SciTech-Mathmatics-RealAnalysis: Cantor-Schröder-Bernstein Theorem

Cantor Set Theory 与 Cantor-Schröder-Bernstein Theory 是 Lebesgue积分及Real Analysis的Kernel。
证明过程有re-mapping two 1-1 relate into one bidirectional 1-1 mapping;

Proof of Cantor-Bernstein Theorem:
说明: $g = f^{'}$ 表示对一对分别来自A与B的元素，有$g$与$f$互为反函数；
∵A = A0∪A1, B = B0∪B1, A 到 B0 有 1-1 的 $f$ 映射, B 到 A0 有 1-1 的 $g$ 映射;
∴ 又因为 A与B 是对称的，所以我们只要证明任一侧出发是成立的，则另一侧同理可证；
1.对给定的任意 $a_1$∈A1:

存在1-1映射的正函数$f$, 使 $f(a_1) = b_1$ 且 $b_1$ ∈B0, 成立;
且有1-1映射的反函数$f^{'}, 使 $f^{'} (b_1) = a_1$成立;
对此类A1集合的元素点, 存在与其对应的, 并由B0包含的子集B1与其1-1对应, 并且此时有 $g = f^{'}$, 不然将破坏1-1映射的规则；
2.因此对给定的 $b_2$∈B0: 有三种相互独立的cases：
第一种case是 $b_2$∈B1, 即与任一A1元素1-1对应的:
第二种case是对 $b_2$，存在 $a_2$∈A0，使得 $g(b_2) = a_2$,
并且此时 $g = f^{'}$;
第三种case是对 $b_2$，存在 $a_2$∈A0，使得 $g(b_2) = a_2$,
但是此时 $g ≠ f^{'}$，因此必然要存在 $b_3$∈B0，使得 $f(a_2) = b_3$;
...
经过n次这种 $g ≠ f^{'}$ 类似光子反射；
要么以 $g ≠ f^{'}$ 进行无限次光子反射；
要么以 $g = f^{'}$ 达成稳定的1-1映射，并且$g$与$f$在最终的一对两点上，互为反函数；

对给定的任意 $b_2$∈A1, 存在1-1映射$f$的像 $b_1$ ∈B0, 使 $f(a_1) = b_1$ 成立;

Schroeder-Bernstein Theorem
The Schroeder-Bernstein Theorem(sometimes called the Cantor-Schroeder-Bernstein Theorem)
is a result from set theory, named for Ernst Schröder and Felix Bernstein.
Informally, it implies that if two cardinalities are both less than or equal to each other, then they are equal.

More specifically, the theorem states that if $A$ and $B$ are sets, and there are injections $f: A \to B$ and $g : B \to A$, then there is a bijection $h : A \to B$.

The proof of the theorem does not depend on the axiom of choice, but only on the classical Zermelo-Fraenkel axioms.

Contents
1 Proof
2 Problems
2.1 Introductory
2.1.1 Problem 1
2.1.2 Problem 2
2.2 Intermediate
2.2.1 Problem 1
3 See Also
Proof
We call an element $b$ of $B$ lonely if there is no element $a \in A$ such that $f(a) = b$. We say that an element $b_1$ of $B$ is a descendent of an element $b_0$ of $B$ if there is a natural number $n$ (possibly zero) such that $b_1 = (f \circ g)^n (b_0)$.

We define the function $h: A \to B$ as follows: [h(a) = \begin{cases} g^{-1}(a), &\text{if } f(a) \text{ is the descendent of a lonely point,} \ f(a) &\text{otherwise.} \end{cases}] Note that $f(a)$ cannot be lonely itself. If $f(a)$ is the descendent of a lonely point, then $f(a) = f \circ g (b)$ for some $b$; since $g$ is injective, the element $g^{-1}(a)$ is well defined. Thus our function $h$ is well defined. We claim that it is a bijection from $A$ to $B$.

We first prove that $h$ is surjective. Indeed, if $b \in B$ is the descendent of a lonely point, then $h(g(b)) = b$; and if $b$ is not the descendent of a lonely point, then $b$ is not lonely, so there is some $a \in A$ such that $f(a) = b$; by our definition, then, $h(a) = b$. Thus $h$ is surjective.

Next, we prove that $h$ is injective. We first note that for any $a \in A$, the point $h(a)$ is a descendent of a lonely point if and only if $f(a)$ is a descendent of a lonely point. Now suppose that we have two elements $a_1, a_2 \in A$ such that $h(a_1) = h(a_2)$. We consider two cases.

If $f(a_1)$ is the descendent of a lonely point, then so is $f(a_2)$. Then $g^{-1}(a_1) = h(a_1) = h(a_2) = g^{-1}(a_2)$. Since $g$ is a well defined function, it follows that $a_1 = a_2$.

On the other hand, if $f(a_1)$ is not a descendent of a lonely point, then neither is $f(a_2)$. It follows that $f(a_1) = h(a_1) = h(a_2) = f(a_2)$. Since $f$ is injective, $a_1 = a_2$.

Thus $h$ is injective. Since $h$ is surjective and injective, it is bijective, as desired. $\blacksquare$

Problems
The Schroeder-Bernstein Theorem can be used to solve many cardinal arithmetic problems. For example, one may wish to show $|S|=\kappa$ for some cardinal $\kappa$. One strategy is to find sets $A,B$ such that $|A|=|B|=\kappa$ with injections from $A$ to $S$ and $S$ to $B$, thus concluding that $|A|=|S|=|B|=\kappa$. More generally, the Schroeder-Bernstein Theorem shows that the relation $|A|\leq|B|$ between cardinals $|A|$ and $|B|$ defined by "there is an injection $f:A\rightarrow B$" is a partial-order on the class of cardinals.

Introductory
Problem 1
Show that $\mathbb{Q}$ is countable.

Problem 2
Let $\kappa$ satisfy $\kappa\cdot\kappa=\kappa$. Show that $\kappa^{\kappa}=2^{\kappa}$.

Intermediate
Problem 1
We say a set of reals $O\subseteq\mathbb{R}$ is open if for all $r\in O$, there is an open interval $I$ satisfying $r\in I\subseteq O$. Show that the following sets are equal in cardinality:

$\mathbb{R}$
$2^{\aleph_{0}}$
$\{O\subset\mathbb{R}\mid O\text{ is open}\}$
$\{f:\mathbb{R}\rightarrow\mathbb{R}\mid f\text{ is continuous}\}$
See Also
Categories: Set theoryTheorems

posted @ 2024-03-12 22:07 abaelhe 阅读(98) 评论(0) 收藏举报

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SciTech-Mathmatics-RealAnalysis: Cantor-Schröder-Bernstein Theorem

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