that the class $(\K_n,\leq)$ has a Hrushovski limit $M_n$, i.e. a countable $L_n$-structure $M_n$ whose strong subsets are exactly the $L_n$-graphs in $\K_n$ and which is homogeneous for strong subsets: if $A,B\leq M_n$ then any isomorphism from $A$ to $B$ extends to an automorphism of $M_n$. Here we say as usual that a subset $A$ of $M_n$ is strong in $M_n$ if $A\cap B\leq B$ for any finite set $B\subset M_n$.

\[p:pseudospace\] The Hrushovski limit $M_n$ is a model of $T_n$.

By construction, $V_i\cup\ldots\cup V_{i+j}$ satisfies $(\Sigma 3)_j$ and $(\Sigma 4)_j$ for any $i,j$. In particular, $M_n$ satisfies $(\Sigma 3)_n$ and $(\Sigma 4)_n$.

[**$(\Sigma 1)_n$:**]{} In order to show that $M$ satisfies $(\Sigma 1)_n$, we first note that $V_i\cup V_{i+1}$ is a free pseudoplane for all $i=0,\ldots n-1$. Assume inductively that $V_j\cup\ldots \cup V_{j+i}$ is a free pseudospace of dimension $i$. To see that $V_j\cup\ldots \cup V_{j+i+1}$ is a free pseudospace of dimension $i+1$, we need only verify $(\Sigma 2)_{i+1}$. Hence we have to show that for $a\in V_j$ the residue $R(a)\cap (V_j\cup\ldots \cup V_{j+i+1})$ is a free pseudospace of dimension $i$. We know by induction that $R(a)\cap (V_j\cup\ldots \cup V_{j+i})$ is a free pseudospace.

Clearly, $$R(a)\cap (V_{j+1}\cup\ldots \cup V_{j+i+1})=\bigcup 