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In NOTE 49:
The definition of Tr should be
\smallskip\noindent
$\forall x\exists u\exists f$ such that $u$ is transitive and $f$
is a function from $x$ one to one and onto $u$.
\smallskip
Delete ``(form 174)". (The statement TR$''$ is not form 174.)
\medskip
In NOTE 94:
\smallskip
The relationship between $\Delta_3$-finite and III-finite was
discovered by Omar de la Cruz, neither implies the other. In the
model $\Cal N3$, $A$, the set of atoms, is not $\Delta_3$-finite, but
since $\Cal P(A)$ is Dedekind finite, $A$ is III-finite. In $\Cal N1$,
let $X$ be the set of all finite subsets of $A$. $X$ is not III-finite
because $\Cal P(X)$ is Dedekind infinite. However, $X$ is
$\Delta_3$-finite because it has no infinite linearly ordered set.\newline
\indent
Suppose $X$ has an infinite linearly ordered set $L$ and suppose $E$ is
a support for $L$ and its linear ordering $R$. There must be an
$n\in\omega$ such that $L' = \{x\in L: |x| = n\}$ is infinite.
Otherwise, we can well order L by
$$x < y \text{ iff } |x| < |y|, \text{ or } |x| = |y| \text{ and } xRy.$$
Since $E$ has only finitely many subsets there are elements $u$ and $v$
in $L'$ such that $u\cap E = v\cap E$.
It is easy to see that there is a permutation $\sigma\in G$ such
that $\sigma(u) = v$ and $\sigma(v) = u$. This contradicts the
assumption that $E$ is a support of the linear ordering $R$.
\bye