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\centerline{Additions to Part III: Models}
\smallskip
\item{I.} In $\Cal M40(\kappa)$ replace the part following:
\smallskip
Since 43 + 14 is equivalent to 345 ([345 A]),
form 345 is true.
\smallskip
with
\smallskip
Since 8 is true and 384 + 8 implies 1, form 384 is
false.
\smallskip\noindent
$\Cal M40(\kappa)\models$ 14, 30, 43, 60,
87($\alpha$) for $\aleph_{\alpha}<\kappa$, 165, 295, and 345, but
51, 91, 144, 152, 163, 253, 286, and 384 are false. References
\ac{Pincus} \cite{1977a}, \cite{1977b}, \cite{1997}, \ac{Brunner}
\cite{1982a}, \ac{Herrlich/Steprans} \cite{1997}, \ac{Morris} \cite{1969},
notes 18, 120(26, 28, 34, 43, 45, and 59) and 121.
\smallskip
\item{II.} A new model
\smallskip\noindent
$\Cal N56$: Howard's model III: Assume the the atoms are indexed as
follows:
$A = \{a(i,j) : i\in{\Bbb Q} \hbox{ and } j\in\omega \}$. For each
$i\in \Bbb Q$, let $A_i = \{a(i,j) : j\in \omega\}$. (Call $A_i$ the
$i$th block.) Let
$$
\aligned
\Cal G = \{ \phi:A\to A : &(\forall i\in\Bbb Q)(\exists i'\in \Bbb Q)
(\phi(A_i) = A_{i'})\hbox{ and }\\
&\{i\in\Bbb Q : (\exists j)(\phi(a(i,j))\ne a(i,j))
\}\hbox{ is bounded}\}.
\endaligned
$$
\noindent Here bounded means bounded in the usual ordering on $\Bbb Q$.
($\Cal G$ is the set of all permutations of $\phi$ of $A$ such that
$\phi$ of any block is a block and the set of blocks on which $\phi$ is
not the identity is bounded.)
$S$ is the set of subsets of $A$ of the form $\bigcup_{i\in E} A_i$
where E is a bounded subset of $\Bbb Q$. Form 40 ($C(WO,\infty)$) is
true using a proof similar to the proof of 40 in $\Cal N33$ given in
\ac{Howard/Rubin/Rubin} \cite{1973}. By note 104, form 3 ($2m=m$) is
false. Form 62 ($C(\infty,<\aleph_0)$) is false since the family of
finite, non-empty subsets of $A$ has no choice function. Form 67 (MC)
is false since the set of blocks has no multiple choice function.
\smallskip\noindent
$\Cal N56\models$ 6, 37, 40, 91, 92, 130, 182, 189, 190, 191, 273,
305, 309, 313, 361, 363, 368, and 369, but 3, 15, 62, 67, 152,
and 192 are false. References \ac{Howard} \cite{1990},
\ac{Howard/Rubin/Rubin} \cite{1973}, and notes 18, 120(46).
\smallskip
\item{III.} Add to toc\_mod.tex
\smallskip
\head 55. $\Cal N55$: Keremedis/Tachtsis Model.\endhead
\head (6, 9, 37, 91, 92, 130, 182, 189, 190, 191, 273, 305, 309,
313, 361, 363, 368, and 369 are true but, 15, 116, 131, 154,
165, and 343 are false.)\endhead
\smallskip
\head 56. $\Cal N56$: Howard's Model III.\endhead
\head (6, 37, 40, 91, 92, 130, 182, 189, 190, 191, 273,
305, 309, 313, 361, 363, 368, and 369, but 3, 15, 62, 67, 152,
and 192 are false.) \endhead
\smallskip
\item{IV.} In $\Cal M2$, in the beginning of the description change
``the Boolean Prime Ideal Theorem (14) is false.'' to
``the countable ultrafilter theorem (385) is false.''
\smallskip
At the end of $\Cal M2$, add 385 to the list of theorems that
are false.
\smallskip
\item{VI.} In $\Cal M1$, right after ``so 277 is true.'' insert the
sentence:
\smallskip
Since $\Bbb R$ cannot be well ordered in this model and form 277 is
true, it follows that 369 ($\Bbb R$ is not decomposable) is false.
\smallskip
Also, add 369 to the list of froms that are false.
\smallskip
\item{VI.} Change the last part of N2 to:
\smallskip
\ac{Keremedis} \cite{1999} has shown that 216 (Every
infinite tree has either an infinite chain or an infinite antichain.)
and 388 ( Every infinite branching poset (a partially
ordered set in which each element has at least two lower bounds) has
either an infinite chain or an infinite antichain.) are both false
in $\Cal N2$.
\smallskip\noindent
$\Cal N2\models$ 6, 37, 65, 67, 83, 99, 130, 163,
191, 273, 305, 313, 334, 361, and 363, but 18, 45($n$), 80, 98,
128, 154, 163, 164, 198, 216, 344, 358, and 388 are false. References
\ac{Brunner} \cite{1982a}, \cite{1983c}, \cite{1984b}, \cite{1984f},
\ac{Fraenkel} \cite{1922}, \ac{Hickman} \cite{1978b}, \ac{Howard}
\cite{1984b}, \ac{Jech} \cite{1973b}, \ac{Keremedis} \cite{1999},
\ac{Pincus} \cite{1972a},\cite{1972b}, notes 18, 93, 95, 105, and
120(2 and 56).
\smallskip
\item{V.} Change the first part of the description of $\Cal M7$ to
the following:
\smallskip\noindent
$\Cal M7$: Cohen's Second Model. There are two denumerable
subsets $U=\{U_i:i\in\omega\}$ and $V=\{V_i:i\in\omega\}$ of
$\Cal P({\Bbb R})$ (neither of which is in the model) such that for
each $i\in\omega$, $U_i$ and $V_i$ cannot be distinguished in the
model. Therefore, in this model there is a denumerable set of pairs
of subsets of $\Bbb R$ whose union has no denumerable subset, so
form 18 is false, and $C(\aleph_0,2,\Cal P({\Bbb R}))$ (389) is
also false.
\smallskip
Also in $\Cal M7$, delete 358 from the list of forms that are false and
add 389 to this list.
\smallskip
\item{VI.} Add as the last part of the descriptions of $\Cal M6$:
\smallskip
Sageev
proves that alephs are preserved in this model so, since form 170
($\aleph_1 \leq 2^{\aleph_0}$ is true in the ground model, it is true
in $\Cal M6$. It also follows that 34 ($\aleph_1$ is regular) is true.
(See \ac{Howard/Keremedis/Rubin/Stanley/Tachtsis} \cite{1999} Lemma 5.)
\smallskip
Also, add 34 and 170 to the list of forms that are true in this model and
add \ac{Howard/Keremedis/Rubin/Stanley/Tachtsis} \cite{1999} to the list
of references.
\smallskip
\item{VII.} In $\Cal M12(\aleph)$ after ``(169 is false).'', insert
the following:
\smallskip
It follows from [0 AB] that every perfect subset of $\Bbb R$
has cardinality $2^{\aleph_0}$. Thus, since 169 is false, form 369
(If $\Bbb R$ is partitioned into two sets, at least one of them has
cardinality $2^{\aleph_0}$.) is true.
\smallskip
Add 369 to the list of forms that are true in this model.
\smallskip
\item{VIII.} At the end of $Cal N1$, add:
\smallskip
Form 390 is false because the set of atoms can
neither be partitioned into two infinite sets nor can it be partitioned
into infinitely many sets, each of which has at least two elements.
\smallskip
Add form 390 to the list of forms that is false in $\Cal N1$.
\bye