\{R(b)\cap (V_{j+1}\cup\ldots \cup V_{j+i+1})\colon b\in V_{j+1}, E(a,b) \}.$$ For each neighbour $b\in V_{j+1}$ of $a$, the set $R(b)\cap (V_{j+1}\cup\ldots \cup V_{j+i+1})$ is a free pseudospaces of dimension $i-1$ by induction. Since $(V_{j+1}\cup\ldots \cup V_{j+i+1})$ is a free pseudospace of dimension $i$, $(\Sigma 2)_{i+1}$ follows from the induction hypothesis. Hence $V_0\cup\ldots \cup V_{n-1}$ and $V_1\cup\ldots \cup V_n$ are free pseudospaces of dimension $n-1$.

[**$(\Sigma 2)_n$:**]{} The proof of $(\Sigma 2)_n$ is similar.

We say that a model $M$ of $T_n$ is $\K_n$-saturated if for all finite $A\leq M$ and strong extensions $C$ of $A$ with $C\in\K_n$ there is a strong embedding of $C$ into $M$ fixing $A$ elementwise. Clearly, by construction, $M_n$ is $\K_n$-saturated.

\[l:K-sat\] A model $M$ of $T_n$ is $\omega$-saturated if and only if $M$ is $\K_n$-saturated.

Let $M$ be an $\omega$-saturated model of $T_n$. To show that $M$ is $\K_n$-saturated, let $A\leq M$ and $A\leq B\in\K_n$. By induction we may assume that $F$ is an elementary strong extension of $A$ and it is easy to see that $F$ can be imbedded over $A$ into $M$.

The theory $T_n$ is complete.

Let $M$ be a model of $ T_n$. In order to show that $M$ is elementarily equivalent to 