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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 $A$?

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    $\begingroup$ A title with information on the subject would be maybe more useful than one just conveying your opinion on the value of the question $\endgroup$
    – YCor
    Commented May 2, 2019 at 22:27
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    $\begingroup$ What is a non trivial lower bound... Are you sure you did not mean "upper bound"? $\endgroup$
    – fedja
    Commented May 2, 2019 at 22:57
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    $\begingroup$ As @fedja says, clearly $A = \emptyset$ will satisfy your condition, so it is hard to imagine a non-trivial lower bound. In fact I find it hard to read the question overall. Should it be: "What is the best upper bound [in terms of $F$? Independent of $F$?] on the cardinality of a subset $A$ of $V$ such that $A + A \cap F = \emptyset$" (where $V$ is your chosen vector space)? $\endgroup$
    – LSpice
    Commented May 3, 2019 at 0:11
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    $\begingroup$ It must mean a lower bound on the maximum size of such an $A$. $\endgroup$ Commented May 3, 2019 at 2:30
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    $\begingroup$ Here is a copy of the question on Mathematics Stack Exchange: Cardinality of certain subsets in vector spaces over finite fields. You can find a very reasonable advice about cross-posting in this answer. $\endgroup$ Commented May 3, 2019 at 8:55

2 Answers 2

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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 $|A|=2^{n-1}$.

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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.

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  • $\begingroup$ The number of lines that have either one or two points in $F$ is exactly $|F|*(|V|-|F|)$, so the sum of (b) and (c) alone is always at least $|V|-1$ before accounting for cancellation. Thus this bound doesn't seem to be particularly useful without more work, though I may have missed something as it's late. $\endgroup$
    – dvitek
    Commented May 3, 2019 at 6:04

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