前段时间太忙,一直没有更新。今天有点空闲,再更新两道题。
陶哲轩实分析 3.4.10 和 3.4.11
3.4.10
(1)
∀ x ∈ ( ∪ α ∈ I A α ) ∪ ( ∪ α ∈ J A α ) \forall x \in (\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in J} A_\alpha) ∀x∈(∪α∈IAα)∪(∪α∈JAα) 有 x ∈ ( ∪ α ∈ I A α ) x \in (\cup_{\alpha \in I} A_\alpha) x∈(∪α∈IAα) 或者 x ∈ ( ∪ α ∈ J A α ) x \in (\cup_{\alpha \in J} A_\alpha) x∈(∪α∈JAα)
分两种情况讨论:
当 x ∈ ( ∪ α ∈ I A α ) x \in (\cup_{\alpha \in I} A_\alpha) x∈(∪α∈IAα) 时, x ∈ ( ∪ α ∈ I ∪ J A α ) x \in (\cup_{\alpha \in I \cup J} A_\alpha) x∈(∪α∈I∪JAα)
当 x ∈ ( ∪ α ∈ J A α ) x \in (\cup_{\alpha \in J} A_\alpha) x∈(∪α∈JAα) 时, x ∈ ( ∪ α ∈ I ∪ J A α ) x \in (\cup_{\alpha \in I \cup J} A_\alpha) x∈(∪α∈I∪JAα)
所以
( ∪ α ∈ I A α ) ∪ ( ∪ α ∈ J A α ) ⊆ ( ∪ α ∈ I ∪ J A α ) (\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in J} A_\alpha) \subseteq (\cup_{\alpha \in I \cup J} A_\alpha) (∪α∈IAα)∪(∪α∈JAα)⊆(∪α∈I∪JAα)
∀ x ∈ ( ∪ α ∈ I ∪ J A α ) \forall x \in (\cup_{\alpha \in I \cup J} A_\alpha) ∀x∈(∪α∈I∪JAα) 有 x ∈ ( ∪ α ∈ I A α ) x \in (\cup_{\alpha \in I} A_\alpha) x∈(∪α∈IAα) 或 x ∈ ( ∪ α ∈ J A α ) x \in (\cup_{\alpha \in J} A_\alpha) x∈(∪α∈JAα)
所以 ∀ x ∈ ( ∪ α ∈ I A α ) ∪ ( ∪ α ∈ J A α ) \forall x \in (\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in J} A_\alpha) ∀x∈(∪α∈IAα)∪(∪α∈JAα)
所以
( ∪ α ∈ I ∪ J A α ) ⊆ ( ∪ α ∈ I A α ) ∪ ( ∪ α ∈ J A α ) (\cup_{\alpha \in I \cup J} A_\alpha) \subseteq (\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in J} A_\alpha) (∪α∈I∪JAα)⊆(∪α∈IAα)∪(∪α∈JAα)
所以
( ∪ α ∈ I ∪ J A α ) = ( ∪ α ∈ I A α ) ∪ ( ∪ α ∈ J A α ) (\cup_{\alpha \in I \cup J} A_\alpha) = (\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in J} A_\alpha) (∪α∈I∪JAα)=(∪α∈IAα)∪(∪α∈JAα)
(2) ∀ x ∈ ( ∩ α ∈ I A α ) ∩ ( ∩ α ∈ J A α ) \forall x \in ( \cap_{\alpha \in I }A_\alpha) \cap ( \cap_{\alpha \in J }A_\alpha) ∀x∈(∩α∈IAα)∩(∩α∈JAα)
表明:
x ∈ ∩ α ∈ I A α x \in \cap_{\alpha \in I }A_\alpha x∈∩α∈IAα 同时 x ∈ ∩ α ∈ J A α x \in \cap_{\alpha \in J }A_\alpha x∈∩α∈JAα
表明:
∀ α ∈ I , x ∈ A α \forall \alpha \in I , x \in A_\alpha ∀α∈I,x∈Aα ,同时 ∀ α ∈ J , x ∈ A α \forall \alpha \in J , x \in A_\alpha ∀α∈J,x∈Aα
所以 ∀ α ∈ I ∪ J , x ∈ A α \forall \alpha \in I \cup J , x \in A_\alpha ∀α∈I∪J,x∈Aα
所以 ∀ x ∈ ( ∩ α ∈ I ∪ J A α ) \forall x \in ( \cap_{\alpha \in I \cup J }A_\alpha) ∀x∈(∩α∈I∪JAα)
所以 ( ∩ α ∈ I A α ) ∩ ( ∩ α ∈ J A α ) ⊆ ( ∩ α ∈ I ∪ J A α ) ( \cap_{\alpha \in I }A_\alpha) \cap ( \cap_{\alpha \in J }A_\alpha) \subseteq ( \cap_{\alpha \in I \cup J }A_\alpha) (∩α∈IAα)∩(∩α∈JAα)⊆(∩α∈I∪JAα)
∀ x ∈ ∩ α ∈ I ∪ J A α \forall x \in \cap_{\alpha \in I \cup J }A_\alpha ∀x∈∩α∈I∪JAα
表明: ∀ α ∈ I ∪ J , x ∈ A α \forall \alpha \in I \cup J, x \in A_\alpha ∀α∈I∪J,x∈Aα
所以:
∀ α ∈ I , x ∈ A α \forall \alpha \in I, x \in A_\alpha ∀α∈I,x∈Aα 同时 ∀ α ∈ J , x ∈ A α \forall \alpha \in J, x \in A_\alpha ∀α∈J,x∈Aα
所以 x ∈ ( ∩ α ∈ I A α ) x \in ( \cap_{\alpha \in I }A_\alpha) x∈(∩α∈IAα) 同时 x ∈ ( ∩ α ∈ J A α ) x \in ( \cap_{\alpha \in J }A_\alpha) x∈(∩α∈JAα)
所以 x ∈ ( ∩ α ∈ I A α ) ∩ ( ∩ α ∈ J A α ) x \in ( \cap_{\alpha \in I }A_\alpha) \cap ( \cap_{\alpha \in J }A_\alpha) x∈(∩α∈IAα)∩(∩α∈JAα)
所以: ( ∩ α ∈ I ∪ J A α ) ⊆ ( ∩ α ∈ I A α ) ∩ ( ∩ α ∈ J A α ) ( \cap_{\alpha \in I \cup J }A_\alpha)\subseteq ( \cap_{\alpha \in I }A_\alpha) \cap ( \cap_{\alpha \in J }A_\alpha) (∩α∈I∪JAα)⊆(∩α∈IAα)∩(∩α∈JAα)
所以 ( ∩ α ∈ I A α ) ∩ ( ∩ α ∈ J A α ) = ( ∩ α ∈ I ∪ J A α ) ( \cap_{\alpha \in I }A_\alpha) \cap ( \cap_{\alpha \in J }A_\alpha) = ( \cap_{\alpha \in I \cup J }A_\alpha) (∩α∈IAα)∩(∩α∈JAα)=(∩α∈I∪JAα)
3.4.11
设 X 是一个集合,I 是一个不空的集合,并且对于每个 α ∈ I \alpha \in I α∈I, A α A_\alpha Aα 是 X X X 的一个子集。证明
X \ ∪ α ∈ I A α = ∩ α ∈ I ( X \ A α ) X \ ∩ α ∈ I A α = ∪ α ∈ I ( X \ A α ) X \backslash \cup_{\alpha \in I} A_\alpha = \cap_{\alpha \in I} (X \backslash A_\alpha)\\ X \backslash \cap_{\alpha \in I} A_\alpha = \cup_{\alpha \in I} (X \backslash A_\alpha) X\∪α∈IAα=∩α∈I(X\Aα)X\∩α∈IAα=∪α∈I(X\Aα)
(1) ∀ x ∈ X \ ∪ α ∈ I A α \forall x \in X \backslash \cup_{\alpha \in I} A_\alpha ∀x∈X\∪α∈IAα 有 x ∈ X x \in X x∈X 同时 ∀ α ∈ I , x ∉ A α \forall \alpha \in I, x \notin A_\alpha ∀α∈I,x∈/Aα
所以: ∀ α ∈ I , x ∈ X \ A α \forall \alpha \in I, x \in X \backslash A_\alpha ∀α∈I,x∈X\Aα
所以: x ∈ ∩ α ∈ I X \ A α x \in \cap_{\alpha \in I} X \backslash A_\alpha x∈∩α∈IX\Aα
所以:
X \ ∪ α ∈ I A α ⊆ ∩ α ∈ I ( X \ A α ) X \backslash \cup_{\alpha \in I} A_\alpha \subseteq \cap_{\alpha \in I} (X \backslash A_\alpha) X\∪α∈IAα⊆∩α∈I(X\Aα)
∀ x ∈ ∩ α ∈ I ( X \ A α ) \forall x \in \cap_{\alpha \in I} (X \backslash A_\alpha) ∀x∈∩α∈I(X\Aα)
有: ∀ α ∈ I , x ∈ X \ A α \forall \alpha \in I, x \in X \backslash A_\alpha ∀α∈I,x∈X\Aα
所以: x ∈ X x \in X x∈X 和 $ \forall \alpha \in I , x \notin A_\alpha$
所以 x ∉ ∪ α ∈ I A α x \notin \cup_{\alpha \in I} A_\alpha x∈/∪α∈IAα
所以 x ∈ X \ ∪ α ∈ I A α x \in X \backslash \cup_{\alpha \in I} A_\alpha x∈X\∪α∈IAα
所以
∩ α ∈ I ( X \ A α ) ⊆ X \ ∪ α ∈ I A α \cap_{\alpha \in I} (X \backslash A_\alpha) \subseteq X \backslash \cup_{\alpha \in I} A_\alpha ∩α∈I(X\Aα)⊆X\∪α∈IAα
所以:
∩ α ∈ I ( X \ A α ) = X \ ∪ α ∈ I A α \cap_{\alpha \in I} (X \backslash A_\alpha) = X \backslash \cup_{\alpha \in I} A_\alpha ∩α∈I(X\Aα)=X\∪α∈IAα
(2) ∀ x ∈ X \ ∩ α ∈ I A α \forall x \in X \backslash \cap_{\alpha \in I} A_\alpha ∀x∈X\∩α∈IAα
有 x ∈ X x \in X x∈X 同时 x ∉ ∩ α ∈ I A α x \notin \cap_{\alpha \in I} A_\alpha x∈/∩α∈IAα
所以 ∃ β ∈ I , x ∉ A β \exists \beta \in I, x \notin A_\beta ∃β∈I,x∈/Aβ
所以 x ∈ X \ A β x \in X \backslash A_\beta x∈X\Aβ
所以 x ∈ ∪ α ∈ I X \ A α x \in \cup_{\alpha \in I} X \backslash A_\alpha x∈∪α∈IX\Aα
所以
X \ ∩ α ∈ I A α ⊆ ∪ α ∈ I ( X \ A α ) X \backslash \cap_{\alpha \in I} A_\alpha \subseteq \cup_{\alpha \in I} (X \backslash A_\alpha) X\∩α∈IAα⊆∪α∈I(X\Aα)
∀ x ∈ ∪ α ∈ I ( X \ A α ) \forall x \in \cup_{\alpha \in I} (X \backslash A_\alpha) ∀x∈∪α∈I(X\Aα)
所以 ∃ β ∈ I , x ∈ X \ A β \exists \beta \in I , x \in X \backslash A_\beta ∃β∈I,x∈X\Aβ
所以 x ∈ X , x ∉ A β x \in X, x \notin A_\beta x∈X,x∈/Aβ
所以 x ∉ ∩ α ∈ I A α x \notin \cap_{\alpha \in I} A_\alpha x∈/∩α∈IAα
所以 x ∈ X \ ∩ α ∈ I A α x \in X \backslash \cap_{\alpha \in I } A_\alpha x∈X\∩α∈IAα
所以
∪ α ∈ I ( X \ A α ) ⊆ X \ ∩ α ∈ I A α \cup_{\alpha \in I} (X \backslash A_\alpha) \subseteq X \backslash \cap_{\alpha \in I} A_\alpha ∪α∈I(X\Aα)⊆X\∩α∈IAα
所以
∪ α ∈ I ( X \ A α ) = X \ ∩ α ∈ I A α \cup_{\alpha \in I} (X \backslash A_\alpha) = X \backslash \cap_{\alpha \in I} A_\alpha ∪α∈I(X\Aα)=X\∩α∈IAα