We assume $e,d,n\in \Bbb Z^+$ and $q\geq 2$. We define the quality $A_q(n,d)$ as the maximum $M$ such that an $(n,M,d)$-code exists. I am trying to solving the following exercise from Roman's book:
If $e\leq d$ we want to prove that $$A_q(n,e)\geq A_q(n,d). $$
Can anyone prove clearly why this claim is true? I really struggle to manupilate the notion of $A_q(n,d)$.
Thank you.
Let me prove a lemma first.
$$\exists (n,M,d+1)_q-code \implies \exists (n,M,d)_q-code$$
Proof: Let $C$ be an $(n,M,d+1)_q$-code.
We apply the trick I also applied in one of your previous questions. Let $c,c' \in C$ with $d(c,c')=d+1$ and $c\neq c'$. Choose a position $i$ with $c_i \neq c_i'$. Then, define $c''$ by
$$c''_j:= \begin{cases}c_j \quad i \neq j\\ c_i' \quad i=j\end{cases}$$
Thus $c''$ is the word $c$ in which we changed the $i$-th cordinate to $c_i'$. Then $d(c',c'') = d$.
Define a new code $\tilde{C}:= (C \setminus \{c\}) \cup \{c''\}$. We show that this is an $(n,M,d)$-code.
Suppose to the contrary it is not. Then it must be a $(n,M,k)$-code with $k < d$. The minimal distance of $\tilde{C}$ must be attained by an element $e \in C \setminus \{c\}$ and $c''$ (by our definition of $\tilde{C}$ and the fact that the minimal distance of $C$ is $d > k$). But then
$$k = d(e,c'')$$
so $$d(e,c) \leq d(e,c'') + d(c'',c) = k +1 \leq d < d+1$$
which contradicts that $C$ has minimal distance $d+1$. $\quad \square$
Using this lemma inductively, we obtain: for an integer $l \geq 0$
$$\exists(n,M,d+l)_q-code \implies \exists(n,M,d)_q-code$$
Now, let us see how this implies the claim.
Consider a code $(n,M,d)$ where $M= A_q(n,d)$. Since $e \leq d$, our generalised lemma implies that there is an $(n,M,e)$-code. But then
$$A_q(n,e) \geq M = A_q(n,d)$$