Two-point rigidity: a chunk system on two points has expected total at most $c_B+jb$

Let $X$ be a metric space consisting of exactly two points $s\\neq t$, and let $C$ be a chunk system with online escapes on $X$ with marked points $s,t$, size floor $0$ and size ceiling $c_B$ (a ChunkSystemB X s t 0 cB T pe mL). Assume the Doob martingale $D_h=\\mathbb E[\\sum_j c_j\\mid\\mathcal F_h]$ of its total mass has the pointwise jump bound $|D_{h+1}-D_h|\\le jb$.

Statement. The expected total size of $C$ is at most one chunk plus one Doob jump:

Why. On two points the offline constraint hopt (an offline evader starting at $s$ pays at most $d(s,t)$ on the whole request sequence) makes the request sequence rigid: once a request avoids $s$, every later request must contain $t$, since otherwise the offline evader would be forced to cross between $s$ and $t$ twice, at cost $2d(s,t)$. Consequently a lazy evader — one that stays at $s$ while the requests allow it and stays at $t$ afterwards — pays nothing on any chunk in which the crossing neither happens nor may happen. The chunk-cost axiom then forces such a chunk to have size $0$. Hence a chunk carries mass only at times $h$ at which the crossing has positive conditional probability, and the first such time $\\sigma$ is a stopping time. Before $\\sigma$ no mass is consumed; on the branch of the atom of $\\sigma$ that crosses, the whole future mass collapses to $0$ in one step, so by the one-step Doob bound the conditional future mass at $\\sigma$ is at most $c_B+jb$. Optional stopping over the time-$0$ atoms turns this into the stated bound on the expected total.

Relevance to the mission. This is a rigidity (no-go) statement: it shows that the BCR chunk machinery cannot produce large mass on a two-point metric space, so a two-point space cannot witness a failure of the regrouping lemmas.