SciTech-Mathmatics-Real Analysis-神奇的Cantor Set Theorem康托尔集 + Bolzono-Weierstrass Theorem + Bernstein Theorem

Cantor Set

• Priciple: 1-1 bi-directional mapping to determine whether two sets(infinite or finite) A and B have the same size.
• Cantor Set:
  • trisect the closed interval [0, 1] in same distance, remove intermediate one of open interval \((\frac{1}{3}, \ \frac{2}{3})\), and got two closed interval \([0, \ \frac{1}{3}]\) and \([\frac{2}{3}, \ \frac{2}{3}]\)
  • recursively process the remain closed intervals.
  • after \(n\) times of such kind of operations, we got \(2^n\) closed intervals with the same length: \(\frac{1}{3^n}\)
  • when n -> ∞, we got an infinte series of points \({x| x = \frac{1}{3^n}, \ n \in Z^+\)
  • Cantor Set 有许多“反直觉”的“神奇”性质：例如，在 Operations->∞ 时:
    0. Operations = \(\large n\) 时，总共有 \(2^n\) 个 长度为\(\large \frac{1}{3^n}\)的Closed Interval闭区间.
    1.[0, 1]闭区间被删去的开区间的“总长度”是1(在 Operations->∞ 时)；
    2.Cantor Set“每一段闭区间”，长度的极限为0(无限短无限近似一个点(长度上))；因此总的被删去的开区间长度为0；
    3. Cantor Set”每一段闭区间”，点数(Labeling)上有无限个连续点，
    因为每一段都有Real line的Continuity连续性(即Completeness完备性)。
    4. 因此Cantor Set每一个点都属于一个连续的Closed Interval闭区间，\(\exists \epsilon > 0, \forall p \in [a_n,\ b_n], \text{ such that }N(p, \epsilon) \in [a_n,\ b_n]\)
    即每一段闭区间都是完备集(“自密的”闭集).
    5. 完备集合(闭区间) 与 空集(删去的开区间) 交替出现组成原来的闭区间\([0,\ 1]\);

0. true: \(Z^+\) ~ \(偶数集\)；无限集合A 是 有其真子集 与 其对等的集合；
1. false: [0, 1] ~ \(Z^+\): 闭区间 [0, 1] 上全部的点作成的集合是 不对等于 \(Z^{+}\) 正整数集 上全部的点作成的集合(Cantor Set is a example of such kind of set)。
2. true: (0, 1) ~ (0, +∞)；Proof: B(c, r) where c=(0, 1) and 由点(0, 1)引射线通过右下方的1/4圆弧投射到(0, +∞).

Bolzano-Weierstrass Theorem

有 无穷多点 的 有界闭区间 必定 至少有一个 的 聚点。
using bisect methodology:
every step, at least one of the two bisected intervals have infinite points.
after n steps, also true.
when n -> +∞, use the \(\epsilon\ -\ \delta\) language to prove the limit ensures the n-Ball \(B(p,\ \sigma)\) has infinite points.

Bernstein Theorem:

If \(A\) ~ \(B^*\) and \(B\) ~ \(A^*\), where \(B^*\ is \ a \ subset\ of\ B\) and \(A^* \ is \ a \ subset \ of \ A\)
then \(A\) ~ \(B\).