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\begin{center}
MATH 776 HOMEWORK 10 \\
{\it Semester II, 2006-2007} \\
{\it Due March 22}
\end{center}


\bigskip

\bigskip
\textbf{47.} Let $\Sigma$ be the theory of infinite abelian groups of 
exponent $5$.  Find $|\TP_n(\Sigma)|$ for $n = 1,2$.
\emph{Remark.}  Since $\Sigma$ is
$\aleph_0$--categorical, its countable model is saturated,
so two $n$--tuples of the same type are connected by an automorphism.
\emph{Hint.} Classify $n$--tuples by the dimension of the subspace space
they span.

\bigskip
\textbf{48.}
If $\AAAA$ is any $\LL$--structure, then 
$\eDiag(\AAAA)$ is its elementary diagram in $\LL_A$
and the space $\TP_1(\eDiag(\AAAA))$ describes all possible single elements
in elementary extensions of $\AAAA$.
Now let $\LL = \{ < \}$ and let $\AAAA = (\RRR, <)$.
Prove that  $\TP_1(\eDiag(\AAAA))$ is the compact LOTS you get
from $\RRR$ by adding $\pm \infty$ and then replacing each
$r \in \RRR$ by a triple $\{r^-, r, r^+\}$.
\emph{Remark.} The derived set  $(\TP_1(\eDiag(\AAAA)))'$  is
exactly the double arrow space of Alexandroff and Urysohn (1929).
For \emph{any} $\AAAA$, the isolated points in $\TP_1(\eDiag(\AAAA))$ 
correspond to elements of $A$ and the limit points correspond to
the possible elements outside of $A$.
To analyze the types, you may use the fact
that the theory has quantifier elimination; that is, every formula
is equivalent to a quantifier-free formula.

\bigskip
\textbf{49.} Let $\kappa$ be any infinite regular cardinal.
Let $A = \{f \in \{-1, 0, +1\}^\kappa : |f^{-1} \{-1, +1\}| < \kappa\}$.
Let $<$ be lexicographic order on $A$.  Prove that
$(A; <)$ is a $\kappa$--saturated model of the theory of dense
total orders without endpoints.  \emph{Remark.} This construction
is essentially that of F. Hausdorff, \emph{Grundz\"uge der
Mengenlehre}, 1914.  Note that $|A| = 2^{<\kappa}$, which is $\kappa$
under GCH. \emph{Hint.} To prove that $A$ is $\kappa$--saturated
in the model-theoretic sense, you may use the fact
that the theory has quantifier elimination. 
This reduces saturated to showing that $A$ is
what Hausdorff called
\selectlanguage{german}
"`eine $\eta_\xi$--Menge"'
\selectlanguage{english}
(where $\kappa = \aleph_\xi$).

\bigskip
\textbf{50.}
(Chang-Keisler Exercise 4.3.14)
Assume that $A_i$ for $i \in I$ are finite non-empty sets
and $\UU$ is an ultrafilter on $I$.
Let $B = \prod_i A_i / \UU$.  Prove that one of the following holds:
\begin{itemizz}
\item[a.] For some $n \in \omega$,
$|B| = n$ and $\{i : |A_i| = n\} \in \UU$.
\item[b.] $|B| \ge 2^{\aleph_0}$ and 
$\{i : |A_i| = n\} \notin \UU$ for all $n \in \omega$.
\end{itemizz}
Conclude from this that in $(\omega, <)^I/\UU$, every non-standard
integer has at least $2^{\aleph_0}$ elements below it.
\emph{Hint.} In case (b), the proof that $|B| \ge 2^{\aleph_0}$ 
is similar to the proof that $|A^I/\UU| \ge 2^{\aleph_0}$ when
$A$ is infinite and $\UU$ is countably incomplete.
\emph{Remark.} If $\UU$ is countably complete, then case (a) must hold.
If $\UU$ is countably incomplete, both cases are possible.

\bigskip
\textbf{51.} (Peano, 1886 (by a different proof)) Consider the ODE:
$$
x'(t) = F(x,t); \ x(0) = c \ \ ,
$$
where
$F \in C(\RRR^2, \RRR)$ and $F$ is bounded.
For each positive integer $n$,
one may try to use the Euler method to approximate a solution on
$[0,\infty )$ :
Let
$X(0) = c$ and let
$X( {k+1 \over n})  = X({k \over n}) +
{1 \over n}F(X({k \over n}), {k \over n})$.
This defines $X({k \over n}  )$ for $k = 0, 1, 2, \ldots $.


Now, in an ultrapower $({}^*\RRR;  {}^*\stuff)$ of $(\RRR; \stuff)$,
\emph{fix} an infinitely large posititive integer, $n$,
and apply the Euler method with that $n$ to get $X$.
Then, for \emph{standard} $t \in [0,\infty )$, define
$x(t) = \st(X( {k \over n} ))$ for
some (\emph{any!}) integer $k$ in the ultrapower
such that $\st({k \over n}  ) = t$.  Prove that this defines an
$x \in C^1([0,\infty ), \RRR)$ which satisfies the ODE.

\emph{Remark}. Since the ultrapower elementarily extends everything,
you don't have to specify exactly what $\LL$ (i.e., \stuff ) is here.

\emph{Note}.  The solution to this ODE need not be unique,
and the one you get could depend on the $n$ chosen, and the
Euler approximations need not converge to anything.
Standard proofs usually use the Arzel\`a-Ascoli Theorem.
An example of non-uniqueness that you can use in Math~222 is
$\;x'(t) = 3x^{2/3}$; $x(0) = -1$; you go through a metastable state
at $t = 1, x=0$.
You could replace $x^{2/3}$ by $\min(100, x^{2/3})$ if you want
a bounded $F$.

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