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\ctrline{The Backgammon Fixed Point Theorem}

\bigskip
Define a function $Y=F(X)$ from $R↑n$ to $R↑n$ by $Y↓i=f↓i(X)$,
where each~$f↓i$ is of one of the four forms:

\smallskip
\disleft 25pt:(1):
$f↓i(X)= {\rm constant}\in (A,B)$.

\smallskip
\disleft 25pt:(2):
$f↓i(X)=\Sigma a↓jX↓j$, $\Sigma a↓j=1$, $a↓j≥0$.

\smallskip
\disleft 25pt:(3):
$f↓i(X)=\max(X')$, where $X'$ is a nonempty subsequence of~$X$.

\smallskip
\disleft 25pt:(4):
$f↓i(X)=\min(X')$, where $X'$ is a nonempty subsequence of~$X$.

\smallskip
Each of the above functions is monotone and continuous. Defining a
partial order on vectors by $X≤X'$ if $X↓i≤X'↓j$ for all~$i$, $F$~is
clearly monotone and continuous.

Defining the sequence $X↑0=\langle A,A,\ldots,A\rangle =\bar{A}$,
$X↑{i+1}=F(X↑i)$, we see that $X↑1≥X↑0$, and then by induction
$X↑{i+1}≥X↑i$. Also by induction, $X↑i≤\langle B,B,\ldots,B\rangle =\bar{B}$.
As a bounded increasing sequence, $X↑i$~has a limit~$X↑∞$,
$X↑i≤X↑∞$, and by continuity $F(X↑∞)=X↑∞$, so $X↑∞$~is a fixed point
of~$F$. By induction, if $X'$ is a fixed point of~$F$, $X↑i≤X'$, so 
$X↑∞≤X'$, and $X↑∞$ is the unique minimal fixed point of~$F$, and
lies between $\bar{A}$ and~$\bar{B}$.

Similarly, starting with $X↑{\ast 0}=\langle B,B,\ldots,B\rangle$,
and proceeding as before, we get $X↑∞$ as a maximal fixed point,
lying between $\bar{A}$ and~$\bar{B}$. Iff $X↑∞=X↑{\ast ∞}$,
the fixed point is unique.

We show that a position $P↓i$ has an expected value under all
strategies iff $X↓i↑∞=X↓i↑{\ast ∞}$. 
[RWF: find a proof.]

\vfill\eject\end