1.2 KiB
Math 4121 Lecture 19
Continue on the "small set"
Cantor set
Theorem: Cantor set is perfect, nowhere dense
Proved last lecture.
Other construction of the set by removing the middle non-zero interval (\frac{1}{n},n>0) and take the intersection of all such steps is called $SVC(n)$
Back to \frac{1}{3} Cantor set.
Every step we delete \frac{2^{n-1}}{3^n} of the total "content".
Thus, the total length removed after infinitely many steps is:
\sum_{n=1}^{\infty} \frac{2^{n-1}}{3^n} = \frac{1}{3}\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n=1
However, the quarter cantor set removes \frac{3^{n-1}}{4^n} of the total "content", and the total length removed after infinitely many steps is:
skip this part, some error occurred.
Monotonicity of outer content
If S\subseteq T, then c_e(S)\leq c_e(T).
Proof:
If C is cover of T, then S\subseteq T\subseteq C, so C is a cover of S. Since c_e(s) takes the inf over a larger set that c_e(T), c_e(S) \leq c_e(T).
QED
Theorem Osgood's Lemma
Let S be a closed, bounded set in \mathbb{R}, and S_1\subseteq S_2\subseteq \ldots, and S=\bigcup_{n=1}^{\infty} S_n. Then \lim_{k\to\infty} c_e(S_k)=c_e(S).