batch 16

Prompt

AUDIT QUESTION (P2EX siege, auditor role: audit SECTOR FACT, the weakest unaudited planar link). Self-contained. Label every claim PROVED (you prove it), SOURCE (exact reference; unchecked?), LAB (from this prompt, status as given), CONJECTURE, OPEN, or RESTATEMENT (a gap reworded). If a definition is ambiguous, state your reading in one line and continue. Say where your argument stops. Take your time and be extremely careful. A. DEFINITIONS PRE-DISPATCH CERTIFICATE (batch 16, microeval channel). (C1) R1 / E_p^-, Step 7. Target: audit SECTOR FACT as stated in B. YES (fatal defect): PER-GAP, ONE-PER-GAP and the one-gap localisation are withdrawn and the pair-carrier map changes. NO (no defect): SECTOR FACT becomes an audited-candidate building block. Neither alone is CLOSED. (C2) Controls: ARC-4^- (two-sided landing at (0,0), decomposable); a disk with a whisker (fails BD); the canonical case E = {xi_0, xi_1}; boundary endpoints of a gap as negative cases. (C3) New input: an independent review (2026-11-08) ranked SECTOR FACT / PER-GAP the weakest acceptance link of the current attack; it has one author answer and no audit. (C4) Duplicate check: no audit task on SECTOR FACT exists. Definitions: a continuum is a nonempty compact connected metric space; subcontinua are closed and connected; nondegenerate means more than one point. Planar means embeddable in R^2, hence in the sphere S^2 = R^2 + {inf}; nonseparating means S^2 minus X is connected. Tree-like: for every eps>0 an onto map to a finite tree with all fibres of diameter < eps (tree-like continua are hereditarily unicoherent, HU: two subcontinua meet in the empty set or a continuum; every subcontinuum of a planar tree-like continuum is tree-like and nonseparating). Indecomposable: not a union of two proper subcontinua; then every proper subcontinuum is nowhere dense, the union of two proper subcontinua that meet is proper, and the composant of x (union of all proper subcontinua containing x) partitions X into uncountably many pairwise disjoint dense composants. For a homeomorphism g: cm(g) means no proper nonempty g-invariant subcontinuum; cm_strong(g) is the same with singletons counted as subcontinua, so g has no fixed point. Per(g), Fix(g): periodic, fixed points. FPP (fixed point property): every continuous self-map has a fixed point; PPF: periodic-point-free; PFPP: the open plane fixed point problem (does every nonseparating plane continuum have FPP?). Cartwright-Littlewood (SOURCE): an orientation-preserving homeomorphism of the plane leaving a nonseparating plane continuum invariant has a fixed point in it (Bell, Brown extensions). Lab vocabulary: R1, E_p^- = lab route labels (setting below); siege 1 = the lab's current attack on P2EX. Statuses: PARTIAL/OBSTRUCTION = author's verdict; UNAUDITED = no audit yet; CONFIRMED-WITH-REPAIRS = audit passed, repairs written in, not yet closed; one GPT agent = one GPT run; never assign VERIFIED/CLOSED. Take your time, be extremely meticulous, I believe in you. SETTING (route R1, branch E_p^-; LAB; package UNAUDITED). X is a nondegenerate planar nonseparating tree-like indecomposable continuum in R^2, U = S^2 minus X (a simply connected domain containing inf), and X = boundary(U) (clause BD; automatic, as planar tree-like X has empty interior). p is an odd prime. b: X -> X is a homeomorphism with cm_strong(b) and anchor cycle A = {a_0,...,a_{p-1}}, b(a_i) = a_{i+1}; the a_i are called anchors. Sources disagree on the spectrum hypothesis: (R-A) Per(b) = A or (R-B) only Fix(b^p) = A; OPEN, state which reading each step uses. F: S^2 -> S^2 is an ORIENTATION-REVERSING homeomorphism with F(X) = X, F(inf) = inf and F|X = b^p; so F^2 preserves orientation, F^2|X = b^{2p}, and b^p, b^{2p} are NOT cm (anchors are fixed singletons): cm(F) is never a substitute for cm(b). Complement PPF: F has no periodic point in R^2 minus X. Prime ends. Let phi map D* = {|z|>1} + {inf} conformally onto U with phi(inf) = inf (normalisation only; F need not commute with phi, so F does NOT map radial rays to radial rays). Prime ends of U correspond to points zeta of the unit circle T. For a prime end e, the impression Imp(e) and the principal set Prin(e) are subcontinua of X with Prin(e) inside Imp(e). SOURCE (Pommerenke, Boundary Behaviour of Conformal Maps, 1992, Theorem 2.16, checked on the full text): Imp(e) is the unrestricted cluster set of phi at zeta, and Prin(e) equals the cluster set of phi along the radius ending at zeta, which also equals the intersection of the cluster sets along all curves in D* ending at zeta and the cluster set in any Stolz angle at zeta. e is accessible iff Prin(e) is one point (its landing point); e is full-principal if Prin(e) = X. F induces a homeomorphism h = h_F of T with Imp(h e) = F(Imp e), Prin(h e) = F(Prin e). Since F reverses orientation, h has exactly two fixed ends xi_0, xi_1, swaps the two open arcs of T between them, and h^2 preserves orientation; every h-periodic end has least period 1 or 2. A 2-cycle is a pair {e, e'} with h(e) = e' != e; its two ends lie on opposite arcs, so {e, e'} and {xi_0, xi_1} are LINKED (alternate on T). P2EX: no 2-cycle exists, i.e. Fix(h^2) = {xi_0, xi_1}. The lab's canonical prime-end law CAN is equivalent to P2EX (LAB, audited). FP-FIX (standing assumption, LAB; its production from the source antecedent is OPEN): the two fixed ends are full-principal, Prin(xi_0) = Prin(xi_1) = X. There is no prime-end action of b itself, only of F = b^p. Oversteegen-Valkenburg Example 5.4 (SOURCE, not rechecked): a planar pseudo-arc embedding with every impression the whole pseudo-arc and a dense second-category set of full-principal ends, so many full-principal ends carry no static contradiction. B. LEMMAS (LAB = earlier lab answers, unaudited unless noted; G = F^2, H = h^2, E = Fix(H), L = landing map from accessible ends to points, Comp(x) = composant of x) (SPECTRA) (R-A): for every x, b^n(x) = x for some n >= 1 iff x in A. (R-B): b^p(x) = x iff x in A. NP2_X: G x = x implies F x = x. R-A implies NP2_X; R-B alone does not (b-periods 2 and 2p are not excluded). (FOUNDATION, LAB) unique access (each point is the landing point of at most one accessible end); accessible ends are dense on T; E is closed, nowhere dense, contains no accessible end other than possibly xi_0, xi_1 (none under NP2_X), every end of E is full-principal under R-A; H is increasing on each gap (component of T minus E) and its orbits there converge to the gap endpoints. (L-SC, LAB, repaired global-sector proof, audit owed) If ends alpha, beta alternate on T with two full-principal ends, no proper nonseparating subcontinuum C of X contains Prin(alpha) u Prin(beta). Proof sketch: collapse C to a point c (Moore: the quotient is a sphere, the map is a homeomorphism off C); the two radial rays to alpha, beta map to arcs forming a Jordan curve J meeting q(X) only at c; the two open angular sectors between alpha and beta go to the two components of q(U) minus J, which lie in distinct components of the sphere minus J (else the other component would lie in q(X), which has empty interior); the radial rays of the two full-principal ends lie in different sectors, so q(X) lies in both closures, hence in J, so q(X) = {c}, contradiction. Fixedness of the separating ends is not used (GENERALIZED L-SC). (KEY-STRONG, Sonnet audit CONFIRMED-WITH-REPAIRS) a proper continuum D containing an accessible point is disjoint from F(D); so Comp(F x) != Comp(x) for accessible x, and no proper continuum through an accessible point contains an F-fixed point. (TAIL EXCLUSION, same audit) under NP2_X no proper continuum contains a forward or backward G-orbit tail of an accessible point. (PAIR EQUIVALENCE, DERIVED) for accessible eta in a gap, with x = L(eta), K = Comp(x): some proper continuum contains x and G x iff G(K) = K. (SECTOR FACT, LAB, one answer, unaudited) if a proper continuum C contains accessible u = L(e_1), v = L(e_2), then after collapsing C the two landing rays form a Jordan curve meeting the image of X only at the collapse point, and the ends on the side that misses X form an open arc V_C of T between e_1 and e_2 with Prin(e) in C for every e in V_C. If e_1 = eta and e_2 = H eta lie in one gap I, V_C is the arc (eta, H eta) inside I (the other arc contains xi_0, xi_1). (PER-GAP, LAB, conditional on SECTOR FACT) if one accessible eta in a gap I has a proper continuum through L(eta), G L(eta), then K = Comp(L eta) is G-invariant, F(K) != K, Prin(e) and Imp(e) lie in K for EVERY end e of the OPEN gap I (the H-translates of the sector V_C cover I), and the b-period of K (as a set) is 2 or 2p. ONE-PER-GAP: two distinct G-invariant accessible composants cannot saturate the same gap. (OPEN CORE) Step 7 under R-A: exclude an end alpha not in {xi_0, xi_1} with H alpha = alpha and Prin(alpha) = Prin(h alpha) = X (FP2). Under bare R-B also exclude SPLIT: Prin(alpha), Prin(h alpha) proper, disjoint, G-invariant, no proper continuum containing both. PROOF TO AUDIT (author sketch). Let C be proper, containing u = L(e_1), v = L(e_2), e_1 != e_2. (1) X minus C is connected (X indecomposable, C proper, so X minus C contains a dense composant and is connected). (2) Collapse C to c (Moore); q(X) minus {c} is connected. (3) The landing rays r_1, r_2 (phi-images of the radii, completed by u, v) map to arcs from q(inf) to c with disjoint interiors; J = their union is a Jordan curve with J meeting q(X) only at c. (4) q(X) minus {c} lies in one component W_1 of the sphere minus J; the other component W_0 meets only q(U). (5) The ends e whose radius lies in W_0 form one of the two open arcs of T between e_1, e_2 (call it V_C). (6) For e in V_C the radial cluster set lies in closure(W_0) intersect q(X) = {c}, so Prin(e) is in C. (7) If e_1 = eta, e_2 = H eta in one gap I, V_C is the arc inside I. C. THE TARGET Audit steps (1)-(7). Pay special attention to: (1) is X minus C connected for every proper subcontinuum of an indecomposable continuum, and is it needed; (3) do the two radial images have disjoint interiors and is J intersect q(X) = {c} correct when u, v lie in C; (5) the passage from 'radius in W_0' to an arc of T (radii are disjoint curves ordered like their angles); (6) closure of W_0 meets q(X) only in c; (7) why the arc containing xi_0, xi_1 cannot be V_C (it contains full-principal ends). NEGATIVE CASES: e_1, e_2 themselves; gap endpoints of I; ARC-4^- with C a small interval around (0,0) and e_1 = 0^+, e_2 = 0^-; C containing an anchor. D. ASK (AUDITOR ROLE) Walk the proof step by step. For each step say OK, GAP (missing argument, give it if you can) or FALSE (give the counterexample). Report the FIRST fatal step with the exact quantifier that breaks, or state that none was found. Test each named negative case explicitly. Separate (R-A) from (R-B) where it matters. No novelty/VERIFIED/CLOSED claims; no helper agents. E. CONTROLS (all must hold) ARC-4^- (lab control): X = [-1,1] x {0}, f the time-one map of the flow of v(x) = (x^3 - x)/(1 + x^4) (increasing homeomorphism of [-1,1], Fix(f) = {-1,0,1}, no other periodic points), F(x,y) = (f(x), -2y) (orientation-reversing; R^2 minus X is PPF). Prime ends: x^+ and x^- (upper and lower sides) for -1 < x < 1 plus one end at each endpoint; h(x^+) = f(x)^-, h(x^-) = f(x)^+; the endpoint ends are the fixed ends (singleton principal sets, NOT full-principal) and {0^+, 0^-} is a genuine 2-cycle landing at (0,0) from two sides. X is decomposable and chainable. Interval reflection: b(t) = -t on [-1,1]; Fix = {0}, all other points have period 2; every symmetric [-c,c] is invariant. Circle reversal: z -> conj(z) has two fixed points and period-2 pairs, so reversal alone cannot give P2EX. Rotating p-od: p arcs meeting at a centre o, b rotating the arms; translates of the terminal half of one arm are pairwise disjoint; the only b-invariant point is o (forbidden only by cm_strong). Pseudo-arc: hereditarily indecomposable chainable; it has FPP, so it is never a PPF host, but it is the static calibration for full-principal ends (Example 5.4). Pseudo-circle and SPIRAL(3) separate the plane; a disk with a whisker fails BD. State each added control's first exit. A proof that works verbatim on ARC-4^-, on a disk with whisker, or on a chainable host is suspect; name the step where indecomposability, tree-likeness, nonseparation or cm_strong(b) is used and show it fails on the control. F. DEAD ENDS (not progress) Restating that SECTOR FACT is unaudited; objections that only drop indecomposability or nonseparation, which are hypotheses. G. FORMAT Line 1 exactly: VERDICT: PROVED | COUNTEREXAMPLE-CANDIDATE | PARTIAL | OPEN-OBSTRUCTION | RESTATEMENT. Then SUMMARY (at most 100 words), the argument (at most 2,000 words), STILL OPEN EXACTLY. Short of effort: line 1 + summary; an honest OPEN is valid. No finite computation as proof; no theorem beyond its hypotheses; UNKNOWN if unsure a cited result exists (give exact theorem numbers or say UNSURE). RED FLAGS: using cm(F) or cm(b^p) in place of cm(b); treating F as commuting with phi or mapping radial rays to radial rays; confusing impression with principal set; assuming Fix(b^{2p}) = A under (R-B); assuming P u P' is b-invariant; dropping the canonical full-principal fixed ends xi_0, xi_1 as a negative test.