Strict asymptotic separation need not give zero finite-block error

The source entropy is $H(P_S)=\ln 2$. For a BSC with crossover $1\mathrm{/}4$ and input $X\sim \mathrm{Bern}(r)$, the output parameter is $q=1\mathrm{/}4+r\mathrm{/}2$, and [F3] gives $I(X;Y)=h(q)-h(1\mathrm{/}4)$. By [F4], the maximum is at $r=1\mathrm{/}2$, so $C(W)=\ln 2-h(1\mathrm{/}4)=\frac{3}{4}\ln 3-\ln 2$.

Fix any $k\ge 1$, any deterministic encoder $f_k:\{0,1{\}}^k\to \{0,1{\}}^{6k}$, and decoder $g_k:\{0,1{\}}^{6k}\to \{0,1{\}}^k$. Choose any output word $y_0$ and then a source word $s^k\mathrm{\ne }{g}_k(y_0)$, which exists because $\mathrm{\mid }\{0,1{\}}^k\mathrm{\mid }\ge 2$. By [A1] and [F1], $\mathrm{Pr}[S^k=s^k,Y^{6k}=y_0]=2^{-k}{W}^{6k}(y_0\mid f_k(s^k))>0$. On this event the decoder is wrong, so the block-error probability is positive for every finite code.

Hence $6C(W)-H(P_S)=\frac{9}{2}\ln 3-7\ln 2>0$, since $3^9=19683>16384=2^{14}$. The asymptotic threshold is strict, so [F5] gives a reliable code sequence at ratio six.

The strict asymptotic threshold in step 2.1 therefore cannot imply a zero-error code at each finite blocklength ratio. This refutes the statement while leaving the asymptotic separation conclusion intact.