How prove this $2^{2012}-\binom{2012}{1006}let $A=\{1,2,3,\cdots,2013\}$,and $A_{1},A_{2},A_{3},\cdots,A_{m}\subset A,A_{i}\neq A_{j},1\le i\le m,1\le j\le n$
and such that:For any $i,j\in \{1,2,3,\cdots,m\},i\neq j$,then 
$$|A_{i}\bigcap A_{j}|\neq 1$$
prove that
$$2^{2012}-\binom{2012}{1006}<m_{\max}<2^{2012}$$
I think this is interesting problem, and This is my frend ask me, I think This problem may be use http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
 A: To prove the lower bound, consider the following sets:
1. The null set.
2. All subsets with at least $1008$ elements.
3. All subsets of $\{1, 2, \ldots , 2012\}$ with at least 1007 elements.
It is clear that the intersection of the null set and any other set has 0 elements.
The intersection of any 2 sets in group 2 have $\geq 1008+1008-2013 = 3$ elements.
The intersection of any 2 sets in group 3 have $\geq 1007 + 1007 - 2012 = 2$ elements.
The intersection of any 2 sets in group 2 and 3 have $\geq 1007 + 1007 - 2012 = 2$ elements.   
There are $1 + \left( 2^{2012} - { 2013 \choose 1007} \right) + {2012 \choose 1007} = 2^{2012} - {2012 \choose 1006} + 1 $ such subsets.

We will prove the a weaker upper bound by strong induction. For any integer $n\geq 2$, $A_n = \{1, 2, \ldots, n \} $, let $M_n$ be the maximum number of sets satisfying the conditions. We will show that $M_n \leq 2^{n-1}$.
The base cases of $n=2, 3$ are easy to deal with. Note that when $n=2$, we have equality. This is required for the induction step.
Induction step: Consider any collection of subsets.
If there is a subset of 1 element, then no other subset can contain it, hence there are less than $M_{n-1}+1 \leq 2^{n-1}$ subsets in this collection. Hence we may assume that any non-empty subset has at least 2 elements.   
If we have 2 non-empty sets with an empty intersection, say $|A_i| = i , |A_j | = j $ with $i+j \leq n$ and $i, j \geq 2$, then by restricting our attention to the sets $|A_k \cap A_i|$ and $|A_k \cap A_j|$, the number of sets is at most
$$M_i \times M_j \times 2^{n-i-j} \leq 2^{i-1} 2^{j-1} 2^{n-i-j} < 2^{n-1}.$$
Otherwise, we may assume that no 2 non-empty sets have an empty intersection. This implies that given any subset (that is not $A_n$ or $\emptyset$) and it's complement, at most one of them is in this collection. If so, this gives us at most $2^{n-1} + 1$. Hence, in order for the induction hypothesis to be false, given any subset, either it or it's complement must be in the set.
However, by considering all the subsets with $\lfloor \frac{n}{2} \rfloor$ and $ \lceil \frac{n}{2} \rceil$ elements, we clearly must have 2 which intersect in at most 1 element.
Now, it remains to show that $2^{n-1}$ cannot be achieved for $n \geq 3$. Do the $n=3$ case separately. 
For $n\geq 4$, consider the arguments in the previous proof. Let $\mathbb{B}$ denote the subsets with at least $\lceil \frac{n}{2} \rceil$ elements. We know that only possibility for $M_n = 2^{n-1}$, if for at most one subset $B \in \mathbb{B}$, such that neither $B$ nor $A_n - B$ is in the collection. Let $\mathbb{C}$ denote the subsets with exactly $\lceil \frac{n}{2} \rceil$ elements. If $n$ is even, we can very easily find 2 subsets whose intersections (or intersections of their complement) has at most 1 element. This is a contradiction, since any 2 non-empty sets must have a non-empty intersection that is not 1 element.
