On two points every chunk system has total at most $d(s,t)$

Suppose the metric space $X$ consists of exactly two points $s\\neq t$. Then every chunk system with online escapes on $X$ with marked points $s,t$ and declared total $T$ satisfies

Why. The offline constraint built into a chunk system says that an offline evader starting at $s$ serves the whole request sequence for at most $d(s,t)$. On two points this forces the sequence to be rigid: once a request avoids $s$, every later request contains $t$, for otherwise the offline evader would have to cross twice. Therefore the one-way lazy evader — which stays at $s$ while every request so far contains $s$ and moves to $t$ for good afterwards — is a legitimate online evader that pays at most one crossing, i.e. at most $d(s,t)$, on every such sequence. Since the expected total size of a chunk system is bounded by the expected cost of any evader, and the declared total is a lower bound for the expected total size, the claim follows.

Consequence. No chunk system on a two-point space can carry more mass than a single crossing. In particular the inductive chunk statements of the Bubeck–Coester–Rabani programme are unsatisfiable over two-point spaces whenever the total they demand exceeds $d(s,t)$, so no two-point instance can witness a failure of the regrouping lemmas. No hypothesis on the size floor, the size ceiling or the martingale of the total is needed.