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's $A$ is a subset of this vector space when the intersection of $A+A$ and $F$ is empty.
The question is this: What is a non trivial lower bound for the cardinality of $A$?
Thank you.
 A: Well, I can give an example where $A$ must be empty. Suppose your vector space is $(\mathbb{F}_2)^n$ and $F$ is any $n$ element subset that includes $0^n$. Then for each $a \in A$ indeed any $a \in (\mathbb{F}_2)^n$ note that $a+a = 0^n$.
In fact something similar would hold if $\mathbb{F}_2$ were replaced by any finite field with characteristic 2.
So your question as asked, the lower bound is 0 and it is tight.

ETA: If you require every element in $F$ to be nonzero, then [for the case where vector space $X$ is $(\mathbb{F}_2)^n$ anyway] $A$ can be as large as $|X|/2$. Indeed, let $F$ be the set of vectors $u$ in $(\mathbb{F}_2)^n$ such that exactly one coordinate in $u$ is nonzero, and let $A$ be the set of vectors $y$ such that $y$ has an even number of coordinates nonzero. Then $|A| = |X|/2$ but $A+A = A$--indeed $A$ is a vector space--and $A$ does not intersect $F$.
A set $A$ can be no larger. Indeed, if $A$ is larger than $|X|/2$ then $A+A$ contains every nonzero element $u$ of $(\mathbb{F}_2)^n$. Indeed,  $u +A$ has cardinality $> |X|/2$, and therefore intersects $A$ itself. So let $a_1,a_2 \in A$ be such that $u+a_1 = a_2$. Then as $u$ is nonzero, it follows that $a_1 \not = a_2$, and $a_1+a_2 = u$ so $u \in A+A$. 
So half the number of elements in the vector space $X$ is an upper bound. And for $X =$ $(\mathbb{F}_2)^n$ it is a tight upper bound.
