Math4121 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:

Every time we remove \frac{1}{4^n} of the remaining intervals. So on each layer, we remove \frac{2^{n-1}}{4^n} of the total "content".

So the total length removed is:


\begin{aligned}
1-\frac{1}{4}-\frac{2}{4^2}-\frac{2^2}{4^3}-\cdots&=1-\frac{1}{4}\sum_{n=0}^{\infty} \left(\frac{2}{4}\right)^n\\
&=1-\frac{1}{4}\cdot\frac{1}{1-\frac{2}{4}}\\
&=1-\frac{1}{4}\cdot\frac{4}{2}\\
&=1-\frac{1}{2}\\
&=\frac{1}{2}
\end{aligned}

Generalized Cantor set (SVC(n))

The outer content of SVC(n) is \frac{n-3}{n-2}.

Monotonicity of outer content

If S\subseteq T, then c_e(S)\leq c_e(T).

Proof of Monotonicity of outer content

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).

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).

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