Prove that for two vectors x,y over GF(q), the number of vectors that are closer to x is the same as the number of vectors that closer to y.

Question

Let $x,y\in\mathbb F_q^n$ be vectors. We'll define:

$X= \{ u\in\mathbb F_q^n \mid d(x,u)<d(y,u)\}$

$Y= \{ u\in\mathbb F_q^n \mid d(y,u)<d(x,u)\}$

Prove that $|X|=|Y|$.

Well. I tried really hard and I really don't know. I tried using the sphere packing bound for proving something, but of course the the code $C=\{x,y\}$ is not perfect, so it won't help.

I tried constructing a linear code $C=\operatorname{span}\{x,y\}$ and use the Lagrange theorem for cosets (That says: All $q^n$ vectors are in one of the coests of $C$; each coset have same number of words; and all not-equal cosets are disjoint), but i don't know how to use it, because the code is $q$-ary, so this code have more words than $x,y$ in this code (so i can't do anything).

Thanks.

Answer 1

Hint:

Proof that $$f : \mathbb F_q^n \to \mathbb F_q^n,\quad v\mapsto v - x + y$$ induces a bijection $X \to Y$.