Theorem 13.1 of Sacks' Saturated Model Theory (1st ed.) says in part, that if every diagram of this type:
can be completed as shown, then $T$ is substructure complete (i.e., $T\cup\text{diag}(A)$ is complete for any substructure $A$ of a model of $T$).
To prove this, he constructs this diagram:
and then argues that on the one hand we can take the limit of the elementary chains, getting $B_0\preceq B_\infty$ and $C_0\preceq C_\infty$, but also we can construct an isomorphism $B_\infty\approx C_\infty$ "as [we did] in 10.4". So $\langle B_0,f(a)\rangle_{a\in A}\equiv \langle C_0,g(a)\rangle_{a\in A}$ and $T\cup\text{diag}(A)$ is complete.
There is a much simpler proof of this result using 10.4, which is the standard test for model completeness: if every embedding $A\to B$ of models of $T$ can be extended to $A\to B\to D$ with $A\preceq D$, then $T$ is model complete. Specializing his diagram for 13.1 by setting $A=C$, we conclude that $T$ is model complete, and just the leftmost part of his diagram ($A$, $B_0$, $B_1$, and $C_0$) is enough to finish the argument.
But I think his claim that we can construct an isomorphism $B_\infty\approx C_\infty$ is invalid. The problem is showing that "up and down" is the same as "straight across", i.e., (for example) $j_{12}k_{01}=t_{12}t_{01}$.
Here's a counter-example (I think): let all the $B_i$'s and $C_i$'s be the disjoint union of $\mathbb{N}$ with one object, say $\bullet$. Let $A=\{ \bullet \}$. Let $f(\bullet)=g(\bullet)=\bullet$. Let all the $h$'s and $t$'s be the identity. Finally, let $j(n)=k(n)=n+1$ for all $n\in\mathbb{N}$, where $j$ is any of the $j$'s and $k$ is any of the $k$'s. Then $B_\infty$ and $C_\infty$ are both just $\mathbb{N}\sqcup\{ \bullet \}$, and the "isomorphism" is actually $n\mapsto n+2$ (and $\bullet\mapsto\bullet$).
I suspect that Sacks was mislead by 10.4's diagram:
where the up-and-down really is the same as the straight-across.
Two related questions:
- Am I missing something?
- If not, was this error corrected in the second edition?
