I came across a sentence in page 196 Chang & Keisler's model theory book that I don't understand. It says: Every complete theory has the joint embedding property.

Def. A theory $T$ has joint embedding property if for every $M,N\models T$ then there is a model $K\models T$ such that $M,N$ are isomorphically embeddable in $K$.

Why is it true? Can anybody guid me with this?


1 Answer 1


HINT: Let $L$ be the language of $T$. Consider the new language $L'=L\sqcup\{c_a: a\in M\}\sqcup\{d_b: b\in N\}$ gotten by adding new constants to $L$ corresponding to the elements of $M$ and $N$ (assuming WLOG that $M$ and $N$ have disjoint domains).

Now let $T'$ be the $L'$-theory consisting of $T$ together with the atomic diagram of $M$, for the $c_a$s, and the atomic diagram of $N$, for the $d_b$s. E.g. if $M\models a_1Ra_2$ then we put "$c_{a_1}Rc_{a_2}$" into $T'$.

Exercise: If $T'$ has a model $K$, then $M, N$ embed into (the reduct of) $K$ (to $L$).

This is immediate.

Exercise: $T'$ has a model.

This uses completeness of $T$, and compactness of first-order logic. By compactness, we just need to show that every finite subset of $T'$ has a model. But we can show that any finite configuration occuring in $M$ also occurs in $N$, since this is witnessed by an existential formula.

  • $\begingroup$ @Liam A finite set of atomic or negated atomic formulas which are satisfied somewhere in the structure. For instance, maybe $M$ contains three elements $x, y, z$ such that $R(x, y), R(y, z), R(x, z)$ all hold (for $R$ a binary relation symbol in $L$); then so does $N$. Another stronger way to phrase this would be: "For each formula $\varphi(\overline{x})$ in some tuple of variables $\overline{x}$, there is a tuple $\overline{a}\in M$ such that $M\models \varphi(\overline{a})$ iff there is a tuple $\overline{b}\in N$ such that $N\models \varphi(\overline{b})$." Do you see why this is true? $\endgroup$ May 6, 2016 at 6:56
  • $\begingroup$ Actually, for a complete theory you can work not only with the atomic diagrams, but with the entire elementary diagram of A and B, right? And the joint embedding are actually elementary then. $\endgroup$
    – mtg
    Feb 2, 2022 at 12:38

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