Minimizing the maximum degree of a set of sets Suppose that ${\cal C}$ is a set of subsets of $\{1,\ldots,n\}$ with the following properties:


*

*$\{1,\ldots,n\}\notin {\cal C}$,

*for all $x,y\in \{1,\ldots, n\}$ there is $A\in {\cal C}$ such that $\{x,y\}\subseteq A$, and

*$|A\cap B| \leq 1$ for all $A\neq B\in{\cal C}$.


For $j\in \{1,\ldots,m\}$ we set the degree of $j$ to be $\text{deg}(j) = |\{A\in {\cal C}: j\in A\}|$ and set $$m({\cal C}) = \max\big\{\text{deg}(j): j\in\{1,\ldots,n\}\big\}.$$
Is it possible that $m({\cal C}) < n-1$? If yes, how small in terms of $n$ can $m({\cal C})$ become? 
(You only need to answer the first question to get the answer accepted; the second question is a bonus question.)
 A: Consider the 9-point affine plane $\ F_3\times F_3,\ $ over the 3-element Galois field $\ F_3.\ $ Then $\ n=9,\ $ while the respective $\ m(C)=4 < n-1$.
Here $\ C\ $ is the set of affine lines, i.e. sets described by the linear equations (homogeneous and non-homogenous, just like in an elementary school).
The explicit set of 4 straight lines which include point (0\ 0) is:
$$ \{(0\ 0)\ \ (0\ 1)\ \ (0\ 2)\}$$
$$ \{(0\ 0)\ \ (1\ 0)\ \ (2\ 0)\}$$
$$ \{(0\ 0)\ \ (1\ 1)\ \ (2\ 2)\}$$
$$ \{(0\ 0)\ \ (1\ 2)\ \ (2\ 1)\}$$

 

Now let us consider a 6-point space which is a union of two parallel lines of the above plane, i.e. now $\ n=6.\ $ Then in family $\ C\ $ consists of the intersections of the straight lines with our 6-point space (all members of this new space are two 3-point sets, and nine 2-point sets). Now $\ m(C) = 4 < n-1. $ This time I believe that this $\ n=6\ $ is minimal.
Thge example is simple enough to forget about geometry--let our space be $\ \{\, 1\ 2\ 3\ 4\ 5\ 6\,\}.\ $ Then let consist of two 3-point sets:
$ \{a+1\ \ a+2\ \ a+3\}\quad $ for $\ a=0\ $ and for $\ a=3$
and nine 2-point sets:
$$ \{a\ b\}\ :\ (a\ b)\in \{1\ 2\ 3\}\times\{4\ 5\ 6\}\ $$.
A: I'll show that

THEOREM $\ n=5\ $ is minimal.

First, observe that it's not possible for all members of $\ C\ $ to
have 2 (or less) points. Indeed, all sets $\ \{0\ x\},\ $ where $\ x\ne 0\, $ would belong to $\ C,\ $ hence $\  m(C)\ge n-1.\  $
Thus, we may assume that there is $\ A\in C\ $ such that $\ \{1\ 2\ 3\}\subseteq A.\ $ If there was only one point $\ b\notin A\ $ then
$\ \{a\ b\}\in C\  $ for every $\ a\in A,\ $ hence $\ m(C)\ge |A|=n-1\ $ -- a contradiction.
We see that the minimal $\ n\ $ is at least $\ 5.\ $

Indeed, we have a 5-point example consisting of complex numbers
$$ -1\qquad -\!i\qquad 0\qquad i\qquad 1 $$
The members of $\ C\ $ are defined as the nontrivial intersections of the Euclidean lines with the given 5-point set (where nontrivial means that the intersection has at least $\ 2\ $ different points). We see like Euclid would that $\ m(C)=3<4 = n-1.\ $ This proves the $\ n=5\ $ is correct and minimal. END of PROOF

REMARK   Whenever we have a system described by Dominic (just don't insist on points being consecutive integers) then the trace of such a system on a subset forms a similar system.

