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@@ -54,7 +54,7 @@ What is the structure of $S=[0,1]\setminus T$? (or Sparse)
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- Topologically (not dense)
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- Measure, for now meaning small or zero outer content.
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## Chapter 4: Nowhere Dense SEts and the Problem with the Fundamental Theorem of Calculus
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## Chapter 4: Nowhere Dense Sets and the Problem with the Fundamental Theorem of Calculus
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### Nowhere Dense Sets
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@@ -1 +1,72 @@
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# Lecture 18
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# Math4121 Lecture 18
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## Continue
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### Small sets
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A set that is nowhere dense, has zero outer content yet is uncountable.
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By modifying this example, we can find similar with any outer content between 0 and 1.
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#### Definition: Perfect Set
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$S\subsetes[0,1]$ is perfect if $S=S'$.
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Example:
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- $[0,1]$ is perfect
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- perfect sets are closed
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- Finite collection of points is not perfect because they do not have limit points.
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- perfect sets are uncountable (no countable sets can be perfect)
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#### Middle third Cantor set
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We construct the set by removing the middle third of the interval.
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Let $C_0=[0,1]$, $C_1=[0,\frac{1}{3}]\cup[\frac{2}{3}]$ ...
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Continuing this process indefinitely, we define the Cantor set as
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$$
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C=\Bigcap_{n=0}^{\infty}C_n
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$$
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1. $C_n\subseteq C_{n-1}$
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2. $\ell(C_n)=\ell(C_{n-1})$
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3. Each $C_n$ is closed.
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> The algebraic expression for $C_n$, where $a\in[0,1]$, we write as a decimal expansion in base $3$.
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>
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> $$ a=\sum_{n=1}^{\infty} \frac{a_n}{3^n}$$, where $a_n\in\{0,1,2\}$.
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>
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> In this case, $C_0\to C_1$ means deleting all numbers with $a_1=1$. (the same as deleting the interval $[\frac{1}{3},\frac{2}{3}]$)
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>
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> $C_1\to C_2$ means deleting all the numbers with $a_2=1$.$
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>
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> So we can write the set as $$C=\left\{\sum_{n=1}^{\infty}\frac{a_n}{3^n},a_n\in\{0,2\}\right\}$$
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#### Proposition 4.1
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$C$ is perfect and nowhere dense, and outer content is 0.
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Proof:
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(i) $c_e(C)=0$
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Let $\epsilon>0$, then $\exists n$ such that $\left(\frac{2}{3}\right)<\epsilon$. Then $C_n$ is a cover of $C$, and $\ell(C_n)<\epsilon$.
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(ii) $C$ is perfect
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Since $C_n$ is closed, $C$ is closed (any intersection of closed set is closed) so $C'\subseteq C$.
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Let $a\in C$, and we need to show $a$ is a limit point. Let $\epsilon>0$, and we need to find $a^*\in C\setminus\{a\}$ and $|a^* - a| < \epsilon$. Suppose $a=\sum_{n=1}^{\infty} \frac{a_n}{3^n}, a_n \in \{0, 2\}$, Notive that if $a^*\in C$ has the expansion as $a$ except the k-th term.
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So $|a^*-a|=\frac{2}{3^k}$, which can be made arbitrarily small by choosing a sufficiently large $k$. Thus, $a$ is a limit point of $C$, proving that $C$ is perfect.
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(iii) $C$ is nowhere dense
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It is sufficient to show $C$ contains no intervals.
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Any open intervals has a real number with 1 in it's base 3 decimal expansion (proof in homework)
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_take some interval in $(a,b)$ we can change the digits that is small enough and keep the element still in the set_
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@@ -20,9 +20,7 @@ export default {
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Math4121_L15: "Introduction to Lebesgue Integration (Lecture 15)",
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Math4121_L16: "Introduction to Lebesgue Integration (Lecture 16)",
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Math4121_L17: "Introduction to Lebesgue Integration (Lecture 17)",
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Math4121_L18: {
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display: 'hidden'
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},
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Math4121_L18: "Introduction to Lebesgue Integration (Lecture 18)",
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Math4121_L19: {
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display: 'hidden'
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},
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