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\begin{document}

\title{Model Theory II Homework 10}
\author{Dakota Blair}
\maketitle

\setcounter{section}{19}

\begin{hw}
Every model of a countable $\aleph_1$-categorical theory is $\omega$-homogeneous.
\end{hw}

\begin{proof}
By the Baldwin-Lachalan theorem $T$ is $\omega$-stable.
Let $\tuple{a}, \tuple{b}, a_0 \in \st{M}$ such that $\tuple{a} \equiv_M \tuple{b}$.
If $\st{M}$ is $\omega$-saturated
then $\tp(a_0/\tuple{a})$ is realized in $\st{M}$ by some element $b_0$.
Thus $\tuple{a}a_0 \equiv \tuple{b}b_0$.
If $\st{M}$ is uncountable then it is $\omega$-saturated by categoricity.
Therefore the only case remaining is when
$\st{M}$ is countable and not $\omega$-saturated.
In this case, $p = \tp(a_0/\tuple{a})$ is isolated.
If $p$ is not isolated then there exists a $\phi$ such that
$p \union \{\phi\}$ and $\{\symnot\phi\}$
are distinct, consistent and not isolated.
Then a binary tree argument shows that there are uncountably many types
over countably many parameters, which contradicts $\omega$-stability of $T$.
Therefore $\st{M},\tuple{a},a_0)$
is a constructible model and therefore $\omega$-homogeneous.
\end{proof}

\begin{hw}
Show there is a natural injection of $M$ into $S_1(B)$ for $M \subset B$.
Identify $M$ with its image and let $\obar{M}$ be its, topological,
closure in $S_1(B)$.
Show that $q \in S_1(B)$ is in $\obar{M}$
if and only if
$q$ is a coheir of $q \restrict M$.
\end{hw}

\begin{proof}
Let $f:M \to S_1(B)$ be defined as $f(m) = \tp(m/B)$.
Note that this map is well defined as $M \subset B \subset \monster$.
Note that if $m_1, m_2 \in M$ are distinct
then $q_1 = \tp(m_1/B) \neq q_2 = \tp(m_2/B)$
since the formula $x = m_1$ is in $q_1$
but $x \neq m_1$ is in $q_2$.
Therefore $f$ is injective.

Let $\obar{M} = \obar{f(M)}$ in $S_1(B)$ and $q \in \obar{M}$.
To show that $q$ is a coheir of $q \restrict M$
we need to see that it is finitely realized in $M$,
that is for each formula such that
$q \models \varphi(x,\tuple{b})$
there is an $m \in M$ such that
$\monster \models \varphi(m,\tuple{b})$.
Let $q \models \forall x \varphi(x, \tuple{b})$,
we need to show $\exists m \in M$ such that $\varphi(m,n) \in q$.
Since $q \models \forall x \varphi(x, \tuple{b})$
there exists a $b \in B$ such that
$U = \stonenabe{\varphi(b, \tuple{b})} \subset q$.
Since $U$ is an open set and $q \in \obar{M}$
we have that $U \intersect M$ is nonempty.
Thus there is an $m \in M$ such that
$\tp(m/B) = f(m) \subset q$.
Therefore $\varphi(m,n) \in q$.
Hence $q$ is a coheir of $q\restrict M$.

For the converse, again
let $b \in B$ and $U = \stonenabe{\varphi(b,\tuple{b})}$
and assume $U \subset q$ and therefore $q \models \varphi(b,\tuple{b})$.
Then since $q$ is a coheir of $q \restrict M$
there is an $m \in M$ such that $p = \stonenabe{\varphi(m,n)} \subset q$.
Because $m \in M$ and $M \subset B$ we have
$q = p \intersect \tp(m/B)$ is nonempty. 
Therefore $q \subset q \restrict M = q \intersect M \subset M$.
Hence any open set contained in $q$ intersects $M$
and consequently $q \in \obar{M}$.
\end{proof}
\end{document}
