Base case of BCR Lemma 12 at the tight escape price

Let $\mathcal M_w$ be the level-$w$ member of the recursive Bubeck--Coester--Rabani space family, with marked points $s_w,t_w$ at distance $d_w=\beta\,3^w$, and let $\mathrm{ChunksTight}(\alpha,\beta,w)$ be the level invariant of KServer.BCRInductiveChunksTight: a chunk system on $\mathcal M_w$ with sizes in $[3^w/2,\,3\cdot3^w/2]$, expected total at least $\alpha\beta w^2 3^w$, escape price $d_w$, exactly $M\ge\lceil\alpha\beta w^2\rceil$ chunks, constant initial information and no empty chunk.

Statement. For every $\alpha\ge0$, every integer $\beta\ge2$ and every level $w$ with

the invariant $\mathrm{ChunksTight}(\alpha,\beta,w)$ holds.

Why this is the easy half. When $\alpha w^2\le1$ the demanded expected total $\alpha\beta w^2 3^w$ is at most $\beta 3^w=d_w$, so no randomness is needed: a deterministic single-file adversary suffices. Level $w$ of the family contains a unit geodesic chain $g$ with $g(0)=s_w$, $g(\beta 3^w)=t_w$ and $d(g(i),g(j))=|i-j|$; sampling it at the multiples of $3^w$ gives $\beta+1$ marked positions at spacing $3^w$, and the singleton requests at those positions, one per chunk, form a chunk system of $\beta$ chunks of size $3^w$. Each chunk forces the evader to advance one step of the chain, and bailing out is never cheaper because the escape price $\beta3^w$ is at least the chunk size $3^w$.

Relation to the existing base case. This is KServer.bcr_induction_base with the escape price lowered from $2d_w$ to $d_w$ and with the two structural clauses that the level steps additionally require: the exact chunk count and nonemptiness of the chunks. The lower price is what makes the invariant chainable, since KServer.level_step_sturdy assumes its input price is at most $d(s,t)$, and it costs nothing here: the cost bound of the single-file system holds against any escape price at least the chunk size. The geodesic chain in the glued family and the single-file system along an arbitrary chain are the two ingredients of the existing proof and can be reused verbatim.