= Finitely many accepting extensions of a rejected stem
Suppose a reservoir rejects $s$ and decides every extension $s\cup\{a\}$. Only finitely many of those one-point extensions can have accepting tails. If infinitely many did, collect their new points into $C$; every infinite subset of $C$ begins with an accepting successor, so $[s,C]$ would lie in the family. That would make $C$ accept $s$, contradicting rejection. This finite-obstruction fact lets a second fusion preserve rejection of every finite stem.
Back to article page
