m51

Prompt

RESEARCH QUESTION (plane continua / the LOCATION wall (renamed W-SPAN-PROPER, the PER core): in an (S)-instance, does g = h^2 fix a composant (world beta) or none (world alpha)? The intrinsic structure of both worlds is known (D2082); which planar law decides?). 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. A. DEFINITIONS Setting (compressed from the lab report api_opus_spanproper_glue ([SPG]), D2082 (Argon 20), the audited UBS lemma and the W-SPAN-PROPER wall row; single-agent statements, UNAUDITED unless a status says otherwise; INFERRED marks the spec author's own reading). - (S): X is a nondegenerate nonseparating plane continuum, U := C \ X (connected), h: X -> X a homeomorphism with g := h^2 continuum-minimal (cm: no nonempty proper subcontinuum K with g(K) inside K; then h, h^-1, g^-1 are cm as well). So h and g have no fixed point, m_1 := min |h(x) - x| > 0, m_2 := min |g(x) - x| > 0. The lab also has perfect plane extensions F of h and G of h^-1 (not homeomorphisms) with unique outchannels E_F != E_G (BFMOT Theorem 6.2.1, SOURCE); a proof may use them. The class of (S)-instances may be empty (that is PER); the question concerns its members only. - Audited or classical consequences: X is indecomposable, int X is empty, X is hereditarily unicoherent (HU). x ~ y iff a proper subcontinuum contains both; the classes are the composants: dense, pairwise disjoint, uncountably many; a proper subcontinuum lies in one composant; every homeomorphism permutes composants. - A composant C is g-fixed if g(C) = C. World alpha: no composant is g-fixed. World beta: some composant is g-fixed. S := {x : x ~ g(x)} is the union of the g-fixed composants. Ov(g) := {K : K a subcontinuum, K cap g(K) nonempty}. World gamma: S = X. - Planar conversion ([SPG], DERIVED from Janiszewski RJ: for continua A, B in S^2 with A cap B nonempty and disconnected, S^2 minus (A union B) is disconnected). For a plane continuum P meeting X let K^(P) := X cap cl(P union the bounded components of C \ (X union P)). Lemma FILL: K^(P) is a subcontinuum of X. Theorem JC: if X union P does not separate the plane, X cap P is a continuum. B. LEMMAS (status; where used) Known (statuses as given). - D2082 (Argon 20, VERIFIED-with-repairs by Sonnet and Grok audits per the lab record; the spec author read the statements in notes/web_queue/manual/argon10/argon20_l2_caseB.tex, not the audits; hypotheses are those of the lab's L2 setting, which the spec author takes to be (S), INFERRED: check it). For f = g: Ov(f) minus {X} is nonempty iff some composant is f-fixed. World alpha: Ov(g) = {X}, so the Whitney-minimal overlapping continuum is X. World beta: every Whitney-minimal K_* in Ov(g) is a proper continuum with empty interior in a g-fixed composant, irreducible between z and g(z) for z in K_* cap g^-1(K_*), and (Tridiagonal Chain) g^i(K_*) cap g^j(K_*) is nonempty iff |i - j| <= 1; finite unions of consecutive links are proper and nowhere dense, while the forward union and the backward union are dense in X. Argon 20's own planar step (a chain link against the barrier laws of the lab's Case B, not restated here) stayed a CONJECTURE. - UBS (VERIFIED building block: one cross-family Sonnet audit, repairs written in the ledger, closure record not checked here). For every r > 0 there is T such that every subcontinuum K with K cap h(K) nonempty has diam K >= m_1 and the union of h^j(K), 0 <= j <= T, is r-dense; a composant contains such a proper K iff it is h-fixed. The audited proof uses only 'no proper nonempty W with h(W) inside W', which for g is cm itself, so UBS holds for g with m_2 (INFERRED). - [SPG] (DERIVED, unaudited; a Sol audit is queued). Theorem NC: an indecomposable continuum has no compact free collar, so X has no open subset homeomorphic to Z x [0, inf) and collar or retraction engines have no input in X. Theorem LOC (an exact RESTATEMENT): g is not cm iff some plane continuum P has K^(P) != X and g(K^(P)) inside cl(P union its holes). Prop SPREAD: in world alpha, K^(P) = X for every plane continuum P containing K union g(K), K proper. [SPG]: assembly of a continuum is free, the wall is this LOCATION inclusion. A DeepSeek pre-check (not an audit) flags the zero-dimensional step of NC(b) as uncited and a clause of SPREAD (closure of the union of holes, not one hole). - Baire dichotomy (spec author, INFERRED, CHECK). A_n := {x : some continuum K contains x and g(x) with Hausdorff distance d_H(K, X) >= 1/n} is closed by compactness of the hyperspace; S is the union of the A_n, g-invariant, and a union of composants. If some A_n has interior, then, composants being dense, S = X; otherwise S is a meager F_sigma. So world beta is world gamma or a meager S. (Same-composant pairs form a dense meager F_sigma of X^2, pro1104h Lemma 7; the graph of g is nowhere dense in X^2, so that lemma does not decide the dichotomy.) - Other, unaudited. pro1104n: if C_0 != C_1 are composants fixed by g with h(C_0) = C_1, every minimal compact g-invariant M meets each C_i in a meagre F_sigma of M, so no nonempty compact g-invariant set lies in C_0 union C_1. pro1104e Lemma 2C: no nonempty compact h-invariant set lies in the union of two proper subcontinua. A201: the dyadic solenoid with a translation g along the dense line is minimal, so cm, has m_1 > 0 and fixes every composant (translation by an element of the dense line, INFERRED); it is not planar. A196, m36: abstract composant labels with cm are consistent with both worlds. C. THE TARGET Decide alpha against beta for every (S)-instance, equivalently whether Ov(g) = {X}. Any proof must use planarity at a named line (the solenoid is not planar): connectedness of U, Janiszewski or FILL, prime ends, accessibility, or E_F != E_G. Acceptable outcomes: ALPHA PROVED; BETA PROVED; one branch of the Baire dichotomy decided; BETA CONSISTENT as Tier 0 data; REDUCED to one lemma (gate G1-G4). What each world gives (INFERRED, [SPG]): alpha makes the overlap laws vacuous (K cap g(K) is empty for proper K) and leaves hyperspace trapping UD/EQ as the only residual of W-SPAN-PROPER; beta localises every LOCATION attempt to one composant carrying the dense chain. Neither closes L2. Not asked: L2 or PER as a whole; assembly of continua (free). D. ASK Part 1 (method). Name ONE method not on the dead-end list and say why it sees where g places a trace-generated continuum, where the failed methods saw only composant labels, categories or hyperspace compactness. Candidates, none claimed to work: (i) BAIRE SPLIT: treat world gamma (S = X) and the meager branch separately; in gamma some open V lies in A_n, so every composant (all meet V) carries a bridge K through a point of V (a continuum containing x and g(x), diameter >= m_2, d_H(K, X) >= 1/n); bridges of different composants are disjoint, so there are uncountably many pairwise disjoint bridges; look for a planar contradiction (Janiszewski on two bridges, filled traces) or a build; (ii) CHAIN AGAINST U: for non-adjacent links K_0, K_3 of the tridiagonal chain take crosscuts of U joining them (an M_F crosscut has trace inside the chain segment between its ends, an E_F crosscut has trace X) and FILL on plane continua through several links; say where connectedness of U enters; (iii) CHANNEL ENDS: compare the forward and backward ends of the dense chain with the two channels theta_F != theta_G (prime-end positions of the limit sets; CHECK that both ends have limit set X); (iv) your own. Part 2 (carry it out). Prove ALPHA (or BETA, or one branch), or a dichotomy 'alpha, or the configuration C' with C excluded by a named planar law or realised, or reduce to one named lemma L with gate G1-G4. Mark each use of planarity. State the result for g = h^2 exactly (not h^4). Part 3 (builder or decisive test). Try to realise world beta in the plane: a planar HU indecomposable X, a homeomorphism h with m_1 > 0 and h^2 cm, and a dense tridiagonal chain in a g-fixed composant; or name the first law (nonseparation, int X empty, HU, m_1 > 0, cm of h^2, the tridiagonal pattern) that every candidate violates. Builder gate: name the TARGET law (B0); secure the background laws first (B1); then test (B2); end with 'KILL: law | TARGET or BACKGROUND | mechanism | what a proof must use' (B3). A finite chain is not a host. What already failed (read DEADENDS first): a new method, even partial with an honest gap, beats a polished restatement. Take your time: check every step, hypothesis and quantifier. This is a hard open problem. E. CONTROLS (all must hold) (C1) The solenoid translation (A201) is cm, fixes every composant (it lies in world gamma) and meets the composant dichotomy of D2082, the Baire dichotomy and the category facts (its chain law needs HU, which the solenoid lacks), so a proof of alpha must use a planar line that the solenoid lacks, named. (C2) Chainable hosts (the Knaster bucket handle) have the fixed point property, so no h with m_1 > 0 lives there: they decide nothing. (C3) Control that the failed methods also pass: the abstract composant labelings (A196, m36), the composant dichotomy of D2082 and the category facts (Lemma 7) are met by the solenoid and give no alpha, so the new input must be planar. (C4) Circle control: X = S^1 with an irrational rotation separates the plane, and with a long arc P the set X union P separates, so JC fails exactly at its nonseparation clause; a use of JC or FILL must show X union P does not separate. (C5) h^2 against h^4 and finite windows: no h^2-fixed composant does not exclude an h^4-fixed one, and a statement for one composant, a finite chain, or a threshold depending on K is not alpha. F. DEAD ENDS (not progress) (1) Assembly and conversion: nested retractions and collar exhaustion ([SPG] NC), Janiszewski filling (JC, FILL): a continuum is always produced, the wall is LOCATION. (2) The W-SPAN-PROPER tried list: orbit-wide pocket iteration (b_opus_pocket_orbit), hyperspace trapping UD/EQ (b_opus_rig_trap), invariant measures and hyperspace operators (S9; stationary form pro1104k), HT with semicontinuity (Argon 68), clopen continuum-span (Argon 69), Helly and UFOS (Argon 71, SPAN-OVERLAP-CONF), grid-polygon separation. (3) Moore triod theorem with Cantor-Bendixson (pro1104i), Whitney levels with invariant measures (pro1103f02), Baire exhaustion and Helly cores (pro1104lr), shape or Cech (pro1104h). (4) Composant bookkeeping and cocycles (A196, m36, m40): consistent with both worlds. (5) R43 (a set-valued transport law for strip-defined sets) and R44, R45 are separate microevals: not repeated. RESTATEMENT GATE (mandatory; answered BEFORE the headline, in its own section). If your verdict reduces the node N to a residual lemma L (with or without other proved pieces), you must give ALL of: (G1) the exact statement of L and of N side by side, with the same standing hypotheses; (G2) a proof that L implies N (the reduction); (G3) a reason L is STRICTLY weaker than N: either an explicit configuration (or consistent abstract data meeting every standing law) where L holds and the conclusion of N is not yet available, or the named hypothesis of L that the negation of N does NOT supply; (G4) the line of your argument where L's own hypothesis (the defect, the witness, the scale) is actually used; if it is never used, L is a restatement. If G3 or G4 fails, the headline word is RESTATEMENT, not REDUCED, and the answer reports the proved building blocks only. A headline REDUCED without G1-G4 counts as a failed answer. G. FORMAT Line 1 exactly: VERDICT: ALPHA PROVED (h^2 fixes no composant in any (S)-instance; each law used named, the planar line marked) | BETA PROVED (an h^2-fixed composant exists in every instance) | SPLIT (one branch of the Baire dichotomy S = X or S meager decided) | BETA CONSISTENT (explicit Tier 0 planar data with an h^2-fixed composant meeting every listed law; first missing law named) | REDUCED (to one named lemma L, gate G1-G4 passed) | PARTIAL (one named sub-claim settled) | OPEN. Then SUMMARY (<= 100 words), the argument (<= 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. RED FLAGS: composant bookkeeping, category or hyperspace laws alone (the solenoid translation passes all of them); assuming F or G is a plane homeomorphism; confusing h^2-fixed with h-fixed or h^4-fixed; presenting assembly or Janiszewski filling as progress (it is free); a finite chain, one composant, or a threshold depending on K; claiming L2, PER, SS-COF or DP-K as proved