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 $|\mathfrak P| \geq 2$?

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    $\begingroup$ You should require $|\mathfrak P| \geq 2$ to avoid $\mathfrak P = \{ S \}$. Condition 4 is superfluous as soon as $|S| \geq n+2$. $\endgroup$ – Tom De Medts Jul 15 '15 at 8:53
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    $\begingroup$ More to the point: the non-existence of a geometry of type $m$ would imply the non-existence of such geometries of type $n$ for all $n>m$. Indeed, if a geometry of type $n$ exists, then one can form the "derived geometry" w.r.t. an element $s \in S$ obtained by considering $S' = S \setminus \{s\}$, and $\mathfrak P' = \{ B \setminus \{s\} \mid B \in \mathfrak{P}, s \in B \}$, which is then a geometry of type $n-1$. $\endgroup$ – Tom De Medts Jul 15 '15 at 8:56
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    $\begingroup$ If I'm not mistaken, the previous argument also shows that, as soon as $n \geq 2$, all members of $\mathfrak P$ have the same number of elements. $\endgroup$ – Tom De Medts Jul 15 '15 at 8:57
  • $\begingroup$ I agree - beautiful argument! $\endgroup$ – Dominic van der Zypen Jul 15 '15 at 9:45
  • $\begingroup$ I have doubts about 3 - ($n-1$)-planes in $\mathbb P^n$ almost never satisfy this, so most likely it is a very severe restriction. $\endgroup$ – მამუკა ჯიბლაძე Jul 15 '15 at 19:11

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.

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    $\begingroup$ Thanks for these ideas! It looks like they work, but I'm still a little uneasy about the statement "So both are satisfied in the limit." Can you elaborate on this a little bit? (Sorry for not accepting your answer yet; I will do it once I understand the limit part.) $\endgroup$ – Dominic van der Zypen Jul 16 '15 at 6:55
  • $\begingroup$ @DominicvanderZypen Every $n$-tuple of elements of $S$ appears in one of the odd numbered step, when the last element appears. Then unless it's already contained in an element of $\mathfrak B$, by the next even numbered step it enters $\mathfrak B$, and it stays contained in an element of $\mathfrak B$ for each subsequence step. Similarly, every pair of elements of $\mathfrak B$ appears at some point, and then the intersection becomes $n-1$, and then it doesn't go down or up. $\endgroup$ – Will Sawin Jul 16 '15 at 12:57
  • $\begingroup$ This does not work for $n=1$ as the resulting set is finite. For other $n$ it also needs to be shown that this process does not terminate. $\endgroup$ – Guntram Jul 16 '15 at 15:47
  • $\begingroup$ @Guntram let me modify the odd-numbered steps to, if no elements would otherwise be added, add one element and not put it in no sets. $\endgroup$ – Will Sawin Jul 16 '15 at 17:06

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