Suppose we have structures $M \subseteq N \subseteq P$ in some first order language. If $M \prec N$ and $M \prec P$, does it follow that $N \prec P$? If not, what is a counterexample?

  • $\begingroup$ It is not true, but some other similar statements are true: $\preceq$ is transitive (so $M\preceq N$ and $N\preceq P$ implies $M\preceq P$) and inclusion between elementary substructures is always elementary (so if $M\preceq P$ and $N\preceq P$ and $M\subseteq N$, then $M\preceq N$). $\endgroup$ – tomasz Mar 1 '15 at 3:12

From $M\preceq N$ and $M\preceq P$ it only follows that $N\equiv_M P$. For a counter example consider the following structures of signature $L=\{<\}$.

$P=\{0,1\}\times {\mathbb Z}$ ordered with the lexicographic order.

$M=\{0\}\times{\mathbb Z}$

$N=\{0\}\times{\mathbb Z}\ \cup\ \{1\}\times 2{\mathbb Z}$, where $2{\mathbb Z}$ is the set of even integers.

Then $M\preceq N$ and $M\preceq P$ as required. But clearly $N\not\preceq P$.

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