 Solution

Write $A_2(N,\delta )$ for the largest number of codewords in a binary length-$N$ code with minimum Hamming distance at least $\delta$. The volume of a Hamming ball does not depend on its centre and isFor the Hamming bound, put $r=\lfloor (\delta -1)/2\rfloor$. The radius-$r$ Hamming balls around distinct codewords are disjoint: a point belonging to two such balls would, by the triangle inequality for Hamming distance, put their centres at distance at most $2r<\delta$. Counting points in the binary cube therefore gives

For the Gilbert–Varshamov bound, choose a code maximal under inclusion among those with minimum Hamming distance at least $\delta$, for example by successively adding any allowable word. Its radius-$(\delta -1)$ Hamming balls cover the cube. Otherwise an uncovered word would be at distance at least $\delta$ from every codeword and could be added, contradicting maximality. ThusThe lower bound is an existence assertion about a suitable code, while the upper bound holds for every such code. Together they give

Let $h_2(x)=-x{\log }_{2}x-(1-x){\log }_2(1-x)$ be the binary entropy. For $\delta =\lfloor \lambda N\rfloor$,The assumed Hamming ball volume exponent applies also to any integer radii $d_N$ with $d_N/N\to \theta \in (0,1/2)$. Indeed, for each small $\epsilon >0$ these radii eventually lie between $\lfloor (\theta -\epsilon )N\rfloor$ and $\lfloor (\theta +\epsilon )N\rfloor$. monotonicity of $v_N$ sandwiches its normalized logarithm between limits $h_2(\theta -\epsilon )$ and $h_2(\theta +\epsilon )$; continuity then lets $\epsilon$ tend to zero. Consequently the asymptotic Gilbert–Varshamov bound and asymptotic Hamming bound have respective exponential scalesEquivalently, the optimal code rate satisfiesThese are bounds on the rate, not a claim that the optimal rate has a known limit or equals either endpoint.