An error correcting code has disjoint Hamming balls.

See Hamming ball, error detection and correction

If $C$ is a $t$ -error correcting code, then for all codewords $x, y \in C$ with $x \neq y$ ,
the Hamming balls $B_{t} (x) \cap B_{t} (y) = \emptyset$

Proof

Suppose $d (C) = d$ , and let $t$ be the largest integer $t < \frac{d}{2}$ .
Suppose by way of contradiction that there exist $x \neq y$ in $C$ such that $B_{t} (x) \cap B_{t} (y) \neq \emptyset$ .
Then there exists some $z \in B_{t} (x) \cap B_{t} (y)$ .
Since $z \in B_{t} (x)$ , $d_{H} (x, z) \leq t$
Similarly, $d_{H} (y, z) \leq t$

By definition, $d = d (C) \leq d_{H} (x, y)$ for any $x, y$ .
By triangle inequality, $d_{H} (x, y) \leq d_{H} (x, z) + d_{H} (y, z)$ if $C$ is $t$ -error correcting.
Therefore,

d = d (C) \leq d_{H} (x, y) \leq d_{H} (x, z) + d_{H} (y, z) \leq t + t = 2 t < d

Thus $d < d$ , a contradiction.

QED