{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-tends-to-relation-01",
  "mko_id": "mko-tends-to-relation",
  "title": {
    "zh-Hant": "趨近關係：x→a 不是 x=a，也不保證函數值收斂",
    "en": "Tends-to relation: x→a is not x=a, and does not guarantee that function values converge"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-tends-to-relation-01",
    "source": {
      "mko_id": "mko-tends-to-relation",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "2fbb3c700378e80e68833bcada18866ed6439803d8b639e304fe76d31903f70e",
      "english_sha256": "bed5d2c178f726aeda66b8705697852f39ce93b2d1a6807b7acbb3c97dfd4ac4"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "metric_def",
      "abs_rules",
      "order_basics",
      "archimedean",
      "seq_limit_def",
      "func_limit_def"
    ],
    "goal": {
      "zh-Hant": "在 (ℝ, d(x,y)=|x−y|) 中：(1) 證明數列 x_n=1+1/n 趨近 1，而每一項都有 x_n≠1；(2) 對 g(x)=1（x>1）、g(x)=0（x≤1），證明沿 x_n 有 g(x_n)→1，但 lim_{x→1} g(x) 不存在。由此說明 x→1 本身既不要求 x 等於 1，也不決定函數值是否收斂。",
      "en": "In (ℝ, d(x,y)=|x−y|): (1) prove that the sequence x_n=1+1/n tends to 1 while every term satisfies x_n≠1; (2) for g(x)=1 (x>1) and g(x)=0 (x≤1), prove that g(x_n)→1 along x_n, but lim_{x→1} g(x) does not exist. This shows that x→1 by itself neither requires x to equal 1 nor decides whether function values converge."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "結構：實數集 ℝ 配上度量 d(x,y)=|x−y|（假設 a1）；目標點 a=1。",
          "en": "Structure: the reals ℝ with the metric d(x,y)=|x−y| (assumption a1); target point a=1."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "數列 x_n=1+1/n，n 取遍所有正整數。",
          "en": "The sequence x_n=1+1/n, with n ranging over all positive integers."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "函數 g:ℝ→ℝ：x>1 時 g(x)=1，x≤1 時 g(x)=0。",
          "en": "The function g:ℝ→ℝ with g(x)=1 for x>1 and g(x)=0 for x≤1."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "\\text{a structure for nearness}",
        "after_tex": "(\\mathbb{R},d),\\ d(x,y)=|x-y|,\\ a=1",
        "relation": "definition",
        "rule": "metric_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 假設 a1，談趨近之前必須先指定能討論鄰近的結構。這裡取 ℝ 與度量 d(x,y)=|x−y|：它非負、只在 x=y 時為 0、對稱且滿足三角不等式（abs_rules）。目標點 a=1。之後「接近 1」都是指 |x−1| 小。",
          "en": "By MKO assumption a1, a structure supporting nearness must be specified before approach is discussed. Take ℝ with the metric d(x,y)=|x−y|: it is nonnegative, zero only when x=y, symmetric, and satisfies the triangle inequality (abs_rules). The target point is a=1. From here on 'close to 1' means that |x−1| is small."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "metric_def",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "此步只指定結構，不推出新事實；換一個度量或拓撲，鄰近的意義也會不同。",
            "en": "This step only fixes the structure and derives no new fact; with another metric or topology, nearness would mean something else."
          }
        },
        "leaf_basis": "metric_def + abs_rules"
      },
      {
        "id": "S02",
        "uses": [
          "G2"
        ],
        "before_tex": "n\\ \\text{a positive integer}",
        "after_tex": "x_n=1+\\tfrac1n>1,\\ x_n\\neq 1",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "n>0"
        ],
        "explanation": {
          "zh-Hant": "對每個正整數 n，1/n>0（正數的倒數為正：若 1/n≤0，乘以 n>0 會得 1≤0，矛盾）。依序的加法性質，x_n=1+1/n>1+0=1，所以 x_n≠1。數列沒有任何一項等於 1。",
          "en": "For every positive integer n, 1/n>0 (the reciprocal of a positive number is positive: if 1/n≤0, multiplying by n>0 would give 1≤0, a contradiction). By the additive order property, x_n=1+1/n>1+0=1, so x_n≠1. No term of the sequence equals 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "需要 n>0；結論對每個 n 成立，因為 n 是任意正整數。只推出 x_n>1，不宣稱反向。",
            "en": "It needs n>0; the conclusion holds for each n because n is an arbitrary positive integer. Only x_n>1 is derived; no converse is claimed."
          }
        },
        "leaf_basis": "order_basics + mul_inverse_nonzero + add_zero + forall_intro"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "x_n\\to 1",
        "after_tex": "\\forall\\varepsilon>0\\ \\exists N\\ \\forall n\\ge N:\\ |x_n-1|<\\varepsilon",
        "relation": "equivalent",
        "rule": "seq_limit_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依數列極限定義展開 x_n→1。注意量詞次序：先任意給定 ε，再找 N（N 可以依賴 ε），然後對所有 n≥N 要求距離小於 ε。定義中沒有任何一處要求 x_n=1。",
          "en": "Unfold x_n→1 by the definition of the limit of a sequence. Note the order of quantifiers: ε is given first and arbitrary, then N is found (N may depend on ε), and then the distance must be less than ε for all n≥N. Nothing in the definition requires x_n=1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "seq_limit_def"
          ],
          "explanation": {
            "zh-Hant": "定義的展開與收回雙向成立。",
            "en": "Unfolding and folding back the definition both hold."
          }
        },
        "leaf_basis": "seq_limit_def + def_unfold"
      },
      {
        "id": "S04",
        "uses": [
          "G2",
          "S02"
        ],
        "before_tex": "|x_n-1|",
        "after_tex": "\\tfrac1n",
        "relation": "equivalent",
        "rule": "abs_rules",
        "conditions": [
          "\\tfrac1n>0"
        ],
        "explanation": {
          "zh-Hant": "x_n−1=(1+1/n)−1=1/n（加法結合律與反元素）。因 1/n>0（S02），依絕對值規則 |1/n|=1/n。所以 x_n 與 1 的距離恰好是 1/n。",
          "en": "x_n−1=(1+1/n)−1=1/n (associativity and inverses for addition). Since 1/n>0 (S02), the absolute-value rule gives |1/n|=1/n. So the distance from x_n to 1 is exactly 1/n."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "add_inverse",
            "abs_rules",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個數；|u|=u 需要 u≥0，S02 已給出 1/n>0。",
            "en": "Both sides are the same number; |u|=u needs u≥0, which S02 supplies as 1/n>0."
          }
        },
        "leaf_basis": "add_assoc + add_comm + add_inverse + add_zero + abs_rules"
      },
      {
        "id": "S05",
        "uses": [
          "S03"
        ],
        "before_tex": "\\varepsilon>0\\ \\text{arbitrary}",
        "after_tex": "N\\in\\mathbb{N},\\ N>\\tfrac1\\varepsilon,\\ N\\ge 1",
        "relation": "implies",
        "rule": "archimedean",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "任意取 ε>0，則 1/ε 是正實數。依阿基米德性質，存在自然數 N>1/ε；因 1/ε>0，N>0，所以 N≥1。N 依賴 ε：ε 越小，需要的 N 越大。例如 ε=1/10 時 1/ε=10，可取 N=11。",
          "en": "Take any ε>0; then 1/ε is a positive real. By the Archimedean property there is a natural number N>1/ε; since 1/ε>0, N>0 and so N≥1. N depends on ε: the smaller ε, the larger the N needed. For example, with ε=1/10 we have 1/ε=10 and may take N=11."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "archimedean",
            "exists_elim"
          ],
          "explanation": {
            "zh-Hant": "需要 ε>0 才有 1/ε；N 的存在不是由試算幾個 ε 得到，而是由阿基米德性質對每個 ε 保證。",
            "en": "ε>0 is needed for 1/ε to exist; the existence of N is not obtained by trying a few values of ε but is guaranteed for every ε by the Archimedean property."
          }
        },
        "leaf_basis": "archimedean + exists_elim + mul_inverse_nonzero"
      },
      {
        "id": "S06",
        "uses": [
          "S05"
        ],
        "before_tex": "n\\ge N\\ge 1",
        "after_tex": "\\tfrac1n\\le\\tfrac1N",
        "relation": "implies",
        "rule": "L_recip_le",
        "conditions": [
          "N>0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_recip_le（在 S06.1 證明），取 u=N、v=n：由 0<N≤n 得 1/n≤1/N。索引越大，距離 1/n 不會變大。",
          "en": "By Lemma L_recip_le (proved in S06.1) with u=N and v=n: from 0<N≤n we get 1/n≤1/N. As the index increases, the distance 1/n does not increase."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_recip_le",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "引理需要 u>0；這由 S05 的 N≥1 提供。",
            "en": "The lemma needs u>0, supplied by N≥1 from S05."
          }
        },
        "leaf_basis": "order_basics + case_analysis + mul_inverse_nonzero"
      },
      {
        "id": "S06.1",
        "uses": [
          "order_basics"
        ],
        "before_tex": "0<u\\le v",
        "after_tex": "\\tfrac1v\\le\\tfrac1u",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "u>0",
          "v>0"
        ],
        "explanation": {
          "zh-Hant": "分兩種情況。若 u=v，則 1/v=1/u。若 u<v：因 u、v>0，1/(uv)>0；兩邊同乘正數 1/(uv) 保持不等號，u·1/(uv)<v·1/(uv)，即 1/v<1/u。兩種情況合起來得 1/v≤1/u，證得引理 L_recip_le。",
          "en": "Two cases. If u=v, then 1/v=1/u. If u<v: since u, v>0, 1/(uv)>0, and multiplying both sides by the positive number 1/(uv) keeps the inequality, u·1/(uv)<v·1/(uv), that is 1/v<1/u. Together the cases give 1/v≤1/u, proving Lemma L_recip_le."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "case_analysis",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "乘以正數保持不等號方向需要 1/(uv)>0；若 u 或 v 可為負，結論可能失敗（例如 u=−1、v=1 時 1/v=1>−1=1/u）。",
            "en": "Keeping the direction needs 1/(uv)>0; if u or v could be negative the conclusion may fail (for example u=−1, v=1 gives 1/v=1>−1=1/u)."
          }
        },
        "leaf_basis": "order_basics + case_analysis + frac_rules + mul_inverse_nonzero"
      },
      {
        "id": "S07",
        "uses": [
          "S05"
        ],
        "before_tex": "N>\\tfrac1\\varepsilon",
        "after_tex": "\\tfrac1N<\\varepsilon",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "\\varepsilon>0",
          "N>0"
        ],
        "explanation": {
          "zh-Hant": "ε/N>0（兩個正數之商）。兩邊同乘正數 ε/N：左邊 N·(ε/N)=ε，右邊 (1/ε)·(ε/N)=1/N，所以 ε>1/N，即 1/N<ε。",
          "en": "ε/N>0 (a quotient of positive numbers). Multiply both sides by the positive number ε/N: the left side gives N·(ε/N)=ε and the right side (1/ε)·(ε/N)=1/N, so ε>1/N, that is 1/N<ε."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "需要 ε/N>0 以保持不等號方向；這一步只用到 S05 的 N>1/ε。",
            "en": "ε/N>0 is needed to keep the direction of the inequality; this step uses only N>1/ε from S05."
          }
        },
        "leaf_basis": "order_basics + frac_rules + mul_inverse_nonzero"
      },
      {
        "id": "S08",
        "uses": [
          "S02",
          "S03",
          "S04",
          "S06",
          "S07"
        ],
        "before_tex": "|x_n-1|=\\tfrac1n\\le\\tfrac1N<\\varepsilon\\ \\ (n\\ge N)",
        "after_tex": "x_n\\to 1,\\ \\text{while}\\ x_n\\neq 1\\ \\text{for every}\\ n",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對每個 n≥N：|x_n−1|=1/n（S04）≤1/N（S06）<ε（S07），依序的傳遞，|x_n−1|<ε。ε 是任意的正數，N 由 S05 給出，依 ∀引入，S03 的右邊成立，所以 x_n→1。結合 S02：x_n 趨近 1，但沒有任何一項等於 1。",
          "en": "For each n≥N: |x_n−1|=1/n (S04) ≤1/N (S06) <ε (S07), so by transitivity of order |x_n−1|<ε. ε was an arbitrary positive number and N comes from S05, so by ∀-introduction the right side of S03 holds and x_n→1. Combined with S02: x_n tends to 1, yet no term equals 1."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S04",
            "S05",
            "S06",
            "S07",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "目標 G：x_n→1。足夠條件 H：對任意 ε>0 有 N 使 n≥N 時 |x_n−1|<ε；H⇒G 由 S03。H 由 S04 到 S07 正向證成，ε 是任意的，沒有把 G 當作前提。",
            "en": "Goal G: x_n→1. Sufficient condition H: for every ε>0 some N gives |x_n−1|<ε for n≥N; H⇒G by S03. H is proved forwards in S04 to S07 with ε arbitrary; G is never used as a premise."
          }
        },
        "leaf_basis": "order_basics + forall_intro + logic_chain"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "G3"
        ],
        "before_tex": "x_n>1\\ \\text{for every}\\ n",
        "after_tex": "g(x_n)=1\\ \\text{for every}\\ n,\\ g(x_n)\\to 1",
        "relation": "implies",
        "rule": "seq_limit_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 S02，每個 x_n>1，依 G3，g(x_n)=1。對任意 ε>0 取 N=1，所有 n 都有 |g(x_n)−1|=|0|=0<ε，所以 g(x_n)→1。沿著這一條趨近 1 的數列，函數值收斂到 1。",
          "en": "By S02 each x_n>1, so by G3, g(x_n)=1. For any ε>0 take N=1: every n has |g(x_n)−1|=|0|=0<ε, so g(x_n)→1. Along this one sequence approaching 1, the function values converge to 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "G3",
            "seq_limit_def"
          ],
          "explanation": {
            "zh-Hant": "結論只關於這一條數列；不宣稱 g 在 1 的極限存在，S14 證明它不存在。",
            "en": "The conclusion is about this one sequence only; it is not claimed that the limit of g at 1 exists, and S14 proves that it does not."
          }
        },
        "leaf_basis": "seq_limit_def + abs_rules + forall_intro"
      },
      {
        "id": "S10",
        "uses": [
          "G3",
          "S01"
        ],
        "before_tex": "\\lim_{x\\to 1}g(x)=L",
        "after_tex": "\\forall\\varepsilon>0\\ \\exists\\delta>0\\ \\forall x:\\ 0<|x-1|<\\delta\\Rightarrow|g(x)-L|<\\varepsilon",
        "relation": "equivalent",
        "rule": "func_limit_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "函數極限需要比裸箭頭 x→1 多得多的資料：目標值 L、量詞次序（先 ε 後 δ），以及去心鄰域條件 0<|x−1|<δ（x=1 本身被排除，所以 g(1) 的值不影響極限）。1 是定義域 ℝ 的極限點，定義適用。",
          "en": "A function limit needs far more data than the bare arrow x→1: a target value L, the order of quantifiers (ε before δ), and the punctured-neighbourhood condition 0<|x−1|<δ (x=1 itself is excluded, so the value g(1) does not affect the limit). 1 is a limit point of the domain ℝ, so the definition applies."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "func_limit_def"
          ],
          "explanation": {
            "zh-Hant": "定義的展開雙向成立。",
            "en": "Unfolding the definition holds both ways."
          }
        },
        "leaf_basis": "func_limit_def + def_unfold"
      },
      {
        "id": "S11",
        "uses": [
          "S10"
        ],
        "before_tex": "\\lim_{x\\to 1}g(x)=L\\ \\text{(assumed)},\\ \\varepsilon=\\tfrac12",
        "after_tex": "\\delta>0:\\ 0<|x-1|<\\delta\\Rightarrow|g(x)-L|<\\tfrac12",
        "relation": "implies",
        "rule": "exists_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "為了反證，暫時假設某個實數 L 使 lim_{x→1} g(x)=L。把 S10 的定義用於 ε=1/2（∀消去），得到某個 δ>0（∃消去，取新名字 δ），使所有滿足 0<|x−1|<δ 的 x 都有 |g(x)−L|<1/2。",
          "en": "For a proof by contradiction, temporarily assume some real L has lim_{x→1} g(x)=L. Apply the definition S10 with ε=1/2 (∀-elimination) to obtain some δ>0 (∃-elimination, with the fresh name δ) such that every x with 0<|x−1|<δ has |g(x)−L|<1/2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "forall_elim",
            "exists_elim"
          ],
          "explanation": {
            "zh-Hant": "這是反證的暫時假設，在 S14 推出矛盾後釋放；δ 是新名字，最後結論不提到它。",
            "en": "This is the temporary assumption of the proof by contradiction, discharged after S14 reaches a contradiction; δ is a fresh name and the final conclusion does not mention it."
          }
        },
        "leaf_basis": "forall_elim + exists_elim"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "G3"
        ],
        "before_tex": "p=1+\\tfrac\\delta2,\\ q=1-\\tfrac\\delta2",
        "after_tex": "0<|p-1|<\\delta,\\ 0<|q-1|<\\delta,\\ g(p)=1,\\ g(q)=0",
        "relation": "implies",
        "rule": "L_half",
        "conditions": [
          "\\delta>0"
        ],
        "explanation": {
          "zh-Hant": "取 p=1+δ/2、q=1−δ/2。|p−1|=δ/2，|q−1|=|−δ/2|=δ/2。依引理 L_half（在 S12.1 證明），0<δ/2<δ，所以 p、q 都在去心鄰域內。又 p>1、q<1，依 G3，g(p)=1、g(q)=0。例如 δ=1/4 時 p=9/8、q=7/8，兩者與 1 的距離都是 1/8<1/4。",
          "en": "Take p=1+δ/2 and q=1−δ/2. |p−1|=δ/2 and |q−1|=|−δ/2|=δ/2. By Lemma L_half (proved in S12.1), 0<δ/2<δ, so p and q both lie in the punctured neighbourhood. Also p>1 and q<1, so by G3, g(p)=1 and g(q)=0. For example, with δ=1/4, p=9/8 and q=7/8, both at distance 1/8<1/4 from 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_half",
            "abs_rules",
            "G3"
          ],
          "explanation": {
            "zh-Hant": "需要 δ>0 才有 0<δ/2<δ；p、q 都不等於 1，所以去心條件成立。|−δ/2|=δ/2 用到 δ/2>0 與絕對值規則。",
            "en": "δ>0 is needed for 0<δ/2<δ; neither p nor q equals 1, so the punctured condition holds. |−δ/2|=δ/2 uses δ/2>0 and the absolute-value rule."
          }
        },
        "leaf_basis": "order_basics + abs_rules + frac_rules"
      },
      {
        "id": "S12.1",
        "uses": [
          "order_basics"
        ],
        "before_tex": "\\delta>0",
        "after_tex": "0<\\tfrac\\delta2<\\delta",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "\\delta>0"
        ],
        "explanation": {
          "zh-Hant": "δ>0 兩邊乘以正數 1/2，得 δ/2>0。又 δ−δ/2=δ/2>0，兩邊加 δ/2 得 δ>δ/2。證得引理 L_half。",
          "en": "Multiplying δ>0 by the positive number 1/2 gives δ/2>0. Also δ−δ/2=δ/2>0, and adding δ/2 to both sides gives δ>δ/2. This proves Lemma L_half."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "只用到乘以正數與加上同一數保持不等號；δ≤0 時結論不成立。",
            "en": "Only multiplication by a positive number and adding the same number to both sides are used; for δ≤0 the conclusion fails."
          }
        },
        "leaf_basis": "order_basics + frac_rules + two_nonzero"
      },
      {
        "id": "S13",
        "uses": [
          "S11",
          "S12"
        ],
        "before_tex": "\\text{S11 applied at}\\ x=p\\ \\text{and}\\ x=q",
        "after_tex": "|1-L|<\\tfrac12,\\ |0-L|<\\tfrac12",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "p、q 都滿足 0<|x−1|<δ（S12），把 S11 的全稱語句分別代入 x=p 與 x=q：|g(p)−L|<1/2 與 |g(q)−L|<1/2；再代入 g(p)=1、g(q)=0。",
          "en": "Both p and q satisfy 0<|x−1|<δ (S12), so instantiate the universal statement of S11 at x=p and at x=q: |g(p)−L|<1/2 and |g(q)−L|<1/2; then substitute g(p)=1 and g(q)=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "S12",
            "forall_elim",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "代入前必須先確認 p、q 滿足 S11 的前提（S12）。",
            "en": "Before instantiating, p and q must be shown to satisfy the hypothesis of S11 (S12)."
          }
        },
        "leaf_basis": "forall_elim + modus_ponens + eq_replace"
      },
      {
        "id": "S14",
        "uses": [
          "S13",
          "L_abs_neg"
        ],
        "before_tex": "|1-L|<\\tfrac12,\\ |0-L|<\\tfrac12",
        "after_tex": "1<1\\ (\\text{contradiction});\\ \\lim_{x\\to 1}g(x)\\ \\text{does not exist}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 S14.1、S14.2 的計算得 1<1，與序的三一律矛盾。所以 S11 的暫時假設不成立：沒有任何實數 L 使 lim_{x→1} g(x)=L，即這個極限不存在。L 是任意的，所以結論對所有候選值成立。",
          "en": "The computations in S14.1 and S14.2 give 1<1, contradicting trichotomy of order. So the temporary assumption of S11 fails: no real L has lim_{x→1} g(x)=L, that is the limit does not exist. L was arbitrary, so the conclusion covers every candidate value."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "contradiction",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "矛盾只來自 S11 的假設，因此被否定的正是「極限存在」；論證沒有對 L 做任何限制。",
            "en": "The contradiction comes only from the assumption of S11, so what is refuted is exactly 'the limit exists'; the argument places no restriction on L."
          }
        },
        "leaf_basis": "contradiction + order_basics + abs_rules + quantifier_negation"
      },
      {
        "id": "S14.1",
        "uses": [
          "abs_rules"
        ],
        "before_tex": "|-u|",
        "after_tex": "|u|",
        "relation": "equivalent",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "|−u|=|(−1)u|=|−1|·|u|（|uv|=|u||v|）。因 −1<0，|−1|=−(−1)=1，所以 |−u|=1·|u|=|u|。證得引理 L_abs_neg；之後用於 |L−0|=|−(0−L)|=|0−L|。",
          "en": "|−u|=|(−1)u|=|−1|·|u| (by |uv|=|u||v|). Since −1<0, |−1|=−(−1)=1, so |−u|=1·|u|=|u|. This proves Lemma L_abs_neg, used next for |L−0|=|−(0−L)|=|0−L|."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "abs_rules",
            "neg_mul",
            "mul_one"
          ],
          "explanation": {
            "zh-Hant": "每一步都是恆等式，兩邊是同一個數，雙向成立。",
            "en": "Each move is an identity and both sides are the same number, so the step holds both ways."
          }
        },
        "leaf_basis": "abs_rules + neg_mul + mul_one"
      },
      {
        "id": "S14.2",
        "uses": [
          "S13",
          "S14.1"
        ],
        "before_tex": "1=|1-0|=|(1-L)+(L-0)|",
        "after_tex": "1\\le|1-L|+|0-L|<\\tfrac12+\\tfrac12=1",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "1=|1−0|，且 1−0=(1−L)+(L−0)（插入 −L+L=0）。依三角不等式，|(1−L)+(L−0)|≤|1−L|+|L−0|；依 L_abs_neg，|L−0|=|0−L|。再用 S13 的兩個不等式：|1−L|+|0−L|<1/2+1/2=1。合起來 1<1。",
          "en": "1=|1−0|, and 1−0=(1−L)+(L−0) (inserting −L+L=0). By the triangle inequality, |(1−L)+(L−0)|≤|1−L|+|L−0|; by L_abs_neg, |L−0|=|0−L|. With the two inequalities of S13: |1−L|+|0−L|<1/2+1/2=1. Altogether 1<1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "S13",
            "S14.1",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "三角不等式是單向的（≤），兩個嚴格不等式相加得到嚴格不等式；整條鏈只需單向。",
            "en": "The triangle inequality is one-way (≤), and adding two strict inequalities gives a strict one; the whole chain only needs one direction."
          }
        },
        "leaf_basis": "abs_rules + add_inverse + order_basics + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S08",
          "S09",
          "S14"
        ],
        "before_tex": "x_n\\to 1,\\ x_n\\neq 1,\\ g(x_n)\\to 1,\\ \\lim_{x\\to 1}g(x)\\ \\text{does not exist}",
        "after_tex": "x\\to 1\\ \\text{neither forces}\\ x=1\\ \\text{nor decides whether}\\ g(x)\\ \\text{converges}",
        "relation": "implies",
        "rule": "mko_tends_to",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整並對照 MKO：(1) x_n→1 而每項 x_n≠1（S02、S08），所以趨近不要求等於；(2) 同樣在 1 附近，沿 x_n 的函數值收斂（S09），但完整的 ε–δ 極限不存在（S14），所以裸箭頭 x→1 本身不決定函數值是否收斂；要說「收斂」，還必須指定目標值、量詞次序與去心鄰域條件（S10）。",
          "en": "Collect and compare with the MKO: (1) x_n→1 while every x_n≠1 (S02, S08), so approaching does not require equality; (2) near the same point 1, the function values converge along x_n (S09) but the full ε–δ limit does not exist (S14), so the bare arrow x→1 by itself does not decide whether function values converge; to claim convergence one must also specify the target value, the order of quantifiers and the punctured-neighbourhood condition (S10)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S08",
            "S09",
            "S14",
            "mko_tends_to",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G 是題目兩部分結論的合取；每一部分都由所列步驟正向證成，此步只做合取並與 MKO 對照，沒有把 G 當作前提。",
            "en": "Goal G is the conjunction of the two parts of the problem; each is proved forwards by the listed steps, and this step only forms the conjunction and compares it with the MKO; G is never used as a premise."
          }
        },
        "leaf_basis": "mko_tends_to + and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在 (ℝ,|x−y|) 中，x_n=1+1/n 趨近 1，而沒有任何一項等於 1；對在 1 處跳躍的 g，沿 x_n 有 g(x_n)→1，但 lim_{x→1} g(x) 不存在（任何候選 L 都會導出 1<1）。所以 x→1 本身既不要求 x=1，也不決定函數值是否收斂；完整的極限語句還需要目標值、量詞次序與去心鄰域條件。",
      "en": "In (ℝ,|x−y|), x_n=1+1/n tends to 1 while no term equals 1; for the function g with a jump at 1, g(x_n)→1 along x_n, yet lim_{x→1} g(x) does not exist (every candidate L leads to 1<1). So x→1 by itself neither requires x=1 nor decides whether function values converge; a full limit statement also needs a target value, the order of quantifiers and the punctured-neighbourhood condition."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "x_n→1，所以某一項 x_n 等於 1",
          "en": "x_n→1, so some term x_n equals 1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "每一項都有 x_n=1+1/n>1（S02）。",
          "en": "Every term has x_n=1+1/n>1 (S02)."
        }
      },
      {
        "claim": {
          "zh-Hant": "沿 x_n 有 g(x_n)→1，所以 lim_{x→1} g(x)=1",
          "en": "g(x_n)→1 along x_n, so lim_{x→1} g(x)=1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "左側的點 q 有 g(q)=0；S11 到 S14 證明極限不存在。",
          "en": "The left-hand point q has g(q)=0; S11 to S14 prove that the limit does not exist."
        }
      },
      {
        "claim": {
          "zh-Hant": "x→1 就是代入 x=1，所以 lim_{x→1} g(x)=g(1)=0",
          "en": "x→1 means substituting x=1, so lim_{x→1} g(x)=g(1)=0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "定義排除 x=1（S10），而且極限不存在（S14）。",
          "en": "The definition excludes x=1 (S10), and the limit does not exist (S14)."
        }
      },
      {
        "claim": {
          "zh-Hant": "存在 N，使對所有 ε>0 與所有 n≥N 都有 |x_n−1|<ε",
          "en": "There is N such that |x_n−1|<ε for all ε>0 and all n≥N"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "取 ε=1/N、n=N，|x_N−1|=1/N 不小於 ε。",
          "en": "Take ε=1/N and n=N; |x_N−1|=1/N is not less than ε."
        }
      },
      {
        "claim": {
          "zh-Hant": "寫出 x→1 就已經說明 g(x) 收斂",
          "en": "Writing x→1 already says that g(x) converges"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "MKO 指出趨近關係本身不指定函數值是否收斂；本例 g 在 1 的極限不存在（S14）。",
          "en": "The MKO says the tends-to relation by itself does not say whether function values converge; here g has no limit at 1 (S14)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "archimedean",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "阿基米德性質：對任意實數 x，存在自然數 n>x；因此對任意 ε>0，存在 n 使 1/n<ε。",
          "en": "Archimedean property: for every real x there is a natural number n>x; hence for every ε>0 there is n with 1/n<ε."
        }
      },
      {
        "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": "func_limit_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數極限：lim_{x→a} f(x)=L 表示對每個 ε>0，存在 δ>0，使所有滿足 0<|x−a|<δ 的定義域點 x 都有 |f(x)−L|<ε（a 須為定義域的極限點）。",
          "en": "Limit of a function: lim_{x→a} f(x)=L means that for every ε>0 there is δ>0 such that |f(x)−L|<ε for every domain point x with 0<|x−a|<δ (a must be a limit point of the domain)."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "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": "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": "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": "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": "exists_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃消去：由 ∃x∈D, P(x)，可取一個新名字 a 表示某個滿足 P(a) 的元素；只要最後結論不含 a，就可用 a 推理。",
          "en": "∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a."
        }
      },
      {
        "id": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "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": "logic_chain",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。",
          "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔."
        }
      },
      {
        "id": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "id": "two_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數中 0<1，故 0<1+1=2，所以 2≠0。",
          "en": "In the reals 0<1, hence 0<1+1=2, so 2≠0."
        }
      },
      {
        "id": "modus_ponens",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "肯定前件：由 P 與 P⇒Q 得 Q。",
          "en": "Modus ponens: from P and P⇒Q infer 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": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "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": "mko_tends_to",
        "kind": "cited",
        "source": {
          "mko_id": "mko-tends-to-relation",
          "sha256": "2fbb3c700378e80e68833bcada18866ed6439803d8b639e304fe76d31903f70e",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "x→a 是極限語言中的方向性關係；它本身尚未指定函數值是否收斂，也不等同於 x 最終必須等於 a。",
          "en": "The relation x→a is directional language used in limits. By itself it does not say whether function values converge, nor does it require x eventually to equal a."
        }
      },
      {
        "id": "L_recip_le",
        "kind": "lemma",
        "proved_in": [
          "S06.1"
        ],
        "statement": {
          "zh-Hant": "對實數 0<u≤v：1/v≤1/u。",
          "en": "For reals 0<u≤v: 1/v≤1/u."
        }
      },
      {
        "id": "L_half",
        "kind": "lemma",
        "proved_in": [
          "S12.1"
        ],
        "statement": {
          "zh-Hant": "對實數 δ>0：0<δ/2<δ。",
          "en": "For a real δ>0: 0<δ/2<δ."
        }
      },
      {
        "id": "L_abs_neg",
        "kind": "lemma",
        "proved_in": [
          "S14.1"
        ],
        "statement": {
          "zh-Hant": "對實數 u：|−u|=|u|。",
          "en": "For a real u: |−u|=|u|."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S05",
        "lhs": "1/(1/10)",
        "rhs": "10"
      },
      {
        "id": "N2",
        "step": "S05",
        "lhs": "11",
        "relation": ">",
        "rhs": "10"
      },
      {
        "id": "N3",
        "step": "S07",
        "lhs": "1/11",
        "relation": "<",
        "rhs": "1/10"
      },
      {
        "id": "N4",
        "step": "S04",
        "vars": [
          "n"
        ],
        "range": [
          1,
          50
        ],
        "lhs": "(1+1/n)-1",
        "rhs": "1/n"
      },
      {
        "id": "N5",
        "step": "S12",
        "lhs": "1+(1/4)/2",
        "rhs": "9/8"
      },
      {
        "id": "N6",
        "step": "S12",
        "lhs": "1-(1/4)/2",
        "rhs": "7/8"
      },
      {
        "id": "N7",
        "step": "S14.2",
        "lhs": "1/2+1/2",
        "rhs": "1"
      },
      {
        "id": "N8",
        "step": "S14.2",
        "lhs": "abs(1-0)",
        "rhs": "1"
      },
      {
        "id": "N9",
        "step": "S06",
        "lhs": "1/12",
        "relation": "<=",
        "rhs": "1/11"
      },
      {
        "id": "N10",
        "step": "S05",
        "lhs": "abs((1+1/3)-1)",
        "relation": ">=",
        "rhs": "1/3"
      },
      {
        "id": "N11",
        "step": "S12",
        "lhs": "abs(7/8-1)",
        "relation": "<",
        "rhs": "1/4"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 假設 a1 之下工作：明示的度量結構 (ℝ,|x−y|)。使用標準的數列極限與 ε–δ 函數極限定義作為前置知識。不建立 Mathlib 的 Filter.Tendsto、nhds 或 atTop 物件，也不證明它們與 ε–δ 定義的等價；不改動 canonical MKO 或英文 edition。",
      "en": "Works under MKO assumption a1: the explicit metric structure (ℝ,|x−y|). The standard definitions of sequence limits and ε–δ function limits are used as prerequisites. It does not build Mathlib's Filter.Tendsto, nhds or atTop objects and does not prove their equivalence with the ε–δ definitions; it does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "x\\to a",
        "趨近關係（方向性語言）",
        "the tends-to relation (directional language)"
      ],
      [
        "x_n",
        "數列 1+1/n",
        "the sequence 1+1/n"
      ],
      [
        "g",
        "在 1 處跳躍的函數",
        "the function with a jump at 1"
      ],
      [
        "\\varepsilon,\\ \\delta",
        "目標端與來源端的容許距離",
        "tolerances on the target side and on the source side"
      ],
      [
        "N",
        "依賴 ε 的索引門檻",
        "an index threshold depending on ε"
      ],
      [
        "d(x,y)=|x-y|",
        "ℝ 上的標準度量",
        "the standard metric on ℝ"
      ]
    ],
    "md_error_controls": [
      {
        "title": "x→1 不是 x=1 / x→1 is not x=1",
        "zh": "MKO 指出常見誤解是把 x→a 直接讀成 x=a。數列 x_n 趨近 1，但每一項 x_n>1（S02、S08）。函數極限的定義更明確排除 x=1（0<|x−1|，S10）：g(1)=0 這個代入值既不是沿 x_n 的極限值 1，也不構成極限，因為極限根本不存在（S14）。",
        "en": "The MKO names the mistake of reading x→a as x=a. The sequence x_n tends to 1, yet every term has x_n>1 (S02, S08). The definition of a function limit even excludes x=1 explicitly (0<|x−1|, S10): the substituted value g(1)=0 is neither the limit 1 along x_n nor a limit at all, since the limit does not exist (S14)."
      },
      {
        "title": "一條路徑不等於完整定義 / One approach path is not the full definition",
        "zh": "MKO 也提醒不要把裸箭頭當成完整的 Filter.Tendsto 或 ε–δ 定義。沿 x_n 檢查得到 g(x_n)→1（S09），只涵蓋從右側趨近的一條數列；ε–δ 定義要求去心鄰域中所有 x，包括左側的 q（S12），因而極限不存在。",
        "en": "The MKO also warns against treating the bare arrow as the full Filter.Tendsto or ε–δ definition. Checking along x_n gives g(x_n)→1 (S09), which covers only one sequence approaching from the right; the ε–δ definition requires every x in the punctured neighbourhood, including the left-hand point q (S12), so the limit does not exist."
      },
      {
        "title": "量詞次序 / The order of quantifiers",
        "zh": "最容易做錯的是把「對每個 ε 存在 N」改成「存在 N 對每個 ε」。後者對 x_n 不成立：給定任何 N，取 ε=1/N、n=N，則 |x_N−1|=1/N 並不小於 ε（N10 以 N=3 為例）。S05 的 N 依賴 ε，這正是定義所允許的。",
        "en": "A common mistake is turning 'for every ε there is N' into 'there is N for every ε'. The latter fails for x_n: given any N, take ε=1/N and n=N; then |x_N−1|=1/N is not less than ε (N10 shows N=3). The N of S05 depends on ε, which is exactly what the definition allows."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿只使用度量 |x−y| 下的 ε–N 與 ε–δ 定義；與 Mathlib 的 Filter.Tendsto、nhds、atTop 之對應未在本稿證明。",
        "en": "This guide uses only the ε–N and ε–δ definitions for the metric |x−y|; their correspondence with Mathlib's Filter.Tendsto, nhds and atTop is not proved here."
      },
      {
        "zh-Hant": "N4 只在 n=1 到 50 核對 (1+1/n)−1=1/n，是有限檢查；一般情形由 S04 的加法規則證明。N1 到 N3、N5 到 N11 只是對具體數值的核對，論證不依賴它們。",
        "en": "N4 checks (1+1/n)−1=1/n only for n from 1 to 50, a finite check; the general case is proved by the addition rules in S04. N1 to N3 and N5 to N11 only check specific numbers, and the argument does not depend on them."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "9cb369c6f946a7068f52b3a1c30a5922a592d1864238416d34c642fcdd014920",
    "original_source_sha256": "9cb369c6f946a7068f52b3a1c30a5922a592d1864238416d34c642fcdd014920",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}