)}\cap
R^{+}_{Y_{\alpha+1}}=\sum\limits^{n}_{\ell=1}R^{+}_{Y_{\alpha+1}\cup
W(\ell)}$, as the right side is included in the left side trivially we have to
show

$\sum^{n}_{\ell=1}{\frac{p_{\ell}}{q_{\ell}}}\in\sum\limits^{n}_{\ell=1}R^{+}_{Y_{\alpha+1}\cup
W(\ell)}$

if ${\frac{p_{\ell}}{q_{\ell}}}\in R^{+}_{Y_{1}\cup W(\ell)}$ and
$\sum\limits^{n}_{\ell=1}{\frac{p_{\ell}}{q_{\ell}}}\in
R^{+}_{Y_{\alpha+1}\cup W}$ which is easy by projections). But
$G^{\prime}_{\alpha}$ is free by induction hypothesis. ∎

The next claim completes the case “$y$ countable”.

######  Claim 2.43.

If $Y\cup\\{x(1),\dotsc,x(n)\\}\subseteq X$, $x(\ell)\in X\setminus Y$
distinct, $G=R^{+}_{W\cup Y}$, $I=\sum\limits_{i}I_{i}$,
$I_{i}=R^{+}_{W(i)\cup Y}$ then $G/I$ is free, when $Y$ is countable.

###### Proof.

It suffices to prove:

  1. (a)

$G/I$ is torsion free

  2. (b)

if $a_{1},\dotsc,a_{k}\in G/I$ are independent, then

$\Big{\\{}m\in\mathbb{Z}^{+}:\text{ there are }\langle
q_{1},\dotsc,q_{k}\rangle\in L\text{ such that }m\text{
divides}\sum\limits_{i=1}^{k}q_{i}a_{i}\text{ in }G/I\Big{\\}}$

is finite, where

$L=\big{\\{}\langle q_{1},\dotsc,q_{k}\rangle:q_{i}\in\mathbb{Z}\text{, not
all zero with no common divisor}\big{\\}}.$

Let $x_{1}(q)\in X$ for $q=1,\dotsc,n$ be new distinct variable and let
$V=\\{x_{1}(1),\dotsc,x_{1}(n)\\}$. For $u\subseteq\\{1,\dotsc,n\\}$ let us
define $h_{u}:R_{V\cup W\cup Y}\rightarrow R_{V\cup Y}$ an isomorphism
$h_{u}(y)=y$ for $y\in Y$, $h_{u}(x(q))=x_{1}(q)$ if $q\in u$,
$h_{u}(x(q))=x(q)$ if $q\notin u$. So let $a_{1}+I,\dotsc,a_{k}+I$ be
independent.

Suppose $\langle q_{1},\dotsc,q_{k}\rangle\in L$,
$m_{0}m_{1}\in\mathbb{Z}\setminus\\{0\\}$, $m_{0}m_{1}$ divides
$\sum\limits_{i}m_{0}q_{i}a_{i}+I$. So for some $s\in R_{W\cup Y}$ and
$p_{\ell}\in I_{\ell}$ for $\ell=1,\dotsc,n$ we have:
$\sum\limits_{i}m_{0}q_{i}a_{i}=m_{0}m_{1}s+\sum\limits_{\ell=1,\dotsc,n}p_{\ell}$.

Let $u$ vary on subsets of $\\{1,\dotsc,n\\}$,
$b_{u}=\sum\limits_{u}(-1){}^{|u|}h_{u}(a_{\ell})\in R_{V\cup W\cup Y}$, so

$\sum\limits_{i}m_{0}q_{i}b_{i}=\sum\limits_{u}\sum_{i}m_{0}q_{i}(h_{u}(a_{i}))=m_{0}m_{1}\sum\limits_{u}h_{u}(s)+\sum\limits_{\ell=1,\dotsc,n}\sum\limits_{u}h_{u}(p_{\ell}).$

However for each $\ell=1,\dotsc,n$ we have $\sum\limits_{u}h_{u}(p_{\ell})$ is
zero (as $x(\ell)$ does not appear in it).

So $\sum\limits_{i}m_{0}q_{i}b_{i}$ is divisible by $m_{0}m_{0}$ in
$R^{+}_{V\cup W\cup Y}$. As $R^{+}_{V\cup W\cup Y}$ is free, it suffices to
prove $\\{b_{i}:i=1,\dotsc,k\\}$ is independent, equivalently they are
linearly independent (over the rationals) in $Z(Y\cup W\cup V)$. But, if not,
we can substitute suitable numbers for $x_{1}(1),\dotsc,x_{1}(n)$ and get
contradiction to “$\\{a_{i}+I:i=1,\dotsc,n\\}$ is independent.”

That is let $R^{\prime}$ be a subring of $R_{V\cup W\cup Y}$ generated by
$\mathbb{Z}[X]\cup\\{\frac{1}{q_{1}},\dotsc,\frac{1}{q_{m}}\\}$ for some
$m,q,\dotsc,q_{\ell}\in\mathbb{Z}[X]$ such that $h_{u}(a_{i})\in R^{\prime}$.
Let $g$ be a homomorphism from $R^{\prime}$ to $R_{W\cup Y}$ which is the
identity on $R_{W\cup Y}$ and maps each $x_{1}(q)$ to an integer (so we
require from $\big{\langle}g(x_{i}(q)):q=1,\dotsc,n\big{\rangle}$ to make some
finitely many polynomials over the integers nonzero which is possible). Now
$\ell\in u\subseteq\\{1,\dotsc,n\\}\Rightarrow h_{u}(a_{i})\in I_{\ell}$. So
it is enough to show that $\langle g(b_{i}):i=1,\dotsc,k\rangle$ is linearly
independent. But $g(b_{i})=\sum\limits_{u}g(h_{u}(a_{i}))\in
gh_{\varnothing}(a_{i})+I=g(a_{i})+I=a_{i}+I$. ∎

(C1) On Shelah compactness of cardinal, IJM, 31(1978), pgs.34-56 and pg.394
(with S. Ben-David)

We deal with the compactness property of cardinals presented by Shelah, who
proved a compactness theorem for singular cardinals. We improve that result in
eliminating axiom I there and show a new application of that theorem together
with a straightforward proof of it for the special case discussed. We discuss
compactness for regular cardinals and show some independence results: one of
them, a part of which is due to A. Litman, is the independence from ZFC + GCH
of the gap—one two cardinal problem for singular cardinals.

(C2) The Shelah $P$-point independence results, IJM, 43(1982), 28-48. [C2],
(E. Wimmers).

Present the proof of some forcing extension there is no $P$-point. We start
with
$\mathbf{V}\models``2^{\aleph_{0}}=\aleph_{1}+2^{\aleph_{1}}=\aleph^{+}_{2}$.”
Consider a list $\langle D_{\alpha}:\alpha<\aleph_{2}\rangle$ of the filter
$D$ on $\mathbb{N}$ such that ${\mathcal{P}}(\mathbb{N})/D$ satisfies the
c.c.c. (and every co-finite subset of $\mathbb{N}$ belongs). We then force a
close enough subset of the CS product of
$\prod\limits_{\alpha}\mathbb{Q}_{D_{\alpha}}$ for appropriate
$\mathbb{Q}_{D_{\alpha}}$.

(C3) The uniformization property for $\aleph_{2}$, IJM, 36(1980), 248-256.
[C3], (C. Steinhorn and J. King).

We deal with the following problem: let $S^{2}_{1}=\\{\delta<\aleph_{2}:{\rm
cf}(\delta)=\aleph_{1}\\}$, $\eta_{\delta}$ an $\omega_{1}$-sequence
converging to $\delta$, $\Phi=\langle\eta_{\delta}:\eta_{\delta}\in
S^{2}_{1}\rangle$. Assuming GCH, can $(\Phi,\aleph_{1})$ have the
uniformization property, at least for some $\Phi$ (see [She77])? The answer is
positive, so not only GCH $\not\rightarrow\diamondsuit_{S^{2}_{1}}$, but also
GCH is consistent with $S^{2}_{1}$ being small.

We start with $V\models GCH$ and use iteration of length $\omega_{3}$, with
support of power $\leq\aleph_{1}$, of the obvious forcings introducing a
uniformizing functions each condition being an initial segment. The proof is
like [She77] taking $N\prec({\mathscr{H}}(\aleph_{1}),E)$, closed under
countable sequences. But trying to build trees as there we run into two
problems:

  1. (a)

even if we get a tree of candidates for being a conditions, we cannot find
$\aleph_{1}$-closed $N_{1}$ with $N_{1}\cap\omega_{2}\notin S^{2}_{1}$ (if we
replace $S^{2}_{1}$ by $S\subseteq S^{2}_{1}$, $S^{2}_{1}\setminus S$
stationary, our life would be much easier).

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