41 lines
1.2 KiB
Markdown
41 lines
1.2 KiB
Markdown
# Math4121 Lecture 19
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## Continue on the "small set"
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### Cantor set
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#### Theorem: Cantor set is perfect, nowhere dense
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Proved last lecture.
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_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 $SVC(n)$_
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Back to $\frac{1}{3}$ Cantor set.
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Every step we delete $\frac{2^{n-1}}{3^n}$ of the total "content".
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Thus, the total length removed after infinitely many steps is:
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$$
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\sum_{n=1}^{\infty} \frac{2^{n-1}}{3^n} = \frac{1}{3}\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n=1
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$$
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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:
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_skip this part, some error occurred._
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#### Monotonicity of outer content
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If $S\subseteq T$, then $c_e(S)\leq c_e(T)$.
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Proof:
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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 $c_e(T)$, $c_e(S) \leq c_e(T)$.
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QED
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#### Theorem Osgood's Lemma
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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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