Generalized geometries

Question

Let $S$ be a non-empty set. A geometry of type $n$ for $n\geq 1$ on $S$ (consisting of at least $n$ elements) is a set ${\mathfrak P}\subseteq {\mathcal P}(S)$ such that

1. all members of $\mathfrak P$ have at least $n$ elements,

2. any $n$ elements of $S$ are contained in exactly one member of $\mathfrak P$,

3. for $l_1\neq l_2 \in \mathfrak P$ we have $|l_1\cap l_2| = n-1$, and

4. there is $T\subseteq S$ with $|T|=n+1$ and $T\notin \mathfrak P$.

Geometries of type $1$ are "traditional" partitions -- they define an equivalence relation on the set $S$.

A geometry of type $2$ is a projective plane.

Question: Is there for every $n\geq 1$ a geometry $\mathfrak P$ of type $n$ on $\omega$ such that $|\mathfrak P| \geq 2$?

Answer 1

Can we construct a free such object?

Step 1: Start with $n+2$ elements.

Step 2: For every set with $n$ elements, put it in $\mathfrak B$.

Step 3: For each pair of sets in $\mathfrak B$ that intersect in $n-k$ elements, add $k-1$ new elements and add those elements to those sets and those two sets.

Step 4: For each set of $n$ elements that is not already contained in a set in $\mathfrak B$, put it in $\mathfrak B$.

Repeat the last two steps infinitely. Axiom 1 is always satisfied, cause each set is created with $n$ elements. The uniqueness part of Axiom 2 is always satisfied, because we never add a set of $n$ elements to $\mathfrak B$ if they are already in a set in $\mathfrak B$. For Axiom 3, $|l_1 \cap l_2| \leq n-1$ is always satisfied, because new sets are created with $n$ elements that do not all intersect any other set, and because we only add new elements up to that limit. Axiom 4 is always satisfied by the orginal set of $n+2$ elements.

The existence part of Axiom 2 is always satisfied after even-numbered steps, and $|l_1 \cap l_2| \geq n-1$ is always satisfied after odd-numbered steps. So both are satisfied in the limit.

Combining these we get Axioms 1, 2, 3, and 4.