参考对偶原理与蕴含定理.csdn逻辑·对偶·蕴含.csdn实数------逻辑基础实数------实数的性质和基于无限小数的构造实数------实数完备性的6个定理实数完备性定理的31个证明(0~6)实数完备性定理的31个证明(7~12)实数完备性定理的31个证明(13~18)实数完备性定理的31个证明(19~24)实数完备性定理的31个证明(25~30)实数------实数完备性定理的应用对偶原理M,R,P不是一个字符,而是项,大项可拆解到原子项小项也可聚合成大项, MRP就是三个小项聚合成的一个大项( M R P ) ’ M ’ R ’ P ’ (MRP)’M’R’P’(MRP)’M’R’P’蕴含定理和对偶蕴含定理→ 和 → ′ \to 和 \to→和→′两测的命题取反移动到另一侧,不影响命题真假约定在公共厕所,男厕所有一个男雕塑,女厕所有个女雕塑,并且雕塑不能移动。其中左边是男厕所,右边是女厕所, 一个人换厕所则需要先变性。蕴含定理如果男厕所里全是男人,那么女厕所里有男人这句话的真假不受人员走动的影响⋀ p i → ⋁ q j \bigwedge p_i \to \bigvee q_j⋀pi​→⋁qj​对偶蕴含定理女厕所里全是女人,并且男厕所全是男人,这句话的真假不受人员走动的影响p → ′ q d f ¬ p ∧ q p \to q \mathrel{_{df}} \neg p \land qp→′qdf​¬p∧q⋁ p i → ′ ⋀ q j \bigvee p_i \to \bigwedge q_j⋁pi​→′⋀qj​用对偶原理求极限的否定原命题∀ ε ∃ N ∀ n ( ( ε 0 ∧ N 0 ∧ n N ) → ∣ a n − A ∣ ε ) \forall \varepsilon \exists N \forall n \, \big( (\varepsilon 0 \land N 0 \land n N) \to |a_n - A| \varepsilon \big)∀ε∃N∀n((ε0∧N0∧nN)→∣an​−A∣ε)否命题对偶原理带入极限定义∃ ε ∀ N ∃ n ( ( ε 0 ∧ N 0 ∧ n N ) ∧ ∣ a n − A ∣ ≥ ε ) \exists \varepsilon \forall N \exists n \, \big( (\varepsilon 0 \land N 0 \land n N) \land |a_n - A| \geq \varepsilon \big)∃ε∀N∃n((ε0∧N0∧nN)∧∣an​−A∣≥ε)受限量词定义∀ P ( x ) Q ( x ) d f ∀ x ( P ( x ) → Q ( x ) ) \forall_{P(x)}Q(x) \mathrel{_{df}} \forall x\big(P(x) \to Q(x)\big)∀P(x)​Q(x)df​∀x(P(x)→Q(x))∃ P ( x ) Q ( x ) d f ∃ x ( P ( x ) ∧ Q ( x ) ) \exists _{P(x)}Q(x) _{df} \exists x(P(x) \land Q(x))∃P(x)​Q(x)df​∃x(P(x)∧Q(x))定理1¬ ∀ P ( x ) Q ( x ) ⟺ ∃ P ( x ) ¬ Q ( x ) \neg\forall_{P(x)}Q(x) \iff \exists_{P(x)}\neg Q(x)¬∀P(x)​Q(x)⟺∃P(x)​¬Q(x)证明¬ ∀ P ( x ) Q ( x ) \neg\forall_{P(x)}Q(x)¬∀P(x)​Q(x)⟺ ¬ ∀ x ( P ( x ) → Q ( x ) ) \iff \neg\forall x\big(P(x)\to Q(x)\big)⟺¬∀x(P(x)→Q(x))\qquad按定义⟺ ∃ x ¬ ( P ( x ) → Q ( x ) ) \iff \exists x\,\neg\big(P(x)\to Q(x)\big)⟺∃x¬(P(x)→Q(x))\qquad对偶原理¬ ∀ x φ ⟺ ∃ x ¬ φ \neg\forall x\,\varphi\iff\exists x\,\neg\varphi¬∀xφ⟺∃x¬φ⟺ ∃ x ¬ ( ¬ P ( x ) ∨ Q ( x ) ) \iff \exists x\,\neg\big(\neg P(x)\lor Q(x)\big)⟺∃x¬(¬P(x)∨Q(x))\qquad蕴含定理 A → B ⟺ ¬ A ∨ B A\to B\iff\neg A\lor BA→B⟺¬A∨B⟺ ∃ x ( P ( x ) ∧ ¬ Q ( x ) ) \iff \exists x\big(P(x)\land\neg Q(x)\big)⟺∃x(P(x)∧¬Q(x))\qquad对偶原理⟺ ∃ P ( x ) ¬ Q ( x ) \iff \exists_{P(x)}\neg Q(x)⟺∃P(x)​¬Q(x)\qquad按定义定理2¬ ∃ P ( x ) Q ( x ) ⟺ ∀ P ( x ) ¬ Q ( x ) \neg\exists_{P(x)}Q(x) \iff \forall_{P(x)}\neg Q(x)¬∃P(x)​Q(x)⟺∀P(x)​¬Q(x)证明¬ ∃ P ( x ) Q ( x ) \neg\exists_{P(x)}Q(x)¬∃P(x)​Q(x)⟺ ¬ ∃ x ( P ( x ) ∧ Q ( x ) ) \iff \neg\exists x(P(x) \land Q(x))⟺¬∃x(P(x)∧Q(x))\qquad按定义⟺ ∀ x ¬ ( P ( x ) ∧ Q ( x ) ) \iff \forall x\,\neg(P(x) \land Q(x))⟺∀x¬(P(x)∧Q(x))\qquad对偶原理¬ ∃ x φ ⟺ ∀ x ¬ φ \neg\exists x\,\varphi\iff\forall x\,\neg\varphi¬∃xφ⟺∀x¬φ⟺ ∀ x ( ¬ P ( x ) ∨ ¬ Q ( x ) ) \iff \forall x\big(\neg P(x) \lor \neg Q(x)\big)⟺∀x(¬P(x)∨¬Q(x))\qquad对偶原理⟺ ∀ x ( P ( x ) → ¬ Q ( x ) ) \iff \forall x\big(P(x) \to \neg Q(x)\big)⟺∀x(P(x)→¬Q(x))\qquad蕴含定理⟺ ∀ P ( x ) ¬ Q ( x ) \iff \forall_{P(x)}\neg Q(x)⟺∀P(x)​¬Q(x)\qquad按定义例1a n ( − 1 ) n 发散 a_n (-1)^n \text{ 发散}an​(−1)n发散证明1构造子列取奇数项子列a 2 k − 1 ( − 1 ) 2 k − 1 − 1 a_{2k-1} (-1)^{2k-1} -1a2k−1​(−1)2k−1−1故lim ⁡ k → ∞ a 2 k − 1 − 1 \lim_{k \to \infty} a_{2k-1} -1limk→∞​a2k−1​−1取偶数项子列a 2 k ( − 1 ) 2 k 1 a_{2k} (-1)^{2k} 1a2k​(−1)2k1故lim ⁡ k → ∞ a 2 k 1 \lim_{k \to \infty} a_{2k} 1limk→∞​a2k​1。导出矛盾根据极限唯一性定理收敛数列的所有子列必收敛于同一极限。由于− 1 ≠ 1 -1 \neq 1−11故假设不成立数列( − 1 ) n (-1)^n(−1)n发散。证明2要证a n a_nan​发散即证∃ ε 0 , ∀ N 0 , ∃ n N , s.t. ∣ a n − A ∣ ≥ ε \exists \varepsilon 0, \forall N 0, \exists n N, \text{ s.t. } |a_n - A| \geq \varepsilon∃ε0,∀N0,∃nN,s.t.∣an​−A∣≥ε取ε 1 \varepsilon 1ε1。对任意N 0 N 0N0取n N n NnN使得a n a_nan​与A AA异号若A ≥ 0 A \geq 0A≥0取奇数n nn若A 0 A 0A0取偶数n nn则∣ a n − A ∣ ∣ a n ∣ ∣ A ∣ 1 ∣ A ∣ ≥ 1 ε |a_n - A| |a_n| |A| 1 |A| \geq 1 \varepsilon∣an​−A∣∣an​∣∣A∣1∣A∣≥1ε故对任意A AA均存在ε 1 \varepsilon 1ε1满足发散条件数列( − 1 ) n (-1)^n(−1)n发散,所以任意实数A AA都不是a n a_nan​的极限。