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


1. $\{1,\ldots,n\}\notin {\cal C}$,
2. for all $x,y\in \{1,\ldots, n\}$ there is $A\in {\cal C}$ such that $\{x,y\}\subseteq A$, and
3. $|A\cap B| \leq 1$ 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.)



Minimizing the maximum degree of a set of sets
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:


$$ \{(0\ 0)\ \ (0\ 1)\ \ (0\ 2)\}$$
$$ \{(0\ 0)\ \ (1\ 0)\ \ (2\ 0)\}$$
$$ \{(0\ 0)\ \ (1\ 1)\ \ (2\ 2)\}$$
$$ \{(0\ 0)\ \ (1\ 2)\ \ (2\ 1)\}$$



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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:


$$ \{a\ b\}\ :\ (a\ b)\in \{1\ 2\ 3\}\times\{4\ 5\ 6\}\ $$.


