{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-real-banach-space-01", "mko_id": "mko-real-banach-space", "title": { "zh-Hant": "實 Banach 空間：(ℝ², ℓ¹) 的範數、誘導度量與完備性", "en": "Real Banach space: norm, induced metric and completeness of (ℝ², ℓ¹)" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-real-banach-space-01", "source": { "mko_id": "mko-real-banach-space", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "1c66f08e81b1ae1e47f887e3890cc179fd0aa8071970f275641c4c720de626ca", "english_sha256": "2a5bb4c0c274231ae79955cc73e08fe30d45fea796fa77c3d8ca5213dc75c4f4" }, "mode": "theorem_proof", "audience_prerequisites": [ "abs_rules", "order_basics", "vector_space_axioms", "completeness_axiom", "numeric_evaluation" ], "goal": { "zh-Hant": "在 V=ℝ² 上取 ℓ¹ 範數 ‖x‖₁=|x₁|+|x₂| 與同一範數誘導的度量 d₁(x,y)=‖x−y‖₁。（一）證明 ‖·‖₁ 是範數、d₁ 是度量，並算出 d₁((3,−1),(1,2))。（二）由實數的 Cauchy 完備性完整證明 (ℝ²,d₁) 完備，所以 (ℝ²,‖·‖₁) 是實 Banach 空間；並引用有限維實賦範空間完備的命名定理。（三）證明 ‖·‖₁ 不滿足平行四邊形律，所以不是由任何內積誘導。", "en": "On V=ℝ² take the ℓ¹ norm ‖x‖₁=|x₁|+|x₂| and the metric induced by that same norm, d₁(x,y)=‖x−y‖₁. (1) Prove that ‖·‖₁ is a norm and d₁ a metric, and compute d₁((3,−1),(1,2)). (2) Prove completely, from the Cauchy completeness of the reals, that (ℝ²,d₁) is complete, so (ℝ²,‖·‖₁) is a real Banach space; also cite the named theorem that finite-dimensional real normed spaces are complete. (3) Prove that ‖·‖₁ violates the parallelogram law and hence is induced by no inner product." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "V=ℝ²，加法與實數倍數逐坐標進行，是實向量空間。", "en": "V=ℝ² with coordinatewise addition and real scalar multiplication is a real vector space." } }, { "id": "G2", "statement": { "zh-Hant": "‖x‖₁=|x₁|+|x₂|；度量取同一範數誘導的 d₁(x,y)=‖x−y‖₁，不使用其他度量。", "en": "‖x‖₁=|x₁|+|x₂|; the metric is the one induced by this same norm, d₁(x,y)=‖x−y‖₁, and no other metric is used." } }, { "id": "G3", "statement": { "zh-Hant": "數值例：x=(3,−1)、y=(1,2)；平行四邊形律的例子 u=(1,0)、v=(0,1)。", "en": "Numerical example: x=(3,−1) and y=(1,2); for the parallelogram law, u=(1,0) and v=(0,1)." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "G2", "mko_banach_def" ], "before_tex": "(\\mathbb{R}^2,\\|\\cdot\\|_1)\\ \\text{Banach}", "after_tex": "\\|\\cdot\\|_1\\ \\text{norm}\\ \\land\\ d_1=\\|x-y\\|_1\\ \\text{metric}\\ \\land\\ (\\mathbb{R}^2,d_1)\\ \\text{complete}", "relation": "definition", "rule": "mko_banach_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO，Banach 結構把同一個載體 ℝ²、它的向量運算、範數 ‖·‖₁、由這個範數誘導的度量 d₁ 與完備性綁在一起。所以要證三件事：‖·‖₁ 是範數（S02 到 S05）、d₁ 是度量（S06）、(ℝ²,d₁) 完備（S08 到 S12）。", "en": "By the MKO, a Banach structure binds the one carrier ℝ², its vector operations, the norm ‖·‖₁, the metric d₁ induced by that norm and completeness. So three things must be proved: ‖·‖₁ is a norm (S02 to S05), d₁ is a metric (S06), and (ℝ²,d₁) is complete (S08 to S12)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_banach_def", "G2" ], "explanation": { "zh-Hant": "只展開定義；G2 固定度量必須來自同一範數，S07 的 ℓ^∞ 對照說明換了度量就不是同一結構。", "en": "It only unfolds the definition; G2 fixes that the metric comes from the same norm, and the ℓ^∞ contrast in S07 shows that a different metric gives a different structure." } }, "leaf_basis": "mko_banach_def + def_unfold" }, { "id": "S02", "uses": [ "G2", "abs_rules" ], "before_tex": "\\|x\\|_1=|x_1|+|x_2|=0", "after_tex": "x=(0,0);\\ \\ \\text{and}\\ \\|x\\|_1\\ge 0\\ \\text{for all}\\ x", "relation": "equivalent", "rule": "abs_rules", "conditions": [], "explanation": { "zh-Hant": "|x₁|≥0、|x₂|≥0，所以和 ≥0。兩個非負數的和為 0 當且僅當兩者都為 0（若其一為正，和就為正），而 |t|=0 當且僅當 t=0，所以 ‖x‖₁=0 ⇔ x₁=x₂=0 ⇔ x=0。", "en": "|x₁|≥0 and |x₂|≥0, so the sum is ≥0. A sum of two nonnegative numbers is 0 exactly when both are 0 (if one were positive the sum would be positive), and |t|=0 exactly when t=0, so ‖x‖₁=0 ⇔ x₁=x₂=0 ⇔ x=0." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "abs_rules", "order_basics" ], "explanation": { "zh-Hant": "反向：x=(0,0) 時 |0|+|0|=0。正向用「非負數之和為 0 ⇒ 各項為 0」，由順序公理得到。", "en": "Backward: for x=(0,0), |0|+|0|=0. The forward direction uses 'a sum of nonnegatives is 0 ⇒ each is 0', from the order axioms." } }, "leaf_basis": "abs_rules + order_basics" }, { "id": "S03", "uses": [ "G1", "G2", "abs_rules" ], "before_tex": "\\|\\lambda x\\|_1=|\\lambda x_1|+|\\lambda x_2|", "after_tex": "\\|\\lambda x\\|_1=|\\lambda|(|x_1|+|x_2|)=|\\lambda|\\,\\|x\\|_1", "relation": "equivalent", "rule": "abs_rules", "conditions": [], "explanation": { "zh-Hant": "λx 的坐標是 λx₁、λx₂（G1）。依 |uv|=|u||v| 得 |λx_k|=|λ||x_k|，再用分配律提出 |λ|。對每個實數 λ 與每個 x 成立。", "en": "The coordinates of λx are λx₁ and λx₂ (G1). By |uv|=|u||v|, |λx_k|=|λ||x_k|, and distributivity pulls out |λ|. This holds for every real λ and every x." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "abs_rules", "distributive" ], "explanation": { "zh-Hant": "所用都是恆等式，兩邊同值；N3 在一個整數格上作有限核對，只是輔助。", "en": "Only identities are used, so both sides have the same value; N3 is a finite check on an integer grid and is only auxiliary." } }, "leaf_basis": "abs_rules + distributive" }, { "id": "S04", "uses": [ "G1", "G2", "abs_rules" ], "before_tex": "\\|x+y\\|_1=|x_1+y_1|+|x_2+y_2|", "after_tex": "\\|x+y\\|_1\\le(|x_1|+|y_1|)+(|x_2|+|y_2|)=\\|x\\|_1+\\|y\\|_1", "relation": "implies", "rule": "abs_rules", "conditions": [], "explanation": { "zh-Hant": "對每個坐標用實數三角不等式 |x_k+y_k|≤|x_k|+|y_k|，兩個不等式相加（順序公理允許同向不等式相加），再重新分組得 ‖x‖₁+‖y‖₁。", "en": "Apply the real triangle inequality |x_k+y_k|≤|x_k|+|y_k| in each coordinate, add the two inequalities (the order axioms allow adding inequalities in the same direction), and regroup to get ‖x‖₁+‖y‖₁." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "abs_rules", "order_basics", "add_assoc", "add_comm" ], "explanation": { "zh-Hant": "這是不等式，只有一個方向；等號何時成立不在本稿宣稱範圍內。N4 是有限格點核對，不是證明。", "en": "This is an inequality and goes one way only; when equality holds is not claimed. N4 is a finite grid check, not a proof." } }, "leaf_basis": "abs_rules + order_basics + add_assoc + add_comm" }, { "id": "S05", "uses": [ "S02", "S03", "S04", "mko_norm_def" ], "before_tex": "\\text{S02},\\ \\text{S03},\\ \\text{S04}", "after_tex": "(\\mathbb{R}^2,\\|\\cdot\\|_1)\\ \\text{is a real normed vector space}", "relation": "implies", "rule": "mko_norm_def", "conditions": [], "explanation": { "zh-Hant": "非負與正定（S02）、齊次（S03）、三角不等式（S04）正是 MKO 範數定義的三項，所以 ‖·‖₁ 是 ℝ² 上的範數。", "en": "Nonnegativity and definiteness (S02), homogeneity (S03) and the triangle inequality (S04) are exactly the three clauses of the MKO's norm definition, so ‖·‖₁ is a norm on ℝ²." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S02", "S03", "S04", "mko_norm_def" ], "explanation": { "zh-Hant": "目標：‖·‖₁ 是範數。足夠條件：三項公理；三項都已正向證得，沒有把目標當作前提。", "en": "Goal: ‖·‖₁ is a norm. Sufficient condition: the three axioms; all three are proved forwards, and the goal is never used as a premise." } }, "leaf_basis": "mko_norm_def + and_intro_elim" }, { "id": "S06", "uses": [ "S05", "G2" ], "before_tex": "d_1(x,y)=\\|x-y\\|_1", "after_tex": "d_1\\ \\text{is a metric on}\\ \\mathbb{R}^2", "relation": "theorem_application", "rule": "L_induced_metric", "conditions": [], "explanation": { "zh-Hant": "依引理 L_induced_metric（任一範數誘導的 d(x,y)=‖x−y‖ 是度量），d₁ 是度量。引理在子步 S06.1 到 S06.3 只用 S02 到 S04 的三項範數公理證明。", "en": "By Lemma L_induced_metric (for any norm, d(x,y)=‖x−y‖ is a metric), d₁ is a metric. The lemma is proved in sub-steps S06.1 to S06.3 using only the three norm axioms of S02 to S04." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S05", "L_induced_metric" ], "explanation": { "zh-Hant": "需要 ‖·‖₁ 是範數（S05）；反向（每個度量都來自範數）不成立，見負向控制。", "en": "It needs ‖·‖₁ to be a norm (S05); the converse (every metric comes from a norm) is false, see the negative controls." } }, "leaf_basis": "metric_def + mko_norm_def" }, { "id": "S06.1", "uses": [ "S02" ], "before_tex": "d(x,y)=\\|x-y\\|", "after_tex": "d(x,y)\\ge 0,\\ \\ d(x,y)=0\\iff x-y=0\\iff x=y", "relation": "equivalent", "rule": "mko_norm_def", "conditions": [], "explanation": { "zh-Hant": "範數非負，所以 d≥0；範數為 0 當且僅當向量為 0，而 x−y=0 當且僅當 x=y（兩邊加 y）。", "en": "A norm is nonnegative, so d≥0; a norm is 0 exactly for the zero vector, and x−y=0 exactly when x=y (add y to both sides)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_norm_def", "vector_space_axioms" ], "explanation": { "zh-Hant": "每個 ⇔ 都由範數公理或向量加法的可逆性給出，雙向成立。", "en": "Each ⇔ comes from a norm axiom or from invertibility of vector addition, so both directions hold." } }, "leaf_basis": "mko_norm_def + vector_space_axioms" }, { "id": "S06.2", "uses": [ "S03" ], "before_tex": "d(y,x)=\\|y-x\\|=\\|(-1)(x-y)\\|", "after_tex": "d(y,x)=|-1|\\,\\|x-y\\|=d(x,y)", "relation": "equivalent", "rule": "mko_norm_def", "conditions": [], "explanation": { "zh-Hant": "y−x=(−1)(x−y)（向量空間公理），由齊次性 ‖(−1)z‖=|−1|‖z‖=‖z‖，所以 d 對稱。", "en": "y−x=(−1)(x−y) (vector space axioms), and by homogeneity ‖(−1)z‖=|−1|‖z‖=‖z‖, so d is symmetric." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_norm_def", "vector_space_axioms", "abs_rules" ], "explanation": { "zh-Hant": "等式鏈每一步都是等值改寫，雙向成立。", "en": "Every link of the chain is an equal rewriting, so both directions hold." } }, "leaf_basis": "mko_norm_def + vector_space_axioms + abs_rules" }, { "id": "S06.3", "uses": [ "S04" ], "before_tex": "x-z=(x-y)+(y-z)", "after_tex": "d(x,z)\\le d(x,y)+d(y,z)", "relation": "implies", "rule": "mko_norm_def", "conditions": [], "explanation": { "zh-Hant": "把 x−z 拆成 (x−y)+(y−z)，再用範數三角不等式。S06.1 到 S06.3 合起來證得引理 L_induced_metric：非負與分離、對稱、三角不等式都成立。", "en": "Split x−z as (x−y)+(y−z) and apply the triangle inequality of the norm. S06.1 to S06.3 together prove Lemma L_induced_metric: nonnegativity and separation, symmetry and the triangle inequality all hold." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_norm_def", "vector_space_axioms" ], "explanation": { "zh-Hant": "只需範數三角不等式與向量加法的結合律；不宣稱反向。", "en": "It needs only the norm triangle inequality and associativity of vector addition; no converse is claimed." } }, "leaf_basis": "mko_norm_def + vector_space_axioms + metric_def" }, { "id": "S07", "uses": [ "G2", "G3" ], "before_tex": "x=(3,-1),\\ y=(1,2)", "after_tex": "d_1(x,y)=|2|+|-3|=5;\\ \\ \\max(|2|,|-3|)=3\\neq 5", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "x−y=(2,−3)，d₁(x,y)=2+3=5。對照：ℓ^∞ 範數給 max(2,3)=3。兩個數不同，說明 ℓ^∞ 度量不是 ‖·‖₁ 誘導的度量；MKO 要求 norm 與 metric 不可來自不同公式。", "en": "x−y=(2,−3), so d₁(x,y)=2+3=5. For contrast, the ℓ^∞ norm gives max(2,3)=3. The two numbers differ, which shows the ℓ^∞ metric is not the metric induced by ‖·‖₁; the MKO forbids taking norm and metric from different formulas." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G2", "numeric_evaluation" ], "explanation": { "zh-Hant": "精確計算（N1、N2）；只說明這兩個度量不同，不推出任何完備性結論。", "en": "Exact computation (N1, N2); it only shows the two metrics differ and implies nothing about completeness." } }, "leaf_basis": "numeric_evaluation + abs_rules" }, { "id": "S08", "uses": [ "G2" ], "before_tex": "x,y\\in\\mathbb{R}^2,\\ k\\in\\{1,2\\}", "after_tex": "|x_k-y_k|\\le d_1(x,y)", "relation": "implies", "rule": "L_coord_le", "conditions": [], "explanation": { "zh-Hant": "完備性證明的關鍵引理：每個坐標的距離都不超過 d₁ 距離（引理 L_coord_le）。證明在子步 S08.1。", "en": "The key lemma for completeness: the distance in each coordinate is at most the d₁ distance (Lemma L_coord_le). The proof is in sub-step S08.1." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_coord_le" ], "explanation": { "zh-Hant": "只宣稱一個方向的不等式；坐標距離小並不推出 d₁ 小，除非兩個坐標都小（S11 處理）。", "en": "Only a one-way inequality is claimed; a small coordinate distance does not make d₁ small unless both coordinates are small (handled in S11)." } }, "leaf_basis": "abs_rules + order_basics" }, { "id": "S08.1", "uses": [ "G2", "abs_rules" ], "before_tex": "d_1(x,y)=|x_1-y_1|+|x_2-y_2|", "after_tex": "d_1(x,y)\\ge|x_1-y_1|\\ \\text{and}\\ d_1(x,y)\\ge|x_2-y_2|", "relation": "implies", "rule": "order_basics", "conditions": [], "explanation": { "zh-Hant": "因 |x₂−y₂|≥0，d₁(x,y)=|x₁−y₁|+|x₂−y₂|≥|x₁−y₁|+0；同理把第一項換成 0 得第二個不等式。這證得引理 L_coord_le。", "en": "Since |x₂−y₂|≥0, d₁(x,y)=|x₁−y₁|+|x₂−y₂|≥|x₁−y₁|+0; replacing the first term by 0 likewise gives the second inequality. This proves Lemma L_coord_le." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "abs_rules", "order_basics" ], "explanation": { "zh-Hant": "只用 |t|≥0 與「u≥0 ⇒ s+u≥s」。", "en": "It uses only |t|≥0 and 'u≥0 ⇒ s+u≥s'." } }, "leaf_basis": "abs_rules + order_basics + add_zero" }, { "id": "S09", "uses": [ "S08", "cauchy_def" ], "before_tex": "(x^{(m)})\\ \\text{Cauchy in}\\ (\\mathbb{R}^2,d_1)", "after_tex": "(x^{(m)}_k)_{m\\ge1}\\ \\text{Cauchy in}\\ \\mathbb{R}\\ \\ (k=1,2)", "relation": "implies", "rule": "cauchy_def", "conditions": [], "explanation": { "zh-Hant": "取任意一個 d₁-Cauchy 序列（不指定具體數值）。給定 ε>0，由 Cauchy 定義取 N，使 m,p≥N 時 d₁(x^{(m)},x^{(p)})<ε；依 L_coord_le，|x^{(m)}_k−x^{(p)}_k|≤d₁(x^{(m)},x^{(p)})<ε。同一個 N 適用於兩個坐標，所以每個坐標序列都是實數 Cauchy 序列。", "en": "Take an arbitrary d₁-Cauchy sequence (no specific values). Given ε>0, the Cauchy definition gives N with d₁(x^{(m)},x^{(p)})<ε for m,p≥N; by L_coord_le, |x^{(m)}_k−x^{(p)}_k|≤d₁(x^{(m)},x^{(p)})<ε. The same N works for both coordinates, so each coordinate sequence is a Cauchy sequence of reals." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S08", "cauchy_def", "order_basics" ], "explanation": { "zh-Hant": "用到 ≤ 與 < 的傳遞性；序列是任意的，S12 才能全稱化。", "en": "It uses transitivity of ≤ and <; the sequence is arbitrary, which lets S12 generalize." } }, "leaf_basis": "cauchy_def + L_coord_le + order_basics" }, { "id": "S10", "uses": [ "S09", "real_cauchy_complete" ], "before_tex": "(x^{(m)}_k)_{m\\ge1}\\ \\text{Cauchy in}\\ \\mathbb{R}", "after_tex": "x^{(m)}_k\\to a_k\\in\\mathbb{R}\\ (k=1,2),\\ \\ a=(a_1,a_2)\\in\\mathbb{R}^2", "relation": "theorem_application", "rule": "real_cauchy_complete", "conditions": [], "explanation": { "zh-Hant": "依實數的 Cauchy 完備性，兩個坐標序列各自收斂到某個實數 a₁、a₂。令 a=(a₁,a₂)；a 是 ℝ² 的點，所以候選極限仍在空間內。這是本稿唯一從外部引入的分析事實。", "en": "By the Cauchy completeness of the reals, the two coordinate sequences converge to some reals a₁ and a₂. Put a=(a₁,a₂); a is a point of ℝ², so the candidate limit stays in the space. This is the only analytic fact the guide imports." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S09", "real_cauchy_complete" ], "explanation": { "zh-Hant": "前提是坐標序列為 Cauchy（S09）；real_cauchy_complete 是前置規則，未在本稿證明，見待處理項。", "en": "The premise is that the coordinate sequences are Cauchy (S09); real_cauchy_complete is a prerequisite not proved here, see the open items." } }, "leaf_basis": "real_cauchy_complete" }, { "id": "S11", "uses": [ "S10", "cauchy_def", "G2" ], "before_tex": "x^{(m)}_1\\to a_1,\\ x^{(m)}_2\\to a_2", "after_tex": "d_1(x^{(m)},a)\\to 0", "relation": "implies", "rule": "cauchy_def", "conditions": [], "explanation": { "zh-Hant": "要證 x^{(m)} 在 d₁ 下收斂到 a。d₁(x^{(m)},a) 是兩個坐標誤差之和，所以讓每個誤差小於 ε/2 即可。子步 S11.1 選出共同門檻，S11.2 完成估計。", "en": "We must show x^{(m)}→a in d₁. Since d₁(x^{(m)},a) is the sum of the two coordinate errors, it suffices to make each error less than ε/2. Sub-step S11.1 picks a common threshold and S11.2 completes the estimate." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10", "cauchy_def" ], "explanation": { "zh-Hant": "需要兩個坐標都收斂；只有一個坐標收斂不足以推出 d₁ 收斂。", "en": "Both coordinates must converge; convergence of one coordinate alone would not give convergence in d₁." } }, "leaf_basis": "cauchy_def + seq_limit_def + abs_rules" }, { "id": "S11.1", "uses": [ "S10", "seq_limit_def" ], "before_tex": "\\varepsilon>0;\\ \\exists N_k:\\ m\\ge N_k\\Rightarrow|x^{(m)}_k-a_k|<\\tfrac{\\varepsilon}{2}", "after_tex": "N=\\max(N_1,N_2):\\ m\\ge N\\Rightarrow|x^{(m)}_1-a_1|<\\tfrac{\\varepsilon}{2}\\ \\land\\ |x^{(m)}_2-a_2|<\\tfrac{\\varepsilon}{2}", "relation": "implies", "rule": "seq_limit_def", "conditions": [ "\\varepsilon>0" ], "explanation": { "zh-Hant": "給定 ε>0，則 ε/2>0；依實數列極限定義，對每個 k 有 N_k 使 m≥N_k 時坐標誤差小於 ε/2。取 N=max(N₁,N₂)，則 m≥N 同時蘊含 m≥N₁ 與 m≥N₂。", "en": "Given ε>0, also ε/2>0; by the definition of the limit of a real sequence, for each k there is N_k such that the coordinate error is below ε/2 for m≥N_k. Take N=max(N₁,N₂); then m≥N implies both m≥N₁ and m≥N₂." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "seq_limit_def", "order_basics" ], "explanation": { "zh-Hant": "需要 ε/2>0（正數除以 2 仍為正）與最大值的性質；只有兩個坐標，所以 max 有限。", "en": "It needs ε/2>0 (a positive number halved is positive) and the property of the maximum; there are only two coordinates, so the max is finite." } }, "leaf_basis": "seq_limit_def + order_basics" }, { "id": "S11.2", "uses": [ "S11.1", "G2" ], "before_tex": "m\\ge N", "after_tex": "d_1(x^{(m)},a)=|x^{(m)}_1-a_1|+|x^{(m)}_2-a_2|<\\tfrac{\\varepsilon}{2}+\\tfrac{\\varepsilon}{2}=\\varepsilon", "relation": "implies", "rule": "order_basics", "conditions": [], "explanation": { "zh-Hant": "兩個嚴格不等式相加得和小於 ε/2+ε/2=ε（N7 核對這個代數恆等式）。ε 是任意正數，依收斂定義 x^{(m)}→a。", "en": "Adding the two strict inequalities gives a sum below ε/2+ε/2=ε (N7 checks this algebraic identity). Since ε is an arbitrary positive number, x^{(m)}→a by the definition of convergence." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S11.1", "order_basics", "cauchy_def" ], "explanation": { "zh-Hant": "用到同向嚴格不等式可相加；ε 未被指定，所以結論對每個 ε 成立。", "en": "It uses that strict inequalities in the same direction add; ε was not specified, so the conclusion holds for every ε." } }, "leaf_basis": "order_basics + frac_rules + cauchy_def" }, { "id": "S12", "uses": [ "S05", "S06", "S09", "S10", "S11", "mko_complete_def", "mko_banach_def" ], "before_tex": "\\text{arbitrary Cauchy}\\ (x^{(m)})\\ \\text{converges in}\\ (\\mathbb{R}^2,d_1)", "after_tex": "(\\mathbb{R}^2,d_1)\\ \\text{complete};\\ (\\mathbb{R}^2,\\|\\cdot\\|_1)\\ \\text{real Banach space}", "relation": "implies", "rule": "forall_intro", "conditions": [], "explanation": { "zh-Hant": "S09 到 S11 對任意 d₁-Cauchy 序列成立，所以依全稱引入，每個 Cauchy 序列都收斂到 ℝ² 中的點：(ℝ²,d₁) 完備。加上 S05 的範數與 S06 的誘導度量，三項都綁定同一結構，依 MKO，(ℝ²,‖·‖₁) 是實 Banach 空間。", "en": "S09 to S11 hold for an arbitrary d₁-Cauchy sequence, so by universal generalization every Cauchy sequence converges to a point of ℝ²: (ℝ²,d₁) is complete. Together with the norm of S05 and the induced metric of S06, all three parts bind one structure, so by the MKO (ℝ²,‖·‖₁) is a real Banach space." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S05", "S06", "S11", "forall_intro", "mko_complete_def", "mko_banach_def" ], "explanation": { "zh-Hant": "目標 G：Banach。足夠條件 H：範數、同一範數誘導的度量、完備；H⇒G 由 MKO 定義。H 由 S05、S06 與 S09 到 S11 正向證得，沒有用到 G，也沒有用到任何序列樣本。", "en": "Goal G: Banach. Sufficient condition H: a norm, the metric induced by that norm, completeness; H⇒G by the MKO's definition. H is proved forwards in S05, S06 and S09 to S11, using neither G nor any sampled sequence." } }, "leaf_basis": "forall_intro + mko_complete_def + mko_banach_def" }, { "id": "S13", "uses": [ "S12", "thm_findim_complete" ], "before_tex": "\\dim\\mathbb{R}^n=n<\\infty,\\ \\ \\|\\cdot\\|\\in\\{\\|\\cdot\\|_1,\\|\\cdot\\|_\\infty\\}", "after_tex": "(\\mathbb{R}^n,\\|\\cdot\\|)\\ \\text{Banach (named-theorem certificate)}", "relation": "theorem_application", "rule": "thm_findim_complete", "conditions": [ "n\\ \\text{finite}" ], "explanation": { "zh-Hant": "MKO 的證書形式是命名定理：每個有限維實賦範空間在其誘導度量下完備。前提是有限維與「度量由同一範數誘導」。S08 到 S12 是這個定理在 n=2、ℓ¹ 時的完整證明，只依賴實數完備性；ℓ^∞ 與一般 n 的情形本稿只引用，不另證（同樣的坐標論證也可行，但未寫出）。", "en": "The MKO's certificate is the named theorem: every finite-dimensional real normed space is complete in its induced metric. Its hypotheses are finite dimension and a metric induced by the same norm. S08 to S12 are a complete proof of the case n=2, ℓ¹, relying only on completeness of the reals; the ℓ^∞ and general-n cases are only cited here, not proved (the same coordinate argument would work but is not written out)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "thm_findim_complete", "S12" ], "explanation": { "zh-Hant": "定理的引用帶 citation_verified；有限維是必要的前提，無限維賦範空間可以不完備（見錯誤控制）。", "en": "The citation carries citation_verified; finite dimension is an essential hypothesis, and infinite-dimensional normed spaces can fail to be complete (see the error controls)." } }, "leaf_basis": "thm_findim_complete" }, { "id": "S14", "uses": [ "G3", "S12" ], "before_tex": "u=(1,0),\\ v=(0,1):\\ \\|u+v\\|_1^2+\\|u-v\\|_1^2=2^2+2^2=8", "after_tex": "8\\neq 4=2\\|u\\|_1^2+2\\|v\\|_1^2\\ \\Rightarrow\\ \\|\\cdot\\|_1\\ \\text{induced by no inner product}", "relation": "implies", "rule": "L_parallelogram", "conditions": [], "explanation": { "zh-Hant": "u+v=(1,1)、u−v=(1,−1)，ℓ¹ 範數都是 2，所以左邊 4+4=8；而 ‖u‖₁=‖v‖₁=1，右邊 2+2=4。依引理 L_parallelogram 的逆否命題，‖·‖₁ 不是由任何內積誘導。MKO 指出：Banach 的指定範數不自動由內積誘導，也不自動成為相容的 Hilbert 結構；不否認底層向量空間另有內積。引理在子步 S14.1、S14.2 證明。", "en": "u+v=(1,1) and u−v=(1,−1) both have ℓ¹ norm 2, so the left side is 4+4=8, while ‖u‖₁=‖v‖₁=1 makes the right side 2+2=4. By the contrapositive of Lemma L_parallelogram, ‖·‖₁ is induced by no inner product. The MKO notes that a Banach space's specified norm need not be induced by an inner product or define a compatible Hilbert structure; its underlying vector space may still carry another inner product. The lemma is proved in sub-steps S14.1 and S14.2." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_parallelogram", "contrapositive", "numeric_evaluation" ], "explanation": { "zh-Hant": "只用引理的必要方向（內積範數 ⇒ 平行四邊形律）；一組反例足以否定（N5、N6）。", "en": "Only the necessary direction of the lemma is used (inner-product norm ⇒ parallelogram law); one counterexample suffices (N5, N6)." } }, "leaf_basis": "contrapositive + numeric_evaluation" }, { "id": "S14.1", "uses": [ "inner_product_def" ], "before_tex": "\\langle u\\pm v,u\\pm v\\rangle", "after_tex": "\\langle u,u\\rangle\\pm 2\\langle u,v\\rangle+\\langle v,v\\rangle", "relation": "equivalent", "rule": "inner_product_def", "conditions": [], "explanation": { "zh-Hant": "用兩槽線性展開：⟨u±v,u±v⟩=⟨u,u⟩±⟨u,v⟩±⟨v,u⟩+⟨v,v⟩；再用對稱性 ⟨v,u⟩=⟨u,v⟩ 合併中間兩項。", "en": "Expand by linearity in both slots: ⟨u±v,u±v⟩=⟨u,u⟩±⟨u,v⟩±⟨v,u⟩+⟨v,v⟩; then symmetry ⟨v,u⟩=⟨u,v⟩ merges the two middle terms." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "inner_product_def" ], "explanation": { "zh-Hant": "展開是恆等式，兩邊同值。", "en": "The expansion is an identity, so both sides have the same value." } }, "leaf_basis": "inner_product_def + distributive" }, { "id": "S14.2", "uses": [ "S14.1" ], "before_tex": "\\langle u+v,u+v\\rangle+\\langle u-v,u-v\\rangle", "after_tex": "2\\langle u,u\\rangle+2\\langle v,v\\rangle,\\ \\text{i.e.}\\ \\|u+v\\|^2+\\|u-v\\|^2=2\\|u\\|^2+2\\|v\\|^2", "relation": "equivalent", "rule": "eq_replace", "conditions": [], "explanation": { "zh-Hant": "把 S14.1 的兩式相加，±2⟨u,v⟩ 互相抵消；再用 ‖z‖²=⟨z,z⟩ 改寫成範數。S14.1 與 S14.2 合起來證得引理 L_parallelogram。", "en": "Add the two expansions of S14.1: the terms ±2⟨u,v⟩ cancel; then rewrite with ‖z‖²=⟨z,z⟩. S14.1 and S14.2 together prove Lemma L_parallelogram." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S14.1", "add_inverse", "inner_product_def" ], "explanation": { "zh-Hant": "相加與改寫都是等值操作，雙向成立。", "en": "Adding and rewriting are equal-value operations, so both directions hold." } }, "leaf_basis": "eq_replace + add_inverse + inner_product_def" }, { "id": "S15", "uses": [ "S06", "S07", "S12", "S13", "S14" ], "before_tex": "\\text{S06},\\ \\text{S07},\\ \\text{S12},\\ \\text{S14}", "after_tex": "(\\mathbb{R}^2,\\|\\cdot\\|_1)\\ \\text{Banach};\\ d_1((3,-1),(1,2))=5;\\ \\|\\cdot\\|_1\\ \\text{not from an inner product}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "合併：‖·‖₁ 是範數，d₁ 是它誘導的度量，d₁((3,−1),(1,2))=5；(ℝ²,d₁) 完備，所以是實 Banach 空間；ℓ¹ 範數違反平行四邊形律，不來自內積。", "en": "Combine: ‖·‖₁ is a norm, d₁ is its induced metric and d₁((3,−1),(1,2))=5; (ℝ²,d₁) is complete, so the space is a real Banach space; the ℓ¹ norm violates the parallelogram law and does not come from an inner product." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S06", "S07", "S12", "S14", "and_intro_elim" ], "explanation": { "zh-Hant": "目標是三個任務的合取；各部分由所列步驟正向證得，H⇒G 由合取引入。", "en": "The goal is the conjunction of the three tasks; each part is proved forwards by the listed steps, and H⇒G by conjunction introduction." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "‖x‖₁=|x₁|+|x₂| 是 ℝ² 上的範數，d₁(x,y)=‖x−y‖₁ 是它誘導的度量，d₁((3,−1),(1,2))=5（ℓ^∞ 度量給 3，是不同的度量）。任意 d₁-Cauchy 序列的坐標都是實數 Cauchy 序列，收斂到 a₁、a₂，且 d₁(x^{(m)},a)→0，所以 (ℝ²,‖·‖₁) 是實 Banach 空間；一般有限維情形由命名定理給出（citation_verified）。ℓ¹ 範數在 u=(1,0)、v=(0,1) 違反平行四邊形律（8≠4），所以不由任何內積誘導。", "en": "‖x‖₁=|x₁|+|x₂| is a norm on ℝ², d₁(x,y)=‖x−y‖₁ is its induced metric, and d₁((3,−1),(1,2))=5 (the ℓ^∞ metric gives 3, a different metric). The coordinates of any d₁-Cauchy sequence are real Cauchy sequences converging to a₁ and a₂, and d₁(x^{(m)},a)→0, so (ℝ²,‖·‖₁) is a real Banach space; the general finite-dimensional case comes from the named theorem (citation_verified). The ℓ¹ norm violates the parallelogram law at u=(1,0), v=(0,1) (8≠4), so it is induced by no inner product." }, "negative_controls": [ { "claim": { "zh-Hant": "在 (ℝ²,‖·‖₁) 中測了五條 Cauchy 序列都收斂，所以它完備", "en": "Five Cauchy sequences in (ℝ²,‖·‖₁) were tested and all converged, so it is complete" }, "expected": "reject", "reason": { "zh-Hant": "結論雖真，但有限抽樣不是證明；c₀₀ 中抽樣也能全部收斂卻不完備。完備性來自 S09 到 S12。", "en": "The conclusion is true, but finite sampling is not a proof; samples in c₀₀ can all converge although c₀₀ is incomplete. Completeness comes from S09 to S12." } }, { "claim": { "zh-Hant": "(ℝ²,‖·‖₁) 配上 ℓ^∞ 度量就是這個 Banach 空間", "en": "(ℝ²,‖·‖₁) equipped with the ℓ^∞ metric is this Banach space" }, "expected": "reject", "reason": { "zh-Hant": "ℓ^∞ 度量不由 ‖·‖₁ 誘導：(3,−1) 與 (1,2) 的距離是 3，而 ‖x−y‖₁=5。", "en": "The ℓ^∞ metric is not induced by ‖·‖₁: the distance between (3,−1) and (1,2) is 3 while ‖x−y‖₁=5." } }, { "claim": { "zh-Hant": "d′(x,y)=min(1,‖x−y‖₁) 是 ℝ² 上的完備度量，所以 (ℝ²,d′) 是 Banach 空間", "en": "d′(x,y)=min(1,‖x−y‖₁) is a complete metric on ℝ², so (ℝ²,d′) is a Banach space" }, "expected": "reject", "reason": { "zh-Hant": "d′ 不由範數誘導：d′(2e₁,0)=1，而齊次性要求 2d′(e₁,0)=2。", "en": "d′ is induced by no norm: d′(2e₁,0)=1, while homogeneity would require 2d′(e₁,0)=2." } }, { "claim": { "zh-Hant": "每個 Banach 空間的範數都由某個內積誘導", "en": "The norm of every Banach space is induced by some inner product" }, "expected": "reject", "reason": { "zh-Hant": "ℓ¹ 範數在 u=(1,0)、v=(0,1) 違反平行四邊形律：8≠4。", "en": "The ℓ¹ norm violates the parallelogram law at u=(1,0), v=(0,1): 8≠4." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "abs_rules", "kind": "prerequisite", "statement": { "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|v 之一；u0 推出 uwvw。", "en": "Order of the reals: exactly one of uv holds; u0 give uwvw." } }, { "id": "vector_space_axioms", "kind": "prerequisite", "statement": { "zh-Hant": "向量空間公理（實數或其他域 F 上）：加法成交換群；純量乘法滿足 a(u+v)=au+av、(a+b)u=au+bu、(ab)u=a(bu)、1u=u。", "en": "Vector space axioms (over ℝ or another field F): vectors form a commutative group under addition; scalar multiplication satisfies a(u+v)=au+av, (a+b)u=au+bu, (ab)u=a(bu) and 1u=u." } }, { "id": "completeness_axiom", "kind": "prerequisite", "statement": { "zh-Hant": "實數完備性：ℝ 的每個非空且有上界的子集都有最小上界（上確界）。", "en": "Completeness of the reals: every nonempty subset of ℝ that is bounded above has a least upper bound (supremum)." } }, { "id": "numeric_evaluation", "kind": "prerequisite", "statement": { "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。", "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable." } }, { "id": "def_unfold", "kind": "prerequisite", "statement": { "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。", "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true." } }, { "id": "distributive", "kind": "prerequisite", "statement": { "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。", "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu." } }, { "id": "add_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。", "en": "Associativity of addition: (u+v)+w=u+(v+w)." } }, { "id": "add_comm", "kind": "prerequisite", "statement": { "zh-Hant": "加法交換律：u+v=v+u。", "en": "Commutativity of addition: u+v=v+u." } }, { "id": "and_intro_elim", "kind": "prerequisite", "statement": { "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。", "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q." } }, { "id": "metric_def", "kind": "prerequisite", "statement": { "zh-Hant": "度量：d:X×X→ℝ 滿足 d(x,y)≥0 且 d(x,y)=0 ⇔ x=y；d(x,y)=d(y,x)；三角不等式 d(x,z)≤d(x,y)+d(y,z)。", "en": "Metric: d:X×X→ℝ with d(x,y)≥0 and d(x,y)=0 ⇔ x=y; d(x,y)=d(y,x); triangle inequality d(x,z)≤d(x,y)+d(y,z)." } }, { "id": "add_zero", "kind": "prerequisite", "statement": { "zh-Hant": "加法單位元：u+0=u。", "en": "Additive identity: u+0=u." } }, { "id": "seq_limit_def", "kind": "prerequisite", "statement": { "zh-Hant": "數列極限：a_n→L 表示對每個 ε>0，存在 N 使所有 n≥N 都有 |a_n−L|<ε。", "en": "Limit of a sequence: a_n→L means that for every ε>0 there is N such that |a_n−L|<ε for all n≥N." } }, { "id": "frac_rules", "kind": "prerequisite", "statement": { "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。", "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0." } }, { "id": "forall_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。", "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain." } }, { "id": "contrapositive", "kind": "prerequisite", "statement": { "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。", "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q." } }, { "id": "eq_replace", "kind": "prerequisite", "statement": { "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。", "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true." } }, { "id": "add_inverse", "kind": "prerequisite", "statement": { "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。", "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)." } }, { "id": "mko_banach_def", "kind": "cited", "source": { "mko_id": "mko-real-banach-space", "sha256": "1c66f08e81b1ae1e47f887e3890cc179fd0aa8071970f275641c4c720de626ca", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "實 Banach 空間必須把同一載體、實向量運算、範數、由該範數誘導的度量及完備性綁成一體。本輪使用有限維 ℝⁿ 配命名 ℓ¹ 或 ℓ^∞ 範數的精確定理證書，不以序列抽樣證完備。", "en": "A real Banach space binds one carrier, its real vector operations, its norm, the metric induced by that norm and completeness into one structure. This slice uses exact named-theorem certificates for finite-dimensional ℝⁿ with ℓ¹ or ℓ^∞, not sampled sequences as completeness proofs." } }, { "id": "mko_norm_def", "kind": "cited", "source": { "mko_id": "mko-real-normed-vector-space", "sha256": "8f2547c8a3312254b34134e96a3083f313e6e7334ab0a87a1ba73cc147931964", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "範數把向量送到非負實數，且 norm(v)=0 當且僅當 v=0、norm(av)=|a| norm(v)、norm(u+v)≤norm(u)+norm(v)。", "en": "A norm sends vectors to nonnegative real numbers, with norm(v)=0 iff v=0, norm(av)=|a| norm(v), and norm(u+v)≤norm(u)+norm(v)." } }, { "id": "mko_complete_def", "kind": "cited", "source": { "mko_id": "mko-complete-metric-space", "sha256": "783e45783b1dc07ccecb89f0c76b128f6c4a9a3da036e1678e908f2250530c73", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "度量空間 (X,d) 完備，若 X 內每個 Cauchy 序列都收斂到某個仍屬 X 的點；抽樣若干序列不足以證一般完備性。", "en": "A metric space (X,d) is complete if every Cauchy sequence in X converges to some point of X; testing a few sequences does not establish general completeness." } }, { "id": "cauchy_def", "kind": "prerequisite", "statement": { "zh-Hant": "在度量空間 (X,d) 中，序列 (x^{(m)}) 是 Cauchy 序列，若對每個 ε>0 存在 N 使所有 m,p≥N 有 d(x^{(m)},x^{(p)})<ε；它收斂到 a，若對每個 ε>0 存在 N 使所有 m≥N 有 d(x^{(m)},a)<ε。", "en": "In a metric space (X,d), a sequence (x^{(m)}) is Cauchy if for every ε>0 there is N with d(x^{(m)},x^{(p)})<ε for all m,p≥N; it converges to a if for every ε>0 there is N with d(x^{(m)},a)<ε for all m≥N." } }, { "id": "real_cauchy_complete", "kind": "prerequisite", "statement": { "zh-Hant": "實數的 Cauchy 完備性：ℝ 中（以 |s−t| 為距離）每個 Cauchy 序列都收斂到某個實數。這是最小上界公理的標準推論，本稿不重證。", "en": "Cauchy completeness of the reals: every Cauchy sequence in ℝ (with distance |s−t|) converges to a real number. This is a standard consequence of the least-upper-bound axiom and is not reproved here." } }, { "id": "inner_product_def", "kind": "prerequisite", "statement": { "zh-Hant": "實內積 ⟨·,·⟩ 是兩槽線性、對稱、正定的實值函數；範數由它誘導，指 ‖x‖²=⟨x,x⟩ 對所有 x 成立。", "en": "A real inner product ⟨·,·⟩ is a real-valued function, linear in each slot, symmetric and positive definite; a norm is induced by it when ‖x‖²=⟨x,x⟩ for every x." } }, { "id": "thm_findim_complete", "kind": "cited", "source": { "reference": "Theo Buehler and Dietmar A. Salamon, Functional Analysis, 11 May 2016 notes, Corollary 1.2.7, p.16", "original_reference": "E. Kreyszig, Introductory Functional Analysis with Applications, Wiley, 1978, Section 2.4, Theorem 2.4-2", "url": "https://math.jhu.edu/~js/Math637/eth.functionalanalysis.pdf#page=24", "locator": "Corollary 1.2.7 and its proof", "checked_date": "2026-10-19", "verification": "verified_replacement", "verification_note": "Finite-dimensional normed spaces over the real or complex numbers; the guide uses real scalars." }, "needs_citation": false, "statement": { "zh-Hant": "每個有限維實賦範向量空間，在由其範數誘導的度量下都是完備的；特別地，對每個正整數 n，(ℝⁿ,‖·‖₁) 與 (ℝⁿ,‖·‖_∞) 都是實 Banach 空間。", "en": "Every finite-dimensional real normed vector space is complete in the metric induced by its norm; in particular, for every positive integer n, (ℝⁿ,‖·‖₁) and (ℝⁿ,‖·‖_∞) are real Banach spaces." } }, { "id": "L_induced_metric", "kind": "lemma", "proved_in": [ "S06.1", "S06.2", "S06.3" ], "statement": { "zh-Hant": "若 ‖·‖ 是實向量空間 V 上的範數，則 d(x,y)=‖x−y‖ 是 V 上的度量。", "en": "If ‖·‖ is a norm on a real vector space V, then d(x,y)=‖x−y‖ is a metric on V." } }, { "id": "L_coord_le", "kind": "lemma", "proved_in": [ "S08.1" ], "statement": { "zh-Hant": "對 x,y∈ℝ² 與 k=1,2：|x_k−y_k|≤d₁(x,y)。", "en": "For x,y∈ℝ² and k=1,2: |x_k−y_k|≤d₁(x,y)." } }, { "id": "L_parallelogram", "kind": "lemma", "proved_in": [ "S14.1", "S14.2" ], "statement": { "zh-Hant": "若範數由實內積誘導，則對所有 u,v：‖u+v‖²+‖u−v‖²=2‖u‖²+2‖v‖²。", "en": "If a norm is induced by a real inner product, then for all u,v: ‖u+v‖²+‖u−v‖²=2‖u‖²+2‖v‖²." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S07", "lhs": "abs(3-1)+abs(-1-2)", "rhs": "5" }, { "id": "N2", "step": "S07", "lhs": "max(abs(3-1),abs(-1-2))", "relation": "!=", "rhs": "5" }, { "id": "N3", "step": "S03", "vars": [ "l", "x1", "x2" ], "range": [ -3, 3 ], "lhs": "abs(l*x1)+abs(l*x2)", "rhs": "abs(l)*(abs(x1)+abs(x2))" }, { "id": "N4", "step": "S04", "vars": [ "x1", "x2", "y1", "y2" ], "range": [ -2, 2 ], "lhs": "abs(x1+y1)+abs(x2+y2)", "relation": "<=", "rhs": "abs(x1)+abs(x2)+abs(y1)+abs(y2)" }, { "id": "N5", "step": "S14", "lhs": "(abs(1)+abs(1))**2+(abs(1)+abs(-1))**2", "rhs": "8" }, { "id": "N6", "step": "S14", "lhs": "2*1**2+2*1**2", "relation": "!=", "rhs": "8" }, { "id": "N7", "step": "S11.2", "vars": [ "e" ], "degree": 1, "lhs": "e/2+e/2", "rhs": "e" }, { "id": "N8", "lhs": "min(1,2)", "relation": "!=", "rhs": "2*min(1,1)" } ] }, "presentation": { "md_scope": { "zh-Hant": "在 MKO 的範圍內工作：明示的有限維 ℝ²、命名的 ℓ¹ 範數，度量必須由同一範數誘導（假設 a1）；完備性以證明加命名定理證書處理，不以有限條 Cauchy 序列抽樣代替（假設 a2）。本稿完整證明 n=2、ℓ¹ 的特例，只假設實數的 Cauchy 完備性；一般有限維情形（含 ℓ^∞）引用命名定理。不處理無限維 Banach 空間或 Hilbert 空間理論，不改動 canonical MKO 或英文 edition。", "en": "Works inside the MKO's scope: the explicit finite-dimensional ℝ² with the named ℓ¹ norm, the metric induced by that same norm (assumption a1); completeness is handled by a proof plus a named-theorem certificate, not by sampling finitely many Cauchy sequences (assumption a2). The guide proves the special case n=2, ℓ¹ completely, assuming only Cauchy completeness of the reals; the general finite-dimensional case (including ℓ^∞) is cited from the named theorem. Infinite-dimensional Banach spaces and Hilbert-space theory are not treated, and the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "\\|x\\|_1=|x_1|+|x_2|", "ℝ² 上的 ℓ¹ 範數", "the ℓ¹ norm on ℝ²" ], [ "d_1(x,y)=\\|x-y\\|_1", "由同一範數誘導的度量", "the metric induced by the same norm" ], [ "\\|x\\|_\\infty=\\max(|x_1|,|x_2|)", "ℓ^∞ 範數，只作對照", "the ℓ^∞ norm, for contrast only" ], [ "(x^{(m)})_{m\\ge 1}", "ℝ² 中的序列，坐標 x^{(m)}_k", "a sequence in ℝ², with coordinates x^{(m)}_k" ], [ "a=(a_1,a_2)", "候選極限", "the candidate limit" ], [ "\\varepsilon,\\ N", "收斂與 Cauchy 定義中的容許誤差與門檻", "the tolerance and threshold in the limit and Cauchy definitions" ] ], "md_error_controls": [ { "title": "抽樣 Cauchy 序列不證完備 / Sampled Cauchy sequences do not prove completeness", "zh": "MKO 指出的誤解是測過幾條 Cauchy 序列就宣稱 Banach。完備性量化所有 Cauchy 序列。反例：有限支撐實數列空間 c₀₀ 配 ℓ¹ 範數，序列 x^{(m)} 的第 k 項在 k≤m 時為 2^{−k}、k>m 時為 0；它是 Cauchy（m

m; it is Cauchy (for m