You are given that $C\subseteq D \subseteq \mathbb{F}^n_q$ where $|C| < |D|$ and $C$ is a perfect code. Show that $d(C) > 2d(D)$.
So that means that $M|S_t(\underline{0})| = q^n$ for $C$.
How can I use the Hamming bound equality, $C\subseteq D$ and $|C| < |D|$ to achieve the result $d(C) > 2d(D)$?
I don't know how to begin! Please can someone give me a hint.
Let $w$ be a word in $D$ that is not in $C$. As $C$ is perfect, there is a unique ball of radius $t = \lfloor \frac{d(C)-1}{2} \rfloor$ and centre $c_0$ that $w$ lies in. Then $d(D) \le d(c_0, w)$ (as $C \subset D$) and $d(c_0, w) < t \le \frac{d(C) - 1}{2}$. That should be enough to finish the argument, I think. Maybe distinguish odd and even $d(C)$ for clarity.