$(a_i)$ and $(b_i)$ having the same type in a structure $N$ is equivalent to $M,(a_i)\equiv M,(b_i)$?

Question

Consider two $k$-uples of elements in a structure $N$, $(a_i)$ and $(b_i)$. I am tempted to consider equivalent the two statements:

1) $(a_i)$ and $(b_i)$ have the same type on $N$

2) $N,(a_i)\equiv N,(b_i)$, i.e. the two expanded structures (with $k$ new constant symbols interepreted by the $a_i$ and the $b_i$ respectively) are elementary equivalent.

Is this true? It seems rather clear to me, but since I'm not finding it anywhere I could have missed an important detail...

Thank you in advance.

Answer 1

Here is a brief proof of the statement:

Assume that $tp(\bar{a}) \neq tp(\bar{b})$. Then $\exists \varphi(\bar{x}) \in tp(\bar{a})$ such that $\neg\varphi(\bar{x}) \in tp(\bar{b})$. Then, $(N, \bar{a}) \not \equiv (N, \bar{b})$ since $(N, \bar{a}) \models \varphi(\bar{a})$ and $(N,\bar{b}) \models \neg \varphi(\bar{b})$.

Assume that $(N, \bar{a}) \not \equiv (N, \bar{b})$. Then, $\exists \psi$ such that $(N, \bar{a}) \models \psi$ and $(N, \bar{b}) \models \neg \psi$. Now, $\psi$ must contain some $a_i$ (otherwise $N \not \equiv N$). Then, we can conclude that $tp(\bar{a}) \not = tp(\bar{b})$.