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\begin{document}

\title{Model Theory II Homework 8}
\author{Dakota Blair}
\maketitle

\setcounter{section}{16}
\begin{hw}
Calculate $I(\Tfcp, \aleph_1)$.
\end{hw}

\begin{proof}[Proof that $I(\Tfcp, \aleph_1) = \aleph_0$]
Any $\aleph_1$ model of $\Tfcp$ must have at least one infinite congruence class.
Further any infinite congruence class
must have cardinality either $\aleph_0$ or $\aleph_1$.
If there are no congruence classes of size $\aleph_1$
then there must be $\aleph_1$ congruence classes of cardinality $\aleph_0$.
This defines a single isomorphism class, say $A_0$, and is characterized by having no 
$\aleph_1$ congruence classses and $\aleph_1$ many $\aleph_0$ classes.

Let $m = \aleph_1$ or $m \in \omega + 1$ such that $m > 0$. 
Similarly let $n = \aleph_1$ or $n \in \omega + 1$ such that $n > 0$.
Note then there are $|\aleph_0 \cdot \aleph_0| = \aleph_0$ choices for $(m,n)$.
There is exactly one corresponding isomorphism class
corresponding to a model with exactly $n$ distinct $\aleph_0$ classes
and exactly $m$ distinct $\aleph_1$ classes.
Assume two models $\st{M}_1$ and $\st{M}_2$
correspond to the same pair $(m,n)$.
Then any partial elementary map $f$ preserving the finite equivalence classes
may be extended to an isomorphism.
Let the $\aleph_0$ and $\aleph_1$ congruence classes of $\st{M}_1$
be $\{a_i\}_{i \in {}^{1}I_0}$ and $\{a_i\}_{i \in {}^{1}I_1}$ respectively.
Similarly let the $\aleph_0$ and $\aleph_1$ congruence classes of $\st{M}_2$
be $\{b_i\}_{i \in {}^{2}I_0}$ and $\{b_i\}_{i \in {}^{2}I_1}$ respectively.
Define ${}^{1}I = {}^{1}I_0 \union {}^{1}I_1$ and
${}^{2}I = {}^{2}I_0 \union {}^{2}I_1$.
Further define $g_1 : \st{M}_1 \to {}^{1}I$ 
be the function which gives the index in ${}^{1}I$
of the congruence class of $\st{M}_1$ containing the given element.
Define $g_2 : \st{M}_2 \to {}^{2}I$ similarly.
Then $F : \st{M}_1 \to \st{M}_2$ will define an isomorphism
when the following condition is satisfied:
if $m \in \st{M}_1, \alpha \in \{0,1\}$ and $g_1(m) \in {}^{1}I_{\alpha}$
then $F(m) = f(m)$ or $g_2(F(m)) = f_\alpha(g_1(m))$.

Note that the isomorphism class of each such pair is distinct from $A_0$.
Further two models $\st{M}_1$ and $\st{M}_2$
such that their corresponding pairs are distinct,
that is $(m_1, n_1) \neq (m_2, n_2)$,
define two distinct isomorphism classes.
This is because any infinite congruence class must map
to a class of the same cardinality.
But if either $m_1 \neq m_2$ or $n_1 \neq_2$ then
$|{}^{1}I_0| \neq |{}^{2}I_0|$ or $|{}^{1}I_1| \neq |{}^{2}I_1|$
which prevents the construction of an elementary bijection
from $\st{M}_1$ to $\st{M}_2$.
Then each pair $(m,n)$ corresponds to a unique isomorphism class, say $A_{(m,n)}$.

Therefore $A_0$ and $A_{(m,n)}$ define $\aleph_0$ distinct
isomorphism classes.
These classes exhaust all $\aleph_1$ models of $\Tfcp$
as any other models would have a different cardinality.
Hence $I(\Tfcp, \aleph_1) = \aleph_0$.
\begin{comment}
%the isomorphism class of models with no $\aleph_1$ classes.
Then for any pair $(n,m) \in \omega^2$ where $m > 0$
there is exactly one corresponding isomorphism class
corresponding to a model with exactly $n$ distinct $\aleph_0$ classes
and exactly $m$ distinct $\aleph_1$ classes.
\end{comment}
\end{proof}
\end{document}
