Manuscript En05-A@1 · En05-A.tex · lemma · f076e703

Lemma: Block geometry

Partitioned dynamics and the exact comparison theorem › The schedule past and the metric envelope · lines 522–535 · proof included or linked

En05-A@1#lem:geometry

If $b \in P_{s}$ meets $I_{s + 1}$ , then $b \subseteq I_{s}$ . Consequently, no block that is not inner can place an output into $I_{s + 1}$ .

Proof. For $z \in b \cap I_{s + 1}$ and $w \in b$ ,

d (w, A) \leq d (w, z) + d (z, A) \leq r + r (T - s - 1) = r (T - s) .

Every output remains in its block.

Displayed formulas in this passage

unlabeled · line 530
$d (w, A) \leq d (w, z) + d (z, A) \leq r + r (T - s - 1) = r (T - s) .$