Suppose a model $A$ is $\aleph_0$-saturated. Show every n-type over T is realized in $A$ We define: $A$ is $\aleph_0$-saturated if for any expanision $A_{c_1,...,c_m}$ of $A$ by finitely many constant symbols $c_1,...,c_m$, every 1-type _consistent with_ $Th(A_{c_1,...,c_m})$ is realised in $A_{c_1,...,c_m}$ > Suppose a model $A$ is $\aleph_0$-saturated. Show every n-type over T is realized in $A$ Thanks!

You can prove it by induction on $n$.

The case $n=1$ follows by the $\aleph_0$-saturation of A. For the inductive case, given an $(n+1)$-type $\Gamma(x_1,\ldots,x_{n+1})$, you can write it as the union of the $n$-type given by all formulas in $\Gamma$ using only variables $x_1,\ldots,x_{n}$ and the rest of the formulas in $\Gamma$ (which are the formulas using also the variable $x_{n+1}$).

By induction, there is a realization $c_1,…,c_n$ in $A$ of the $n$-type, and by hypothesis, the new type $\pi(x_{n+1}):=\Gamma(c_1,\ldots,c_n,x_{n+1})$ is a $1$-type consistent with $Th(A_{c_1,\ldots,c_n})$, so there is $d\in A$ realizing $\pi$.

Notice that the tuple $(c_1,\ldots,c_n,d)$ is a realization in $A$ of the original $(n+1)$-type $\Gamma(x_1,\ldots,x_{n+1})$.