Cardinality of certain subsets in vector spaces over finite fields Assume that you have an $n$-dimensional vector space over a finite field (therefore the number of elements in the vector space is finite) and $F$ is a subset of this vector space which contains $m$ elements. Let $A$ be a subset of this vector space such that the intersection of $A+A$ and $F$ is empty. 
The question is: What is a non trivial lower bound for the maximal possible  cardinality of such an $A$?
 A: Consider the addition Cayley graph $\Gamma$ induced by $F$ on $\mathbb Z_2^n$ (which is the graph with the vertex set $\mathbb Z_2^n$, with the vertices $u,v\in\mathbb Z_2^n$ adjacent whenever $u+v\in F$). A set $A$ with $(A+A)\cap F=\emptyset$ is an independent set in $\Gamma$. Since $\Gamma$ is regular of degree $|F|$, it has an independent set of size at least $2^n/(|F|+1)$. 
In general, this bound is best possible: it is attained when $F$ is a subgroup of $\mathbb Z_2^n$ with the zero element removed (so that $|F|=2^k-1$, where $k$ is the rank of $F$), and $A$ contains a unique element from each $F$-coset.
Better bounds can be given if some information about the set  $F$ is available. Say, if $F=\{f_1,\dotsc, f_m\}$ is an independent set, then one can find a subgroup $H<\mathbb Z_2^n$ such that $\mathbb Z_2^n=\langle F\rangle\oplus H$, and  take 
  $$ A := \{ c_1f_1+\dotsb+c_mf_m+h\colon c_1+\dotsb+c_m=0,\ h\in H \} $$ 
to have $|A|=2^{n-1}$. 
A: For $V$ a vector space over ${\mathbb Z}/2{\mathbb Z}$, the non-zero points of $V$ form a projective space in which every line has three points, and the sum of any two points (thought of as elements of $V$) is equal to the third point on the line they share.
So for $A+A$ to avoid $F$, you can take the union of a) all lines that avoid $F$, b) either of the two remaining points from any line that has exactly one point in $F$, and c) the remaining  point from any line that has exactly two points in $F$, and d) the zero vector if it's not in $F$.  You might have to prune this down further, but clearly you can't do any better, so this at least gives an upper bound. I don't claim it's necessarily a very good one.
