{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-finite-product-topology-01", "mko_id": "mko-finite-product-topology", "title": { "zh-Hant": "四點乘積的全部開集", "en": "Every open set in a four-point product" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-finite-product-topology-01", "source": { "mko_id": "mko-finite-product-topology", "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f", "sha256": "9650dd3cf37e29ce4d8b2c6600ead8cb5c0ac577af1f7b69a0aa987341b02d5a", "english_sha256": "0010f081607335f5f72fb187f7825140a790a02f8d03038feafa566885063421" }, "mode": "worked_example", "audience_prerequisites": [ "sets", "topology", "definition_here" ], "goal": { "zh-Hant": "由兩個 Sierpinski 因子生成全部乘積開集並核投影。", "en": "Generate every product open from two Sierpinski factors and audit projections." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "X=Y={0,1}，兩因子拓撲皆 {空集,{0},{0,1}}；a=(0,0),b=(0,1),c=(1,0),d=(1,1),P={a,b,c,d}。", "en": "X=Y={0,1}, both topologies are {empty,{0},{0,1}}; a=(0,0),b=(0,1),c=(1,0),d=(1,1), P={a,b,c,d}." } } ], "steps": [ { "id": "S01", "uses": [ "G1" ], "before_tex": "\\varnothing\\subset\\{0\\}\\subset\\{0,1\\}", "after_tex": "\\tau_X,\\tau_Y\\text{ are topologies}", "relation": "implies", "rule": "topology", "conditions": [], "explanation": { "zh-Hant": "三項鏈含兩端且聯交封閉，兩因子各自合法。", "en": "Each three-member chain contains endpoints and is union/intersection closed, qualifying both factors." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1" ], "explanation": { "zh-Hant": "有序點對的兩座標型別固定。", "en": "Both coordinate types of the ordered pairs are fixed." } }, "leaf_basis": "topology" }, { "id": "S02", "uses": [ "G1" ], "before_tex": "P=X\\times Y", "after_tex": "P=\\{a,b,c,d\\},\\quad |P|=2\\cdot2=4", "relation": "definition", "rule": "definition_here", "conditions": [], "explanation": { "zh-Hant": "完整 Cartesian 載體包含四個有序對。", "en": "The complete Cartesian carrier contains four ordered pairs." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1" ], "explanation": { "zh-Hant": "(0,1) 與 (1,0) 不合併。", "en": "(0,1) and (1,0) remain distinct." } }, "leaf_basis": "definition_here" }, { "id": "S03", "uses": [ "G1", "S01", "S02" ], "before_tex": "U\\times V,\\quad U\\in\\tau_X,V\\in\\tau_Y", "after_tex": "\\mathcal B=\\{\\varnothing,\\{a\\},\\{a,b\\},\\{a,c\\},P\\}", "relation": "definition", "rule": "definition_here", "conditions": [], "explanation": { "zh-Hant": "九個矩形去重成五項；任一因子空給空，四個非空組合如列。", "en": "Deduplicate nine rectangles into five sets: an empty factor gives empty, and the four nonempty combinations are listed." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "S01", "S02" ], "explanation": { "zh-Hant": "尚未把矩形族直接稱完整拓撲。", "en": "The rectangle family is not yet called the complete topology." } }, "leaf_basis": "definition_here" }, { "id": "S04", "uses": [ "S03" ], "before_tex": "\\tau_P=\\{\\bigcup\\mathcal C:\\mathcal C\\subseteq\\mathcal B\\}", "after_tex": "\\tau_P=\\{\\varnothing,\\{a\\},\\{a,b\\},\\{a,c\\},\\{a,b,c\\},P\\}", "relation": "definition", "rule": "definition_here", "conditions": [], "explanation": { "zh-Hant": "兩個不可比較矩形的聯新增 {a,b,c}；其餘聯給已列項，故六項完整。", "en": "The union of the two incomparable rectangles adds {a,b,c}; every other union is already listed, giving exactly six opens." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03" ], "explanation": { "zh-Hant": "特別核新增項並非矩形，避免漏聯。", "en": "The added member is not a rectangle, protecting against omitted unions." } }, "leaf_basis": "definition_here" }, { "id": "S05", "uses": [ "S04" ], "before_tex": "\\{a,b\\}\\cap\\{a,c\\}=\\{a\\},\\quad \\{a,b\\}\\cup\\{a,c\\}=\\{a,b,c\\}", "after_tex": "\\tau_P\\text{ is a topology}", "relation": "implies", "rule": "topology", "conditions": [], "explanation": { "zh-Hant": "除這對不可比較集外皆按包含取交聯；與新增三點集相交聯也留在清單，端點均在。", "en": "Except for the incomparable pair, inclusion determines unions/intersections; operations with the added three-point set also stay listed, and endpoints are present." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04" ], "explanation": { "zh-Hant": "有限族的任意聯可重複二元聯完成，已涵蓋全部。", "en": "Every arbitrary union in this finite family reduces to repeated binary unions." } }, "leaf_basis": "topology" }, { "id": "S06", "uses": [ "G1", "S04", "definition_here" ], "before_tex": "\\pi_1^{-1}(\\{0\\})=\\{a,b\\},\\quad\\pi_2^{-1}(\\{0\\})=\\{a,c\\}", "after_tex": "\\pi_1,\\pi_2\\text{ are continuous}", "relation": "implies", "rule": "definition_here", "conditions": [], "explanation": { "zh-Hant": "另兩個因子開集為空與全體，逆像是空與 P，所以全部逆像開。", "en": "The remaining factor opens are empty and whole, with preimages empty and P; every required preimage is open." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "S04", "definition_here" ], "explanation": { "zh-Hant": "每個投影的全部三個目標開集已核完。", "en": "All three target opens have been checked for each projection." } }, "leaf_basis": "definition_here" }, { "id": "S07", "uses": [ "S03", "S04", "definition_here" ], "before_tex": "h(z)=(f(z),g(z)),\\quad f,g\\text{ continuous}", "after_tex": "h^{-1}(U\\times V)=f^{-1}(U)\\cap g^{-1}(V)\\text{ open}", "relation": "implies", "rule": "definition_here", "conditions": [ "f:Z\\to X,\\ g:Z\\to Y\\text{ continuous}" ], "explanation": { "zh-Hant": "對任意有拓撲的有限來源 Z，連續座標的逆像交開；再對任意矩形聯取逆像聯。", "en": "For a finite topological source Z, continuous coordinates give open preimage intersections; preimages commute with arbitrary rectangle unions." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S04", "definition_here" ], "explanation": { "zh-Hant": "需先核 Z 的拓撲與兩座標連續，這是條件命題。", "en": "The topology of Z and continuity of both coordinates are premises of this conditional argument." } }, "leaf_basis": "definition_here" }, { "id": "S08", "uses": [ "S06", "S07", "definition_here" ], "before_tex": "h\\text{ continuous}", "after_tex": "f=\\pi_1\\circ h,\\quad g=\\pi_2\\circ h\\text{ continuous}", "relation": "implies", "rule": "definition_here", "conditions": [ "h:Z\\to P\\text{ continuous}" ], "explanation": { "zh-Hant": "連續映射合成連續，因逆像是連續逆像的合成；結合前步得座標判準。", "en": "Composition preserves continuity because preimages compose; together with the previous step this gives the coordinate criterion." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "S07", "definition_here" ], "explanation": { "zh-Hant": "判準針對完整六項乘積拓撲，不是未封閉矩形清單。", "en": "The criterion uses the complete six-open product topology, not the unclosed rectangle list." } }, "leaf_basis": "definition_here" } ], "result": { "zh-Hant": "乘積拓撲恰六項，兩投影連續；有限來源的成對映射具座標連續判準。", "en": "The product has exactly six opens and continuous projections; paired maps from finite sources satisfy the coordinate criterion." }, "negative_controls": [ { "claim": { "zh-Hant": "全部開集恰為五個矩形。", "en": "The five rectangles are all opens." }, "expected": "reject", "reason": { "zh-Hant": "{a,b,c} 是兩矩形之聯且不是矩形。", "en": "{a,b,c} is a union of two rectangles and is not itself a rectangle." } }, { "claim": { "zh-Hant": "Cartesian 對不分順序。", "en": "Cartesian pairs are unordered." }, "expected": "reject", "reason": { "zh-Hant": "b 與 c 是不同點。", "en": "b and c are different points." } }, { "claim": { "zh-Hant": "有限 product=box 表示無限亦相同。", "en": "Finite product=box implies equality for infinite products." }, "expected": "reject", "reason": { "zh-Hant": "在可數無限乘積 R^ℕ（所有實數序列），box 開集 ∏_{n∈ℕ}(-1,1) 非 product-open：任何含零的基本 product 鄰域只限制有限座標，可把未限制座標改成 2。", "en": "In the countably infinite product of R indexed by the natural numbers (the real-sequence space), the box-open product with every factor (-1,1) is not product-open: any basic product neighborhood of zero restricts only finitely many coordinates, so an unrestricted coordinate can be changed to 2." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "mko_definition", "kind": "cited", "statement": { "zh-Hant": "二因子finite models先驗每個factor topology，再建立完整Cartesian carrier與全部U×V basic rectangles，取任意聯集生成product topology。投影preimages必為open；對有限source Z，map z↦(f(z),g(z))連續恰當兩個coordinate maps連續。有限乘積與box topology相同，不能把這點外推到無限乘積；任意一族rectangles本身通常不是完整topology。", "en": "In the two-factor exact families qualify each complete finite factor topology, build the full typed Cartesian carrier and every open rectangle U×V, then close under arbitrary unions to obtain the product topology. Each projection has open preimages. A map z↦(f(z),g(z)) from a finite source is continuous exactly when both coordinates are continuous. Finite product and box topologies agree here, but not by an automatic inference for infinite products. A family of basic rectangles is usually not itself the complete topology." }, "source": { "mko_id": "mko-finite-product-topology", "sha256": "9650dd3cf37e29ce4d8b2c6600ead8cb5c0ac577af1f7b69a0aa987341b02d5a", "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f" } }, { "id": "sets", "kind": "prerequisite", "statement": { "zh-Hant": "集合的聯、交、補集、包含與函數逆像以逐元素定義計算。", "en": "Compute unions, intersections, complements, inclusion and preimages element by element." } }, { "id": "topology", "kind": "prerequisite", "statement": { "zh-Hant": "拓撲含空集與全集，對任意聯及有限交封閉；有限集合族可完整核對。", "en": "A topology contains empty and whole sets and is closed under arbitrary unions and finite intersections; a finite family admits exhaustive checks." } }, { "id": "definition_here", "kind": "prerequisite", "statement": { "zh-Hant": "Cartesian 載體由有序對構成；乘積開集為任意開矩形聯；投影取指定座標；連續表示每個目標開集逆像開。", "en": "The Cartesian carrier consists of ordered pairs; product opens are arbitrary unions of open rectangles; a projection selects its designated coordinate; continuity requires every target-open preimage to be open." } } ], "revision_log": [], "numeric_checks": [ { "id": "product-size", "lhs": "2*2", "rhs": "4", "step": "S02" } ] }, "presentation": { "md_scope": { "zh-Hant": "固定兩個非空有限因子；不處理零因子或外推無限 box 拓撲。", "en": "Two nonempty finite factors; no zero-factor convention or infinite-box extrapolation." }, "md_symbols": [ [ "P", "Cartesian 載體", "Cartesian carrier" ], [ "\\mathcal B", "矩形基底", "rectangle basis" ], [ "\\pi_i", "座標投影", "coordinate projection" ] ], "md_error_controls": [ { "title": "聯閉包 / Union closure", "zh": "先建矩形，再取全部聯，不能停在基底。", "en": "Build rectangles and then all their unions, not just the basis." }, { "title": "型別與方向 / Types and direction", "zh": "投影逆像來自對應座標開集。", "en": "Projection preimages use the corresponding coordinate's opens." } ], "md_open_items": [] }, "revision": { "id": "ocme-new142-reader-adoption-20261021", "collection": "coverage-20261021", "source_commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f", "adopted_source_sha256": "b45b64af72af44b80a818bcda104e7319061c6713cd7fd49a7869539b4038479", "original_source_sha256": "b45b64af72af44b80a818bcda104e7319061c6713cd7fd49a7869539b4038479", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "reader_review": "142_of_142_source_bound", "wording_review": "original_Lead_decisions_resolved", "glossary_reader_paraphrases": [], "independent_AI_concurrence": "SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW", "issue_ids": [], "parent_manifest_sha256": "d29f4cae8229ebd24f60d496cb1d3cd4ad8c0a4dca9d44daab0c3245c544283a", "scope": "Owner-adopted reader revision. Lead and bilingual review concern the recorded text and premises; no new universal or Lean proof, canonical promotion or historical-label rewrite." } }