Joint Embedding Property

Question

I want to show that any complete theory has JEP,
And that JEP does not imply comleteness.

I have trouble showing it, and I think I'm missing sometiong here.

And another question:
If $T$ is model complete - why compeleteness is equivalent to JEP?

Answer 1

To see that completeness implies joint embedding property, use compactness. For the other direction, consider the empty theory (in any language).

For the other question, this is just unfolding the definition.

Answer 2

Let T be complete, $A,B \models T$. Note that JEP is equivalent to satisfiability of $$\Gamma := T \cup Diag(A) \cup Diag(B)$$

Let $\Delta$ be a finite subset of $Diag(B)$, $\phi$ the conjunction of formulas from $\Delta$, $c_1 \dots c_n$ the new constants occuring in $\phi$. Replace these with unused variables $v_1 \dots v_n$ to get $\phi'$. By completeness: $$B\models \exists \overline{v}\phi' \Rightarrow A\models \exists \overline{v}\phi'$$

So by interpreting $c_1 \dots c_n$ by suitable witnesses $A\models \Delta$. By compactness $\Gamma$ is satisfiable.

Let T be model complete and $ A,B \models T$, let $C \models T$ s.t. A and B embed into C. By modelcompleteness these embeddings are elementary, so for a sentence $\phi$

$$A \models \phi \Leftrightarrow C \models \phi \Leftrightarrow B \models \phi$$

Hence T is complete.

Also the empty theory over the the language containing no non-logical symbols has JEP, since two models both embed into their union, but the sentence $\exists x \forall y : y=x$ is not decided.