Standing approximation failure tests

The following remain scoped tests, not sufficient premises for extraction CheungXXIIB,CheungXXIID.

• Boundary generation versus attainment. For

every nonzero finite conical combination is interior to ${\mathcal{L}}_m$, by the strict triangle inequality, but $\overline{\mathrm{cone}G}={\mathcal{L}}_m$ by density. A generated null boundary need not contain an attained generator or implemented protocol.

• Finite-stencil delay. If all step velocities lie in a fixed finite $S\subseteq \{\Vert s{\Vert }_2\le 1\}$, with $0\in S$, choose a unit vector $u\notin S$. Then $h_S(u)={\max }_{s\in S}u\cdot s<1$. An endpoint within ${\delta }_n$ of $ru$ satisfies

by projection onto $u$. When $h_S(u)>0$, vanishing spatial error forces an arrival lower bound $r/h_S(u)>r$. When it is zero, sufficiently accurate arrival is impossible.

• Moving arrival jumps. Compare a channel returning a failure flag before $\tau$ and identity with a success flag after $\tau$, with the corresponding jump at ${\tau }_n>\tau$. Even if ${\tau }_n\to \tau$, orthogonal flags between the two arrival times give uniform half-diamond distance one. Endpoint topology does not assert uniform convergence through such an unmatched jump.

• Accumulation is model dependent. For the qubit replacement channel ${\mathcal{R}}_0(X)=\mathrm{Tr}(X)|0⟩⟨0|$, let ${\mathcal{F}}_e=(1-e)\mathrm{id}+e{\mathcal{R}}_0$. Idempotence gives

Thus $ke\to a>0$, with $e\to 0$, can leave error $1-e^{-a}$. Failure of a telescoping upper bound to vanish, by itself, is not a lower bound for every noise or correction model.

• Tolerance versus exact influence. Channels uniformly within $ϵ$ of one intervention-independent channel have signalling strength at most $2ϵ$, not necessarily zero. The family $(1-\delta ){\mathcal{R}}_0+\delta \mathrm{id}$ is at distance $\delta$ from replacement and transmits a bit with distinguishability $\delta >0$. A tolerance graph is not the exact graph (eq:influence).