Coding Theory Set Problem

Question

Supose that $\mathbb{A}$ is a finite set and take $\overline{u},\overline{v} \in \mathbb{A}^n$. Let:

$$X=\{\overline{x} \in \mathbb{A}^n\mid d(\overline{u},\overline{x})<d(\overline{v},\overline{x})\}$$

$$Y=\{\overline{y} \in \mathbb{A}^n\mid d(\overline{u},\overline{y})>d(\overline{v},\overline{y})\}$$

Prove that $Card(X)=Card(Y)$

I am really stuck on this problem. I thought about creation some bijection between both sets. But I don't really know how to do that, what makes me feel it's not the good path. Maybe I have to use some propiety about distances.

Answer 1

To each $i=1,2,\ldots,n,$ let $\pi_i:\Bbb{A}\to\Bbb{A}$ be the permutation defined by $\pi_i(u_i)=v_i$, $\pi_i(v_i)=u_i$, and $\pi_i(a)=a$ for all $a\in\Bbb{A}\setminus\{u_i,v_i\}$. Then define $\phi:\Bbb{A}^n\to\Bbb{A}^n$ by the recipe $$ \phi:(x_1,x_2,\ldots,x_n)\mapsto (\pi_1(x_1),\pi_2(x_2),\ldots,\pi_n(x_n)). $$ Show that

1. $\phi$ is an isometry. In other words $d(\overline{x},\overline{y})=d(\phi(\overline{x}),\phi(\overline{y}))$ for all $\overline{x},\overline{y}\in\Bbb{A}^n$.
2. $\phi(\overline{u})=\overline{v}$ and $\phi(\overline{v})=\overline{u}$.
3. $\phi(X)=Y$ and $\phi(Y)=X$.

Answer 2

It's not necessary that the underlying set is a field (or abelian group). Just show that the cardinality of the set $\{x\in A^n\mid d(u,x)=i\}$ with $i\geq 0$ fixed is independent of the choice of $u\in A^n$.

Then we can write $d_i = |\{x\in A^n\mid d(u,x)=i\}|$ for $i\geq 0$.