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pages/Math4111/Math4111_L22.md

@@ -1 +1,175 @@
 # Lecture 22
+
+## Review
+
+Let $(a_n)$ be a sequence, and let $b_n = a_{f(n)}$ be a rearrangement, where $f$ is given by the following:
+
+| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | $\dots$ |
+| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
+| $f(n)$ | 1 | 2 | 4 | 3 | 6 | 8 | 5 | 10 | 12 | $\dots$ |
+
+(The pattern is "odd,even,even,") Defined the partial sums $s_n = \sum_{k=1}^n a_k$ and $t_n = \sum_{k=1}^n b_k$.
+
+1. In terms of $a_1,a_2,\ldots$, determine $s_n-t_n$ for $n=1,2,3,4,5,6,7$. (For example, $s_3-t_3 = a_3-a_4$.)  
+    $s_1 - t_1 = a_1 - a_1 = 0$  
+    $s_2 - t_2 = a_2 - a_2 = 0$  
+    $s_3 - t_3 = a_3 - a_4$  
+    $s_4 - t_4 = a_4 - a_4 = 0$  
+    $s_5 - t_5 = a_5 - a_6$  
+    $s_6 - t_6 = a_6 - a_8$  
+    $s_7 - t_7 = a_7 - a_8$  
+
+2. What is the smallest $n$ so that $s_n - t_n$ does not contain any of the terms $a_1,\dots, a_5$?  
+    $n=7$
+3. What is the smallest $n$ so that $s_n - t_n$ does not contain any of the terms $a_1,\dots, a_{10}$?  
+    $n=13$
+
+## New Material
+
+### Absolute Convergence
+
+#### Theorem 3.55
+
+Let $(a_n)$ be a sequence in $\mathbb{C}$ such that $\sum_{n=1}^\infty |a_n|$ converges. If $(b_n)$ is a rearrangement of $(a_n)$, then $\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty b_n$.
+
+Proof:
+
+Let $f:\mathbb{N}\to\mathbb{N}$ be a bijection and let $b_n = a_{f(n)}$.
+
+Let $I(n)=\{1,2,\dots,n\}$, $J(n)=\{f(1),f(2),\dots,f(n)\}$.
+
+Then $s_n = \sum_{k\in I(n)} a_k$ and $t_n = \sum_{k\in J(n)} a_k$.
+
+$$
+\begin{aligned}
+s_n - t_n &= \sum_{k=1}^n a_k - \sum_{k=1}^n b_k \\
+&= \sum_{k\in I(n)} a_k - \sum_{k\in J(n)} a_{f(k)} \\
+&= \sum_{k\in I(n)\backslash J(n)} a_k - \sum_{k\in J(n)\backslash I(n)} a_{f(k)} \\
+|s_n - t_n|&\leq \sum_{i\in (I(n)\backslash J(n))\cup(J(n)\backslash I(n))} |a_i|
+\end{aligned}
+$$
+
+We will show that $\lim_{n\to\infty} |s_n - t_n| = 0$.
+
+Let $\epsilon > 0$.
+
+By the Cauchy criterion applied to $\sum_{n=1}^\infty |a_n|$, there exists $N$ such that if $m,n\geq N$, then $\sum_{k=m+1}^n |a_k| < \epsilon$.
+
+Then we choose $p\in\mathbb{N}$ such that $I(n)\subset J(p)$ ($\{1,2,\dots,N\}\subset\{f(1),f(2),\dots,f(p)\}$). $p=\max\{f^{-1}(1),f^{-1}(2),\dots,f^{-1}(N)\}$.
+
+Note: This implies that $p$ is at least $N$.
+
+If $n\geq p$, then $I(n)\subset J(p)\subset I(n)\cap J(n)$ so $s_n = t_n$.
+
+So,
+
+$$
+|s_n - t_n| \leq \sum_{i=N+1}^{\max J(n)} |a_i| < \epsilon
+$$
+
+> Back to the example of the review question.
+>
+> $I(9)=\{1,2,\dots,9\}$, $J(9)=\{1,2,4,3,6,8,5,10,12\}$, $I(9)\backslash J(9)=\{7,9\}$, $J(9)\backslash I(9)=\{10,12\}$.
+>
+>$$
+|s_9 - t_9| \leq |a_7|+|a_9|+|a_{10}|+|a_{12}| \leq \sum_{k=7}^{12} |a_k| < \epsilon
+$$
+
+This proves that $\lim_{n\to\infty} |s_n - t_n| = 0$.
+
+Since $\lim_{n\to\infty} s_n$ exists, $\lim_{n\to\infty} s_n = \lim_{n\to\infty} t_n$.
+
+EOP
+
+#### Theorem 3.54
+
+Special case of Riemann rearrangement theorem
+
+Suppose $a_n\in \mathbb{R}$, $\sum_{n=1}^\infty a_n$ converges conditionally. (i.e. $\sum_{n=1}^\infty a_n$ converges but $\sum_{n=1}^\infty |a_n|$ diverges.) Then $\forall \alpha\in\mathbb{R}$, there exists a rearrangement $(b_n)$ of $(a_n)$ such that $\sum_{n=1}^\infty b_n = \alpha$.
+
+## Chapter 4 Continuity
+
+### Limits of Functions
+
+#### Definition 4.1
+
+Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and $E\subset X$, $p\in E'$, $q\in Y$. Let $f:E\to Y$. We say that $\lim_{x\to p} f(x) = q$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that $\forall x\in E$, if $0 < d_X(x,p) < \delta$, then $d_Y(f(x),q) < \epsilon$.
+
+This is same as:
+
+If $\lim_{x\to p} f(x)=q$, then
+
+$$
+\forall \epsilon > 0, \exists \delta > 0, \forall x\in E, 0 < d_X(x,p) < \delta \implies d_Y(f(x),q) < \epsilon
+$$
+
+In many definitions, $E=X$
+
+#### Theorem 4.2
+
+$\lim_{x\to p} f(x) = q \iff$ forall sequence $(p_n)$ in $E\backslash\{p\}$ with $p_n\to p$, $f(p_n)\to q$.
+
+Proof:
+
+$\implies$
+
+Suppose $\lim_{x\to p} f(x) = q$.
+
+Let $(p_n)$ be a sequence in $E\backslash\{p\}$ with $p_n\to p$.
+
+Let $\epsilon > 0$.
+
+By definition of limit of function, $\exists \delta > 0$ such that if $0 < d_X(x,p) < \delta$, then $d_Y(f(x),q) < \epsilon$.
+
+Since $p_n\to p$, $\exists N$ such that if $n\geq N$, then $0 < d_X(p_n,p) < \delta$. So $f(p_n)\in B_\epsilon(q)$.
+
+Thus, we showed $\forall \epsilon > 0$, $\exists N$ such that if $n\geq N$, then $f(p_n)\in B_\epsilon(q)$.
+
+$\impliedby$
+
+We proceed by contrapositive.
+
+Suppose $\lim_{x\to p} f(x) \neq q$, then $\exists$ sequence $(p_n)$ in $E\backslash\{p\}$ with $p_n\to p$ such that $f(p_n)\cancel{\to} q$.
+
+Suppose $\lim_{n\to\infty} f(p_n) \neq q$, then $\exists \epsilon > 0$ such that for all $\delta > 0$, there exists $x\in E\cap B_\delta(p)\backslash\{p\}$ such that $f(x)\notin B_\epsilon(q)$.
+
+For $n\in\mathbb{N}$, if we apply this with $\delta = \frac{1}{n}$, we get $p_n\in E\cap B_{\frac{1}{n}}(p)\backslash\{p\}$ such that $f(p_n)\notin B_\epsilon(q)$.
+
+Then: $(p_n)$ is a sequence in $E\backslash\{p\}$ with $d_X(p_n,p) = \frac{1}{n}\to 0$ so that as $n\to\infty$, $f(p_n)\notin B_\epsilon(q)$.
+
+So $\lim_{n\to\infty} f(p_n) \neq q$.
+
+EOP
+
+With this theorem, we can use the properties of limit of sequences to study limits of functions.
+
+To prove things about limits of functions, we can use sequences.
+
+- If $\lim_{x\to p} f(x) = q$, and $\lim_{x\to p} f(x)=q'$, then $q=q'$.
+- If $\lim_{x\to p} f(x) = A$ and $\lim_{x\to p} g(x) = B$, then $\lim_{x\to p} f(x) + g(x) = A+B$.
+
+### Continuous functions
+
+#### Definition 4.5
+
+Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and $E\subset X$, $p\in E$. Let $f:E\to Y$. We say that $f$ is continuous at $p$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that $f(E\cap B_\delta(p))\subset B_\epsilon(f(p))$.
+
+- We say that $f$ is continuous on $E$ if $f$ is continuous at every $p\in E$.
+
+> Remark: For $p\in E$, there are two possibilities.
+>
+> - $p$ is an isolated point of $E$.
+> - $p$ is a limit point of $E$.
+>
+> In the first case, $f$ is automatically continuous at $p$. ($E\cap B_\delta(p)=\{p\}$ for all $\delta > 0$.)
+>
+> In the second case, $f$ is continuous at $p$ if and only if $\lim_{x\to p} f(x) = f(p)$.
+
+#### Theorem 4.8
+
+A function $f:E\to Y$ is continuous at $p\in E$ if the pre-image of every open set is open.
+
+Two consequences if $f:E\to Y$ is continuous:
+
+- Image of compact set is compact. (Implies Extreme Value Theorem)
+- Image of connected set is connected. (Implies Intermediate Value Theorem)

pages/Math4111/_meta.js

@@ -21,9 +21,7 @@ export default {
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     Math4111_L20: "Lecture 20",
     Math4111_L21: "Lecture 21",
-    Math4111_L22: {
-        display: 'hidden'
-    },
+    Math4111_L22: "Lecture 22",
     Math4111_L23: {
         display: 'hidden'
     },