initial commit

next.config.mjs

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+import nextra from 'nextra'
+ 
+const withNextra = nextra({
+  theme: 'nextra-theme-blog',
+  themeConfig: './theme.config.jsx'
+})
+ 
+export default withNextra()
+ 
+// If you have other Next.js configurations, you can pass them as the parameter:
+// export default withNextra({ /* other next.js config */ })

package.json

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+{
+    "scripts": {
+        "dev": "next",
+        "build": "next build",
+        "start": "next start"
+    },
+    "dependencies": {
+        "next": "^15.0.3",
+        "react": "^18.3.1",
+        "react-dom": "^18.3.1",
+        "nextra": "^3.2.3",
+        "nextra-theme-blog": "^3.2.3"
+    }
+}

pages/Math4111/Math4111_E2.md

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+# Math 4111 Exam 2 review
+
+$E$ is open if $\forall x\in E$, $x\in E^\circ$ ($E\subset E^\circ$)
+
+$E$ is closed if $E\supset E'$
+
+Then $E$ closed $\iff E^c$ open $\iff \forall x\in E^\circ, \exists r>0$ such that $B_r(x)\subset E^c$
+
+$\forall x\in E^c$, $\forall x\notin E$
+
+$B_r(x)\subset E^c\iff B_r(x)\cap E=\phi$
+
+## Past exam questions
+
+$S,T$ is compact $\implies S\cup T$ is compact
+
+Proof:
+
+Suppose $S$ and $T$ are compact, let $\{G_\alpha\}_{\alpha\in A}$ be an open cover of $S\cup T$
+
+(NOT) $\{G_\alpha\}$ is an open cover of $S$, $\{H_\beta\}$ is an open cover of $T$.
+
+...
+
+EOP
+
+## K-cells are compact
+
+
+We'll prove the case $k=1$ and $I=[0,1]$ (This is to simplify notation. This same ideas are used in the general case)
+
+Proof:
+
+That $[0,1]$ is compact.
+
+(Key idea, divide and conquer)
+
+Suppose for contradiction that $\exists$ open cover $\{G_a\}_{\alpha\in A}$ of $[0,1]$ with no finite subcovers of $[0,1]$
+
+**Step1.** Divide $[0,1]$ in half. $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$ and at least one of the subintervals cannot be covered by a finite subcollection of $\{G_\alpha\}_{\alpha\in A}$
+
+(If both of them could be, combine the two finite subcollections to get a finite subcover of $[0,1]$)
+
+Let $I_1$ be a subinterval without a finite subcover.
+
+**Step2.** Divide $I_1$ in half. Let $I_2$ be one of these two subintervals of $I_1$ without a finite subcover.
+
+**Step3.** etc.
+
+We obtain a seg of intervals $I_1\subset I_2\subset \dots$ such that
+
+(a) $[0,1]\supset I_1\supset I_2\supset \dots$  
+(b) $\forall n\in \mathbb{N}$, $I_n$ is not covered by a finite subcollection of $\{G_\alpha\}_{\alpha\in A}$  
+(c) The length of $I_n$ is $\frac{1}{2^n}$
+
+By (a) and **Theorem 2.38**, $\exists x^*\in \bigcap^{\infty}_{n=1} I_n$.
+
+Since $x^*\in [0,1]$, $\exists \alpha_0$ such that $x^*\in G_{\alpha_0}$
+
+Since $G_{\alpha_0}$ is open, $\exist  r>0$ such that $B_r(x^*)\subset G_{\alpha_0}$
+
+Let $n\in \mathbb{N}$ be such that $\frac{1}{2^n}<r$. Then by $(c)$, $I(n)\subset B_r(x^*)\subset G_{\alpha_0}$
+
+Then $\{G_{\alpha_0}\}$ is a cover of $I_n$ which contradicts with (b)
+
+EOP
+
+## Redundant subcover question
+
+$M$ is compact and $\{G_\alpha\}_{\alpha\in A}$ is a "redundant" subcover of $M$.
+
+$\exists \{G_{\alpha_i}\}_{i=1}^n$ is a finite subcover of $M$.
+
+We define $S$ be the $x\in M$ that is only being covered once.
+
+$$
+S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)
+$$
+
+We claim $S$ is a closed set.
+
+$G_{\alpha_i}\cap G_{\alpha_j}$ is open.
+
+$\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)$ is closed
+
+$S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)$ is closed.
+
+So $S$ is compact, we found another finite subcover yeah!
+

pages/Math4111/Math4111_E3.md

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+# Exam 3 Review session
+
+## Relations between series and topology (compactness, closure, etc.)
+
+Limit points $E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\backslash\{x\}\cap E\neq\emptyset\}$
+
+Closure $\overline{E}=E\cup E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\cap E\neq\emptyset\}$
+
+$p_n\to p\implies \forall \epsilon>0, \exists N$ such that $\forall n\geq N, p_n\in B_\epsilon(p)$
+
+### Some interesting results
+
+#### Lemma
+
+$p\in \overline{E}\iff \exists (p_n)\subseteq E$ such that $p_n\to p$
+
+$p\in E'\iff \exists (p_n)\subseteq E\backslash\{p\}$ such that $p_n\to p$ (you cannot choose $p$ in the sequence)
+
+#### Bolzano-Weierstrass Theorem
+
+Let $E$ be a compact set and $(p_n)$ be a sequence in $E$. Then $\exists (p_{n_k})\subseteq (p_n)$ such that $p_{n_k}\to p\in E$.
+
+Rudin Proof:
+
+Rudin's proof uses a fact from Chapter 2.
+
+If $E$ is compact, and $S\subseteq E$ is infinite, then $S$ has a limit point in $E$ ($S'\cap E\neq\emptyset$).
+
+## Examples of Cauchy sequence that does not converge
+
+> Cauchy sequence in $(X,d),\forall \epsilon>0, \exists N$ such that $\forall m,n\geq N, d(p_m,p_n)<\epsilon$
+
+Let $X=\mathbb{Q}$ and $(p_q)=\{1,1.4,1.41,1.414,1.4142,1.41421,\dots\}$ The sequence is Cauchy but does not converge in $\mathbb{Q}$.
+
+This does not hold in $\mathbb{R}$ because compact metric spaces are complete.
+
+Fact: Every Cauchy sequence is bounded.
+
+## Proof that $e$ is irrational
+
+> $e=\sum_{n=0}^\infty \frac{1}{n!}$
+
+Let $s_n=\sum_{k=0}^n \frac{1}{k!}$
+
+So $e-s_n=\left(\sum_{k=n+1}^\infty \frac{1}{k!}\right)<\frac{1}{n!n}$
+
+If $e$ is rational, then $\exists p,q\in\mathbb{Z}$ such that $e=\frac{q}{p}$ and $q!s_q\in\mathbb{Z}$, $q!e=q!\frac{p}{q}\in \mathbb{Z}$, so $q!(e-s_q)\in\mathbb{Z}$
+
+$0<q!(e-s_q)<\frac{1}{n!n}$ leads to contradiction.
+
+## $\limsup$ and $\liminf$
+
+Let $(a_n)=(-1)^n$
+
+$\limsup a_n=1$ and $\liminf a_n=-1$
+
+Let $(a_n)\to a$
+
+$\limsup a_n=\liminf a_n=a$
+
+### Facts about $\limsup$ and $\liminf$
+
+#### Convergence of subsequence
+
+_$\limsup$ is the largest value that subsequence of $a_n$ can approach to._
+
+_$\liminf$ is the smallest value that subsequence of $a_n$ can approach to._
+
+#### Elements of sequence
+
+$\forall x>s^*,\{n:a_n>x\}$ is finite. $\exists N$ such that $\forall n\geq N, a_n\leq x$
+
+$\forall x<s^*,\{n:a_n>x\}$ is infinite.
+
+One example is $(a_n)=(-1)^n\frac{n}{n+1}$
+
+$\limsup a_n=1$ and $\liminf a_n=-1$
+
+So the size of set of elements of $a_n$ that are greater than any $x<1$ is infinite. and the size of set of elements of $a_n$ that are greater than any $x>1$ is finite.
+
+#### $\limsup(a_n+b_n)\leq \limsup a_n+\limsup b_n$
+
+One example for smaller than is $(a_n)=(-1)^n$ and $(b_n)=(-1)^{n+1}$
+
+$\limsup(a_n+b_n)=0$ and $\limsup a_n+\limsup b_n=2$
+
+## ($\forall n,s_n\leq t_n$) $\implies \limsup s_n\leq \limsup t_n$
+
+One example of using this theorem is $(s_n)=\left(\sum_{k=1}^n\frac{1}{k!}\right)$ and $(t_n)=\left(\frac{1}{n}+1\right)^n$
+
+## Rearrangement of series
+
+Will not be tested.
+
+_infinite sum is not similar to finite sum. For infinite sum, the order of terms matters. But for finite sum, the order of terms does not matter, you can rearrange the terms as you want._
+
+## Ways to prove convergence of series
+
+### n-th term test (divergence test)
+
+If $\lim_{n\to\infty}a_n\neq 0$, then $\sum a_n$ diverges.
+
+### Definition of convergence of series (convergence and divergence test)
+
+If $\sum a_n$ converges, then $\lim_{n\to\infty}\sum_{k=1}^n a_k=0$.
+
+Example: Telescoping series and geometric series.
+
+### Comparison test (convergence and divergence test (absolute convergence))
+
+Let $(a_n)$ be a sequence in $\mathbb{C}$ and $(c_n)$ be a non-negative sequence in $\mathbb{R}$. Suppose $\forall n, |a_n|\leq c_n$.
+
+(a) If the series $\sum_{n=1}^{\infty}c_n$ converges, then the series $\sum_{n=1}^{\infty}a_n$ converges.  
+(b) If the series $\sum_{n=1}^{\infty}a_n$ diverges, then the series $\sum_{n=1}^{\infty}c_n$ diverges.
+
+### Ratio test (convergence and divergence test (absolute convergence))
+
+> $$ \left|\frac{a_{n+1}}{a_n}\right| \leq \alpha \implies |a_n|\leq \alpha^n$$
+
+Given a series $\sum_{n=0}^{\infty} a_n$, $a_n\in\mathbb{C}\backslash\{0\}$.
+
+Then
+
+(a) If $\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$, then $\sum_{n=0}^{\infty} a_n$ converges.  
+(b) If $\left|\frac{a_{n+1}}{a_n}\right| \geq 1$ for all $n\geq n_0$ for some $n_0\in\mathbb{N}$, then $\sum_{n=0}^{\infty} a_n$ diverges.
+
+
+### Root test (convergence and divergence test (absolute convergence))
+
+> $$ \sqrt[n]{|a_n|} \leq \alpha \implies |a_n|\leq \alpha^n$$
+
+Given a series $\sum_{n=0}^{\infty} a_n$, put $\alpha = \limsup_{n\to\infty} \sqrt[n]{|a_n|}$.
+
+Then
+
+(a) If $\alpha < 1$, then $\sum_{n=0}^{\infty} a_n$ converges.  
+(b) If $\alpha > 1$, then $\sum_{n=0}^{\infty} a_n$ diverges.  
+(c) If $\alpha = 1$, the test gives no information
+
+
+### Cauchy criterion
+
+### Geometric series
+
+### P-series
+
+
+(a) $\sum_{n=0}^{\infty}\frac{1}{n}$ diverges.  
+(b) $\sum_{n=0}^{\infty}\frac{1}{n^2}$ converges.
+
+### Cauchy condensation test (convergence test)
+
+Suppose $(a_n)$ is a non-negative sequence. The series $\sum_{n=1}^{\infty}a_n$ converges if and only if the series $\sum_{k=0}^{\infty}2^ka_{2^k}$ converges.
+
+### Dirichlet test (convergence test)
+
+Suppose
+
+(a) the partial sum $A_n$ of $\sum a_n$ form a bounded sequence.  
+(b) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)  
+(c) $\lim_{n\to\infty}b_n=0$.
+
+Then $\sum a_nb_n$ converges.
+
+Example: $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ converges.
+
+### Abel's test (convergence test)
+
+Let $(b_n)^\infty_{n=0}$ be a sequence such that:
+
+(a) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)
+(b) $\lim_{n\to\infty}b_n=0$
+
+Then if $|z|=1$ and $z\neq 1$, $\sum_{n=0}^\infty b_nz^n$ converges.

pages/Math4111/Math4111_L1.md

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+# Lecture 1
+
+## Introduction 
+
+Reading is not recommended before class, it;s hard.
+
+## Chapter 1: The real number and complex number systems
+
+* Natural numbers: $\mathbb{N}=\{1,2,3,4....\}$ note by some conventions, $0$ is also a natural number
+
+* IntegersL $\mathbb{Z}=\{...,-2,-1,0,1,2,...\}$
+
+* Rational numbers: $\mathbb{Q}=\{\frac{m}{n}:m,n\in\mathbb{Z}\ and\ n\neq 0\}$
+
+* Real numbers: $\mathbb{R}$ the topic of chapter
+
+* Complex numbers: $\mathbb{C}=\{a+bi:a,b\in \mathbb{R}\}$
+
+### Theorem ($\sqrt{2}$ is irrational)
+
+$\exist p\in \mathbb{Q},p^2=2$ is false.
+
+$\equiv\cancel{\exist} p\in \mathbb{Q}, p^2=2$
+
+$\equiv p\in \mathbb{Q},p^2\neq 2$
+
+#### Proof
+
+Suppose for contradiction, $\exist p\in \mathbb{Q}$ such that $p^2=\mathbb{Q}$.
+
+Let $p=\frac{m}{n}$, where $m,n \in \mathbb{Z}$ are not both even. (reduced form)
+
+$p^2=2$ and $p=\frac{m}{n}$, so $m^2=2n^2$, so $m^2$ is even, $m$ is even.
+
+So $m^2$ is divisible by 4, $2n^2$ is divisible by 4.
+
+So $n^2$ is even. but they are not both even.
+
+EOP
+
+### Theorem (No closest rational for a irrational number)
+
+Let $A=\{p\in \mathbb{q}, p>0\ and\ p^2\leq 2\}$, Then $A$ does not have a largest element.
+
+i.e. $\exist p\in A$ such that $\forall q\in A, q\leq p$ is false.
+
+> Remark: The book give a very slick proof trying to lean from these kinds of proofs takes some effort. (It is perfectly fine to write that solution this way...)
+
+#### Thought process
+
+Let $p\in A,p\in \mathbb{Q}$, $p>0, p^2<2$.
+
+We want a $\delta\in\mathbb{Q}$ such that $\delta>0$ and $(p+\delta)^2<2$.
+
+$$
+\begin{aligned}
+    
+(p+\delta)^2&<2\\
+p^2+2p\delta+\delta^2&<2\\
+\delta(2p+\delta)&< 2-p^2\\
+\delta&<\frac{2-p^2}{2p-\delta}
+\end{aligned}
+$$
+
+From $(p+\delta)^2<2$, we know $\delta<2$ (this is a crude bound, $\delta<\sqrt{2}$).
+
+So one choice can be $\delta=\frac{2-p^2}{2p+2}$
+
+#### Proof
+
+$\forall p\in A$, we can find a $\delta=\frac{2-p^2}{2p+2}$ which is greater than zero ($p^2<2,2-p^2>0,2p+2>0,\delta>0$) and construct a new number $(p+\delta)^2$ such that $p^2<(p+\delta)^2<2$.
+
+_Here we construct a formula for approximate $\sqrt{2}=\lim_{i\to \infty}p_0=1,p_{i+1}=p_i+\frac{2-p_i^2}{2p_i+2}$_
+
+Interesting...
+
+We can also further optimize the formula by changing the bound of $\delta$ to $\delta< 2-p$, since $(p+\delta)^2<2,p+\delta<2$
+
+```python
+def sqrt_2(acc):
+    if acc==0: return 1
+    c=sqrt_2(n-1)
+    return c+((2-c**2)/(2*c+2))
+```
+
+### Definition and notations for sets
+
+Some set notation
+
+$\Pi\in \mathbb{R}$
+
+use $\subset,\subsetneq$ in this class.
+
+* $A\subset B$, $\forall x\in A, x\in B$
+* $A=B$, $A\subset B$ and $B\subset A$
+* $A\subsetneq$ means $A\subset B$ and $A\neq B$
+

pages/Math4111/Math4111_L10.md

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+# Lecture 10
+
+## Review
+
+Recall: If $K\subset \cup_{\alpha\in A} G_{\alpha}$, then we say $\{G_\alpha\}_{\alpha\in A}$ is a cover of $K$. If, in addition, each set $G_{\alpha}$ is open, then we say $\{G_{\alpha}\}_{\alpha\in A}$ is an open cover of $K$. If $\alpha_1,...,\alpha_n\in A$ are such that $K\subset \bigcup _{i=1}^n G_{\alpha_i}$, then we say $\{G_{\alpha_i}\}_{i=1}^n$ is a finite subcover.
+
+Let $X=\mathbb{R}$. Come up with some examples of covers of $\mathbb{R}$. Try to find a few satisfying each of the following:
+
+1. A cover of $\mathbb{R}$ which is not an open cover.
+    $\{[x,x+1]:x\in \mathbb{Z}\}$
+2. An open cover of $\mathbb{R}$ which does have a finite subcover.
+    $\{\mathbb{R}\}$ is an open cover with finite subcover, itself $\{\mathbb{R}\}$. AND, $\{\mathbb{Q},\mathbb{R}\backslash\mathbb{Q}\}$ is not a subcover of $\{\mathbb{R}\}$ since we need to select subcover from cover set. And not taking the element of sets in the open cover.
+3. An open cover of $\mathbb{R}$ which does not have a finite subcover.
+    $\{(x,x+2):x\in \mathbb{Z}\}$ No finite subcover, infinitely many sets.
+
+    Proof: we proceed by contradiction, suppose we take $\{(n_i,n_i+2):i=1,...,k\}$. The union does not contain $max\{n_1,...,n_k\}+2$
+
+    $\{\{x\in\mathbb{R}:x<n\}:n\in \mathbb{Z}\}$
+
+    And some stupid set we have is $\{\{x\in\mathbb{R}:x<n\}:n\in \mathbb{Z}\}\cup \{\mathbb{R}\}$ with finite subcover $\{\mathbb{R}\}$
+
+## New
+
+### Compact set
+
+$K$ is compact if $\forall$ open cover, $\exists \{G_{\alpha_i}\}_{i=1}^n$ that is a finite subcover.
+
+$\mathbb{R}$ is not compact since we can build a open cover $\{(x,x+2):x\in \mathbb{Z}\}$ such that we cannot find a finite subcover of $\mathbb{R}$
+
+$\{1,2\}$ is compact let $ $\{G_{\alpha}\}_{\alpha\in A}$ be an open cover of $\{1,2\}$
+
+Ironically, $[0,1]$ is compact. This will follow from Theorem 2.40.
+
+#### Theorem 2.33
+
+Let $(X,d)$ be a metric space, and $K\subset Y \subset X$. Then $K$ is compact relative to $X$ ($K$ is open in $X$ ) $\iff K$ is compact relative to $Y$. _This implies that compactness is an absolute property_
+
+Proof:
+
+$\implies$ Suppose $K$ is compact relative to $X$. 
+
+Let $\{V_{\alpha}\}_{\alpha\in A}$ be an open cover of $K$ relative to $Y$.
+
+By Theorem 2.30, for each $\alpha$, $\exist G_{\alpha}$ open in $X$ such that $V_{\alpha}=G_{\alpha}\cap Y$. Then $\{G_\alpha\}_{\alpha}$ is an open cover of $K$ relative to $X$. Since $K$ is compact relative to $X$, $\{G_{\alpha}\}_{\alpha}$ has a finite subcover $\{G_{\alpha_i}\}_{i=1}^n$. Then $\{V_{\alpha_i}\}^n_{i=1}$ is a finite subcover of $\{V_{\alpha}\}_{\alpha}$ of $K$
+
+$\impliedby$ Suppose $K$ is compact relative to $Y$. Let $\{G_\alpha\}_{\alpha\in A}$ be an open cover of $K$ relative to $X$. By **Theorem 2.30**, $\{G_\alpha\cap Y\}_\alpha$ is an open cover of $K$ relative to $X$
+
+Since $k$ is compact relative to $Y$, $\{G_\alpha\cap Y\}_\alpha$ has a finite subcover $\{G_{\alpha_i}\cap Y\}_{i=1}^n$. Then $\{G_{\alpha_i}\}_{i=1}^n$ is a finite subcover of $\{G_\alpha\}_{\alpha}$ of $K$.
+
+EOP
+
+#### Theorem 2.24
+
+Let $(X,d)$ be a metric space, $K\subset X$ is compact $\implies K$ is closed in $X$.
+
+Proof:
+
+Suppose $K$ is compact. We'll show that $K^c$ is open, i.e $\forall p\in K^c$, $\exists r>0$ such that $B_r(p)<K^c$. Let $p\in K^c$ ($p$ is fixed for the remainder of the $q$)
+
+For each $q\in K$, Let $V_q=B_{\frac{1}{2}d(p,q)}(p),W_q=B_{\frac{1}{2}d(p,q)}(q)$ By triangle inequality, $V_q\cap W_q=\phi$
+
+Since $K\subset \bigcup_{q\in K}W_q$ and $\forall q\in K, V_q\cap W_q=\phi$. By set theory, $\left(\bigcap_{q\in K}V_q\right)\cap K=\phi$, and $\left(\bigcap_{q\in K}V_q\right)\subset K^c$
+
+Since $\{W_q\}_{q\in K}$ is an open cover of $K$, so it has a finite cover $\{W_{q_i}\}_{i=1}^n$. So $\left(\bigcap_{q\in K}V_q\right)=\left(\bigcap_{i=1}^n V_{q_i}\right)\subset K^c$.
+
+Then let $r=min\{\frac{1}{2}d(p_i,q_i)\}$, $\bigcap_{i=1}^n V_{q_i}=B_r(p)$

pages/Math4111/Math4111_L11.md

@@ -0,0 +1,104 @@
+# Lecture 11
+
+## Recall
+
+$K$ is **compact** if $\forall$ open cover $\{G_{\alpha}\}_{\alpha\in A}$ of $K$, $\exists$ a finite subcover $\{G_{\alpha_i}\}_{i=1}^n$ (We can only start proof from the cover of our desired set)
+
+Let $(X,d)$ be a metric space. Consider the following statement: If $K$ is compact and $p\in X$, then $K\cup \{p\}$ is compact.
+
+1. To give a proof of the statement, we start with "Suppose $K$ is compact and $p\in X$." What MUST be the next step in the proof be?  
+    **Let $\{G_{\alpha}\}_{\alpha\in A}$ be any open cover of $K\cup \{p\}$**. (Since we want to show $K\cup \{p\}$ is compact)
+2. Complete the proof of the statement.  
+    Since $K\subset K\cup \{p\} \subset \{G_{\alpha}\}_{\alpha\in A}$ and $K$ is compact, then $\exists$ a finite subcover $\{G_{\alpha_i}\}_{i=1}^n$ relative to $X$.  
+    And $p\in K\cup \{p\}$, $\exists \beta \in A$ such that $p\in\beta$, such that $C=\{G_{\alpha_i}\}_{i=1}^n+\beta$. And $C$ is a be So $K\cup \{p\}$ is compact.  
+    _Any sets has an open cover_
+3. Suppose $F\subset K\subset X,K$ is compact, and $F$ is closed (in $X$). Prove that $F$ is compact. [Hint: the proof structure is similar to 2.]  
+    By **Theorem 2.33**
+
+## New Materials
+
+### Compact sets
+
+#### Theorem 2.35
+
+If $F\subset K\subset X$, and $F$ is closed (relative to $X$), and $K$ is compact, then $F$ is compact.
+
+Proof:
+
+Let $\{G_{\alpha}\}_{\alpha\in A}$ be an open cover of $F$. (Since we want to show $F$ is compact)
+
+And $(\bigcup_{\alpha\in A} G_\alpha)\cup F^c=X\supset K$ is an open cover of $K$. Since $K$ is compact, then this open cover has a finite subcover $\Phi\subset (\bigcup_{\alpha\in A} G_\alpha)\cup F^c$ of $K$.
+
+Since $F\subset K$, So $\Phi$ is a cover of $F$ then $\Phi\backslash \{F^c\}$ is a finite subcover of $\{G_\alpha\}_{\alpha \in A}$ of $F$.
+
+EOP
+
+##### Corollary 2.35
+
+If $F$ is closed and $K$ is compact, then $F\cap K$ is closed.
+
+#### Corollary 2.36 From Theorem 2.36
+
+If $K_1\supset K_2\supset K_3$ is sequence of nonempty compact sets, Then $\bigcap ^{\infty}_1 K_n$ is not empty.
+
+Proof:
+
+We proceed by contradiction. Suppose $\bigcap ^{\infty}_1 K_n=\phi$
+
+Then $\bigcap^{\infty}_{n=1} K^c_n=(\bigcup^{\infty}_{n=1} K_n)^c=X\supset K_i$, Since $K_n$ is compact, $K_n$ is closed. $K_n^c$ is open.
+
+So $\{K_n^c\}_{n\in \mathbb{N}}$ is an open cover of $K_n$. By compactness of $K_1$, $\exists$ a finite subcover $\{K^c_{n_1},...,K^c_{n_m}\}$ of $K_1$. So
+
+$$
+K_1\subset \bigcup^m_{i=1} K_{n_i}^c =\left( \bigcap^{m}_{i=i} K_{n_i} \right)^c
+$$
+
+Let $l=\max\{n_1,...,n_m\}$, then $K_l=\bigcap^{m}_{i=i} K_{n_i}$. So $K_1\subset K_l^c$, so $K_1\cap K_l=\phi$
+
+which contradicts with $K_l\subset K_1$
+
+EOP
+
+#### Theorem 2.37
+
+If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
+
+Proof:
+
+We'll prove the following equivalent statement (contrapositive).
+
+If $K$ is compact, $E\subset K$, and $E'\cap K=\phi$, then $E$ is finite
+
+Suppose $K$ is compact, $E\subset K$ , $E'\cap K=\phi$.
+
+For each $q\in K, q\notin E'$, so $\exists$ neighborhood $\forall q\in B_{r_q}(q)$ such that $V_q\cap E\backslash \{q\}=\phi$ (By $E'\cap K=\phi$).
+
+Then $\{V_q\}_{q\in K}$ is an open cover of $K$ , so it has a finite subcover $\{V_{q_i}\}^n_{n=1}$. Then $E\subset K\subset \bigcup_{i=1}^n$, and $\forall i,V_{q_i}\cap E\subset \{q_i\}$, then $E\subset\{q_1,...,q_n\}$
+
+EOP
+
+#### Theorem 2.38
+
+If $I_1,I_2,...$ is a sequence of closed and bounded intervals and $I_1\subset I_2\subset I_3\subset ...$, then $\bigcap{}^{\infty}_{n=1} I_n\neq \phi$
+
+Proof:
+
+Let $E=\{a_n:n\in \mathbb{N}\}$, $E$ is non empty (Since $a_i\in E$)
+
+We claim $\forall m\in \mathbb{N}$, $b_m$ is an upper bound of $E$, i.e. $\forall m\in \mathbb{N},\forall n\in \mathbb{N},a_n\leq b_m$.
+
+To see this:
+
+$$
+a_{n}\leq a_{n+m}\leq b_{n+m}\leq b_{n}
+$$
+
+Let $x=sup(E)$, we claim $x\in \bigcap_{n=1}^{\infty} I_n$, i.e $\forall n\in \mathbb{N}, a_n\leq x\leq b_n$
+
+Fix $n\in \mathbb{N}$. 
+
+Since $x$ is an upper bound of $E$ and $a_n\in E$, then $a_n\leq x$.
+
+Since $x$ is the least upper bound of $E$, and $b_n$ is an upper bound of $E$, then $x\leq b_n$. $x\in E,E\neq \phi$
+
+EOP

pages/Math4111/Math4111_L12.md

@@ -0,0 +1,134 @@
+# Lecture 12
+
+## Review Questions
+
+For a metric space $(X,d)$, we say a subset $S\subset X$  si bounded if there exists $p\in X$ and $r>0$ such that $S\subset B_r(p)$.
+
+Consider the following statement: If a set $S\subset X$ is compact, the its is bounded.
+
+1. Will the proof of this statement involve an arbitrary open cover (one that you, the prover, do not get to choose) or a specific open cover (one that you can choose)?
+    We should choose a specific cover so that we can construct cover that have a set that is a superset of $S$.
+2. Give a proof of the statement. [Suggestion: If you prefer, you could try proving the contrapositive. Both a direct proof and a proof by contrapositive are roughly of the same difficulty.]
+
+### Continue on compact sets
+
+#### Lemma
+
+If $S\in X$ is compact, then $S$ is bounded.
+
+Proof:
+
+Fix $p\in X$, then $\{B_n(p)\}_{n\in \mathbb{N}}$ (specific open cover) is an open cover of $S$ (Since $\bigcup_{n\in \mathbb{N}}=X$). Since $S$ is compact, then $\exists$ a finite subcover ${n\in \mathbb{N}}_{i=1}^k=S$, let $r=max(n_1,...n_k)$, Then $S\subset B_r(p)$
+
+EOP
+
+#### Definition k-cell
+
+A 2-cell is a set of the form $[a_1,b_1]\times[a_2,b_2]$
+
+#### Theorem 2.39 (K-dimension of Theorem)
+
+Theorem 2.38 replace with "closed and bounded intervals" to "k-cells".
+
+Idea of Proof:
+
+Apply the Theorem to each dimension separately.
+
+#### Theorem 2.40
+
+Every k-cell is compact.
+
+We'll prove the case $k=1$ and $I=[0,1]$ (This is to simplify notation. This same ideas are used in the general case)
+
+Proof:
+
+That $[0,1]$ is compact.
+
+(Key idea, divide and conquer)
+
+Suppose for contradiction that $\exists$ open cover $\{G_a\}_{\alpha\in A}$ of $[0,1]$ with no finite subcovers of $[0,1]$
+
+**Step1.** Divide $[0,1]$ in half. $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$ and at least one of the subintervals cannot be covered by a finite subcollection of $\{G_\alpha\}_{\alpha\in A}$
+
+(If both of them could be, combine the two finite subcollections to get a finite subcover of $[0,1]$)
+
+Let $I_1$ be a subinterval without a finite subcover.
+
+**Step2.** Divide $I_1$ in half. Let $I_2$ be one of these two subintervals of $I_1$ without a finite subcover.
+
+**Step3.** etc.
+
+We obtain a seg of intervals $I_1\subset I_2\subset \dots$ such that
+
+(a) $[0,1]\supset I_1\supset I_2\supset \dots$  
+(b) $\forall n\in \mathbb{N}$, $I_n$ is not covered by a finite subcollection of $\{G_\alpha\}_{\alpha\in A}$  
+(c) The length of $I_n$ is $\frac{1}{2^n}$
+
+By (a) and **Theorem 2.38**, $\exists x^*\in \bigcap^{\infty}_{n=1} I_n$.
+
+Since $x^*\in [0,1]$, $\exists \alpha_0$ such that $x^*\in G_{\alpha_0}$
+
+Since $G_{\alpha_0}$ is open, $\exist  r>0$ such that $B_r(x^*)\subset G_{\alpha_0}$
+
+Let $n\in \mathbb{N}$ be such that $\frac{1}{2^n}<r$. Then by $(c)$, $I(n)\subset B_r(x^*)\subset G_{\alpha_0}$
+
+Then $\{G_{\alpha_0}\}$ is a cover of $I_n$ which contradicts with (b)
+
+EOP
+
+#### Theorem 2.41
+
+If a set $E$ in $\mathbb{R}^k$ (only in $\mathbb{R}^k$, not for general topological space or metric spaces.) has one of the following three properties, then it has the other two:
+
+(a) $E$ is closed and bounded.  
+(b) $E$ is compact.  
+(c) Every infinite subset of $E$ has a limit point in $E$.
+
+We'll prove $(a)\implies (b)\implies (c)\implies (a)$
+
+Proof:
+
+$(a)\implies (b)$
+
+Suppose $(a)$ holds, then $\exists$ k-cell $I$ such that $E\in I$.By **Theorem 2.40**, $I$ is compact. By **Theorem 2.35**, $E$ is compact.
+
+$(b)\implies (c)$
+
+Follows from **Theorem 2.37**
+
+$(c)\implies (a)$
+
+We will proceed by contrapositive, which says that $\neg (a)\implies \neg (c)$
+
+$\neg (a)$: $E$ is not closed or not bounded.
+
+$\neg (c)$: $\exists$ infinite subcover $S\subset E$ such that $S'\cup E=\phi$
+
+Suppose $(a)$ does not hold. There are two cases to consider
+
+Case 1: $E$ is not bounded. Then $\forall v\in \mathbb{N},\exists x_n\in E$ such that $|x_n|\geq n$
+
+Let $S=\{x_n,...,n\in\mathbb{N}\}$, then $S'=\phi$ (by **Theorem 2.20**)
+
+Case 2: $E$ is not closed. Then $z\in E'\backslash E$.
+
+Since $z\in E'$, $\forall n\in \mathbb{N},\exists x_n\in E$ such that $|x_n-z|<\frac{1}{n}$
+
+Let $S=\{x_n:n\in \mathbb{N}\}$, we claim $S'\subset \{z\}$. (In fact $s'\in\{z\}$, but we don't need this in the proof)
+
+We'll show if $y\neq z$, then $y\notin  S'$
+
+$\forall w\in B_r(y)$
+
+$$
+\begin{aligned}
+    d(w,z)&\geq d(y,z)-d(y,w)\\
+    &>d(y,z)-r\\
+    &=d(y,z)-\frac{1}{2}d(y,z)\\
+    &=\frac{1}{2}d(y,z)
+\end{aligned}
+$$
+
+So $B_r(y)\cap S$ is finite. By **Theorem 2.20**, $y\notin  S$, this proves the claim so $S'\cap E=\phi$
+
+EOP

pages/Math4111/Math4111_L13.md

@@ -0,0 +1,143 @@
+# Lecture 13
+
+## Review
+
+Consider the metric space $X=\mathbb{R}$ (with the usual metric $d(x,y)=|x-y|$). Let $E=(0,1)$.
+
+1. Find several examples of sets $Y\subset \mathbb{R}$ such that $E\subset Y$ and $E$ is closed in $Y$.  
+    Example:
+    1. $Y=E$, $E$ is closed in $Y$.  
+        We can prove this using normal ways, or
+        **Theorem 2.23** $E$ is closed in $Y\iff E^c$ is open in $Y$  
+        $Y\iff E^c=\phi$ and it's open.
+    2. $Y=\mathbb{R}\backslash\{0,1\}$  
+        $Y\backslash E=(-\infty,0)\cup (1,\infty)$
+        **Theorem 2.30** $E\subset Y\subset X$, $E$ is open in $Y\iff$ $\exists G$ open in $X$ such that $G\cap Y=E$  
+        $G\cap Y=Y\backslash E$
+        And we know $Y\backslash E$ is open in $Y$. By **Theorem 2.23** $E$ is closed in $Y$.
+
+2. If $Y$ is as in part 1, we can conclude that $E$ is closed and bounded in $y$. Part of **Theorem 2.41** says: "If a set is closed and bounded, then it is compact." Why doesn't that theorem apply here.  
+    The set is not closed in $\mathbb{R}^k$.
+
+## New stuffs
+
+### Connected sets
+
+#### Definition 2.45
+
+$A,B\subset X$, we say $A$ and $B$ are separated in $X$ if $A\cup \overline{B}=\phi$ and $\overline{A}\cup B=\phi$
+
+- $E\subset X$ **disconnected** in $X$ if $\exists$ nonempty separated $A,B\subset X$ such that $E=A\cup B$
+- $E\subset X$ is **connected** in $X$ if it is not disconnected.
+
+Example 2.46
+
+$(0,1),(1,2)$ are separated [so $(0,1)\cup (1,2)$ is disconnected]
+
+$[0,1],(1,2)$ are not separated [so $[0,1]\cup (1,2)=\{1\}$] So this doesn't tell us where $[0,1]\cup (1,2)$ is connected or not.
+
+#### Theorem 2.47
+
+Suppose $E\subset \mathbb{R}$
+
+$E$ is connected $\iff \forall (x,y,z)$ with $x,y\in E,x<z<y$ such that $z\in E$.
+
+By negating, this is equivalent to
+
+$E$ is disconnected $\iff \exists (x,y,z)$ with $x,y\in E,x<z<y$ such that $z\notin E$.
+
+Proof:
+
+$\impliedby$
+
+Suppose $\exists (x,y,z)$ with $x,y\in E,x<z<y$ such that $z\notin E$.
+
+Let $A=(-\infty,z)\cap E,B=(z,\infty)\cap E$
+
+**Lemma:** $E\subset F\implies \overline{E}\subset \overline{F}$
+
+Since $A\subset (-\infty,z), \overline{A}\subset (-\infty,z] $
+
+Since $\overline{A}\cap B=\phi$, similarly, $A\cap \overline{B}=\phi$. So $A,B$ are separated. Also they are non empty ($x\in A,y\in B$) and $E=A\cap B$. So $E$ is disconnected.
+
+$\implies$
+
+Suppose $\exists$ nonempty separated $A,B\subset X$ such that $E=A\cup B$.
+
+Our goal is to find $x,y\in E,x<z<y$ such that $z\notin E$.
+
+$A\neq \phi,B\neq \phi\implies \exists x\in A,y\in B$
+
+Without loss of generality, assume $x<y$.
+
+Let $w=sup(A\cup [x,y])$, $w\in \overline{A\cup [x,y]}$ (by **Theorem 2.28**) This implies $w\subset \overline{A}$ and $w\in [x,y]$
+
+Since $A,B$ are separated, $w\notin B$ ($\overline{A}\cup B=\phi$).
+
+Since $y\in B$, $w\in [x,y)$
+
+Consider 2 cases,
+
+**Case 1.** $w\notin A$.
+
+let $z=w$, and $x,y,z$ satisfy the desired properties
+
+**Case 2.** $w\in A$
+
+Since $A,B$ are separated, $w\notin \overline{B}$ ($A\cup \overline{B}=\phi$).
+
+Thus, $\exists r>0$ such that $(w-r,w+r)\cap B=\phi$.
+
+Let $z=w+\frac{r}{2}$, then $x,y,z$ satisfy the desired properties.
+
+EOP
+
+## Chapter 3: Numerical Sequences and Series
+
+### Numerical Sequences
+
+#### Notations
+
+Rudin use $\{p_n\}$ to denote a sequence $p_1,p_2$.
+
+To avoid confusion with sets, we use $(p_n)_{n=1}^\infty$ or $(p_n)$
+
+#### Definition 3.1
+
+Let $(X,d)$ be a metric space. Let $(p_n)$ be a sequence in $X$.
+
+Let $p\in X$. We say $(p_x)$ **converges** to $p$ if $\forall \varepsilon>0,\exists N\in\mathbb{N}$ such that $\forall n\geq N$, $d(p_n,p)<\varepsilon$. ($p_n\in B_\varepsilon (p)$)
+
+Notation $\lim_{n\to \infty} p_n=p$, $p_n\to p$
+
+We say $(p_n)$ converges if $\exists p\in X$ such that $p_n\to p$.
+
+i.e. $\exists p\in X$ such that $\forall\varepsilon>0,\exists N\in\mathbb{N}$ such that $\forall n\geq N,d(p_n,p)<\varepsilon$
+
+We say $(p_n)$ **diverges** if $(p_n)$ doesn't converge.
+
+i.e. $\forall p\in X$, $p_n\cancel{\to} p$
+
+i.e. $\forall p\in X$ such that $\exists \varepsilon>0,\forall N\in\mathbb{N}$ such that $\exists n\geq N,d(p_n,p)\geq\varepsilon$
+
+#### Definition 3.2
+
+We say a sequence $(p_n)$ is bounded if $\exists x\in X$, $\forall r>0$ such that $\forall n\in \mathbb{N},p_n\in B_r(x)$
+
+Example:
+
+$X=\mathbb{C}$, $s_n=\frac{1}{n}$
+
+Then $s_n\to 0$ i.e. $\forall \varepsilon>0 \exists N\in \mathbb{N}$ such that $\forall n\geq N$, $|s_n-0|<\varepsilon$.
+
+Proof:
+
+Let $\varepsilon >0$ (arbitrary)
+
+Let $N\in \mathbb{N}$ be greater than $\frac{1}{\varepsilon}$ (by Archimedean property) e.g. $N=\frac{1}{\varepsilon}+1$ (we choose $N$)
+
+Let $n\geq N$ (arbitrary)
+
+Then $|s_n-q|=\frac{1}{n}\leq \frac{1}{N}\leq \varepsilon$
+
+EOP

pages/Math4111/Math4111_L14.md

@@ -0,0 +1,159 @@
+# Lecture 14
+
+## Review
+
+Consider the following statement: If sequence $(p_n)$ converges, then its bounded.
+
+1. Will the proof involve an arbitrary $\epsilon>0$ (one that you, the prover, do nto get to choose) or a specific $\epsilon>0$ (on that you can choose)  
+    We can choose, for example $\epsilon=1$.
+2. Give a proof of the statement.
+
+## Continue on sequence
+
+### Convergence
+
+#### Theorem 3.2(c)
+
+$(p_n)$ converges $\implies(p_n)$ is bounded.
+
+Proof:
+
+Suppose $(p_n)$ converges, then $\exists p\in X$ such that $p_n\to p$. Let $\epsilon=1$, then $\exists N\in \mathbb{N}$ such that $\forall n\geq N,d(p_n,p)<1$. Let $r=1+max\{1,d(p_1,p),d(p_2,p),\dots,d(p_{N-1},p)\}$.
+
+Then $\forall n\in \mathbb{N}, d(p_n,p)\leq r$.
+
+#### Theorem 3.2
+
+Let $(p_n)$ be a sequence in $(X,d)$
+
+(a) $p_n\to p\iff \forall r>0,\{n\in \mathbb{N},p_n\notin B_r(p)\}$ is finite  
+(b) $p_n\to p; p_n\to p'\implies p=p'$ (converging point is unique)  
+(c) $(p_n)$ converges $\implies(p_n)$ is bounded.  
+(d) If $E\subset X$ and $p\in \overline{E}$, then $\exist (p_n)\in E$ such that $p_n\to p$.
+
+Proof:
+
+(a) We need to show:
+
+$\forall \epsilon>0 \in N$, $\forall n\geq N,d(p_n,p)<\epsilon$ if and only if $\forall r>0, \{n\in \mathbb{N}:p\notin B_r(p)\}$ is finite.
+
+$\implies$
+
+Suppose $\forall \epsilon>0 \in N$, $\forall n\geq N,d(p_n,p)<\epsilon$.
+
+**We start with arbitrary $r>0$.** and choose $\epsilon=n$
+
+$\exists N$ such that $\forall n\geq \mathbb{N},d(p_n,p)<r$.
+
+Then $\{n\in \mathbb{N}:p\notin B_r(p)\}<\{1,2,\dots,N-1\}$ is finite.
+
+$\impliedby$
+
+Suppose $\forall r>0, \{n\in \mathbb{N}:p\notin B_r(p)\}$ is finite. Choosing $r=\epsilon$. We choose $r=\epsilon$. $\{n\in \mathbb{N}:p\notin B_\epsilon(p)\}<\{1,2,\dots,N-1\}$.
+
+Let $N=1+max\{n\in \mathbb{N},p_n\notin B_\epsilon(p)\}$
+
+Then $\forall n\geq \mathbb{N},p_n\leq  B_\epsilon(p)$
+
+(b) We'll prove $\forall \epsilon>0,d(p,p')<2\epsilon$ to prove it, let $\epsilon >0$. Then
+
+$p_n\to p\implies \exists N$ such that $\forall n\geq \mathbb{N},d(p_n,p)<\epsilon$  
+$p_n\to p'\implies \exists N'$ such that $\forall n\geq \mathbb{N},d(p_n,p')<\epsilon$
+
+Let $n_0=max\{N,N'\}$, then
+
+$$
+d(p,p')\leq d(p_n,p_{n_0})+d(p_{n_0},p')<2\epsilon
+$$
+
+And $\forall \epsilon>0,d(p,p')<2\epsilon\implies d(p,p')=0$. So $p=p'$
+
+> Remark: We can also prove this with contradiction. Idea $p\neq p'\implies d(p,p')>0$, let $\epsilon=\frac{1}{2}d(p,q')\dots$
+
+(d) Suppose $p\in \overline{E}$. Then $\forall  n\in \mathbb{N}, B_{\frac{1}{n}}(p)\cap E\neq \phi$. So $\forall n\in \mathbb{N}$, $\exists p_n\in B_{\frac{1}{n}}(p)\cap E$. We'll show $p_n\to p$.
+
+Let $\epsilon>0$. Choose $N\in \mathbb{N}$ such that $N>\frac{1}{\epsilon}$. Then if $n\geq N$, $d(p_n,p)<\frac{1}{n}\leq \frac{1}{N}\leq \epsilon$
+
+EOP
+
+#### Theorem 3.3
+
+Let $(s_n), (t_n)$ be sequence in $\mathbb{C}$. Suppose $s_n\to s,t_n\to t$
+
+(a) $s_n+t_n\to s+t$  
+(b) $cs_n\to cs,c+s_n\to c+s$  
+(c) $s_nt_n\to st$  
+(d) If $\forall n\in \mathbb{N},s_n\neq 0,s\neq 0$, then $\frac{1}{s_n}\to \frac{1}{s}$
+
+Proof:
+
+(a) We want to prove $\forall \epsilon>0, \exists N$ such that $\forall n\geq N, |(s_n+t_n)-(s+t)|<\epsilon$
+
+Let $\epsilon >0$
+
+$s_n\to s\implies \exist N_s$ such that $\forall n\geq N_s,|s_n-s|<\frac{\epsilon}{2}$  
+$t_n\to t\implies \exist N_t$ such that $\forall n\geq N_t,|t_n-t|<\frac{\epsilon}{2}$
+
+Let $N=\max\{N_t,N_s\}$, then if $n\geq N$,
+
+$$
+\begin{aligned}
+|(s_n+t_n)-(s+t)|&=|(s_n+s)-(t_n-t)|\\
+&\leq |s_n-s|+|t_n-t|\\
+&< \frac{\epsilon}{2}+\frac{\epsilon}{2}\\
+&<\epsilon
+\end{aligned}
+$$
+
+(b) exercise  
+
+(c) First we'll prove a special case.
+
+$$
+s_n\to 0 \textup{ and }t_n\to 0\implies s_nt_n\to 0
+$$
+
+Suppose $s_n\to 0$ and $t_n\to 0$.
+
+Let $\epsilon >0$
+
+$s_n\to 0\implies \exist N_s$ such that $\forall n\geq N_s,|s_n-s|<\sqrt{\epsilon}$  
+$t_n\to 0\implies \exist N_t$ such that $\forall n\geq N_t,|t_n-t|<\sqrt{\epsilon}$
+
+Let $N=\max\{N_t,N_s\}$, then if $n\geq N$,
+
+$$
+|s_n t_n|< \sqrt{\epsilon}^2=\epsilon
+$$
+
+Now we prove the general case.
+
+$$
+s_n\to s \textup{ and }t_n\to t\implies s_nt_n\to st
+$$
+
+Since
+
+$$
+s_n t_n=(s_n-s)(t_n-t)+s(t_n-t)+t(s_n-s)
+$$
+
+So
+
+$$
+\lim_{n\to \infty}(s_nt_n-st)=\lim_{n\to \infty}(s_n-s)(t_n-t)+\lim_{n\to \infty}s(t_n-t)+\lim_{n\to \infty}t(s_n-s)
+$$
+
+$\lim_{n\to \infty}(s_n-s)(t_n-t)=0$ by special case
+
+$\lim_{n\to \infty}s(t_n-t)=0$ by (b)
+
+$\lim_{n\to \infty}t(s_n-s)=0$ by (b)
+
+Thought process for (d)
+
+$$
+\left|\frac{1}{s_n}-\frac{1}{s}\right|=\frac{|s_n-s|}{|s_n||s|}< \epsilon
+$$
+
+If $n$ is large enough, then... 

pages/Math4111/Math4111_L15.md

@@ -0,0 +1,162 @@
+# Lecture 15
+
+## Review
+
+Let $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ be sequence in $\mathbb{R}$. Let $x_n=(a_n,b_n)\in \mathbb{R}^2$, so $(x_n)_{n=1}^\infty$ be a sequence in $\mathbb{R}^2$. Consider the following statement:
+
+$$
+a_n\to a\textup{ and }\quad b_n\to b\iff x_n\to (a,b)
+$$
+
+1.Prove the $\impliedby$ direction. That means you should prove the two things:  
+    (a) If $x_n\to (a,b)$, then $a_n\to a$. (The proof of this begins: Suppose $x_n\to (a,b)$. Let $\epsilon>0$ be arbitrary. Then $\exists N$ such that $\forall n\geq N$)  
+    We begins (with the goal $\forall \epsilon>0,\exists N$ such that $\forall n\geq N,|a_n-a|<\epsilon$).  
+    Proof:  
+    Let $\epsilon>0$ be arbitrary, then $\exists N$ such that $\forall n\geq N,|a_n-a|<\epsilon$.
+    Then if $n\geq N$, $|a_n-a|\leq \sqrt{|a_n-a|^2}\leq\sqrt{|a_n-a|^2+|b_n-b|^2}=|x_n-(a,b)|<\epsilon$.  
+    EOP  
+    (b) If $x_n\to (a,b)$, then $b_n\to b$.
+    This follows from the same argument from (a)
+2. Prove the $\implies$ direction.
+    Goal: $\forall \epsilon>0,\exists N$ such that $\forall n\geq N,|a_n-a|<\epsilon$.
+    Proof:
+    Let $\epsilon>0$ be arbitrary.  
+    Since $a_n\to a$, $\exists N_1$ such that $\forall n\geq N_1,|a_n-a|<\epsilon$.  
+    Since $b_n\to b$, $\exists N_2$ such that $\forall n\geq N_2,|b_n-b|<\epsilon$.  
+    Let $N=\max\{N_1,N_2\}$. Then if $n\geq N$, $|a_n-a|<\epsilon$ and $|b_n-b|<\sqrt{2}\epsilon$.  
+    **Same as last time, we can choose any smaller epsilon.**  
+    Since $a_n\to a$, $\exists N_1$ such that $\forall n\geq N_1,|a_n-a|<\frac{\epsilon}{\sqrt{2}}$.  
+    Since $b_n\to b$, $\exists N_2$ such that $\forall n\geq N_2,|b_n-b|<\frac{\epsilon}{\sqrt{2}}$.  
+    Let $N=\max\{N_1,N_2\}$. Then if $n\geq N$, $|a_n-a|<\epsilon$ and $|b_n-b|<\sqrt{\frac{\epsilon^2}{2}+\frac{\epsilon^2}{2}}=\epsilon$.  
+    EOP  
+
+## New Materials
+
+Continue from **Theorem 3.3**
+
+Suppose $(s_n),(t_n)$ are sequences in $\mathbb{C}$ and $s_n\to s,t_n\to t$. Then
+
+(a) $s_n+t_n\to s+t$  
+(b) $cs_n\to cs$, $c+s_n\to c+s$  
+(c) $s_nt_n\to st$  
+(d) If $\forall n\in \mathbb{N},s_n\neq 0, s\neq 0$, then $\frac{1}{s_n}\to \frac{1}{s}$
+
+Thought process for (d):
+
+$$
+\left|\frac{1}{s_n}-\frac{1}{s}\right|=\left|\frac{s-s_n}{s_ns}\right|=\frac{|s-s_n|}{|s||s_n|}
+$$
+
+We choose large enough $N$ such that $\forall n\geq N,|s_n-s|<\frac{|s|}{2}$. Then by triangle inequality, $|s_n|>\frac{|s|}{2}$.
+
+$$
+\begin{aligned}
+|s|&=|s-s_n+s_n|\\
+|s|&\leq |s-s_n|+|s_n|\\
+|s|&<\frac{|s|}{2}+|s_n|\\
+\frac{|s|}{2}&< |s_n|
+\end{aligned}
+$$
+
+So $\frac{|s_n-s|}{|s||s_n|}<\frac{2|s_n-s|}{|s|^2}$.
+
+We choose $n$ large enough such that
+
+$$
+\frac{2|s_n-s|}{|s|^2}<\epsilon
+$$
+
+Then $|s_n-s|<\frac{\epsilon|s|^2}{2}$.
+
+Proof:
+
+Let $\epsilon>0$, since $s_n\to s$
+
+$\exists N$ such that $\forall n\geq N,|s_n-s|<\frac{|s|}{2}$.
+
+$\exists N$ such that $\forall n\geq N,|s_n-s|<\frac{\epsilon|s|^2}{2}$.
+
+Let $N=\max\{N_1,N_2\}$. Then if $n\geq N$,
+
+$$
+\left|\frac{1}{s_n}-\frac{1}{s}\right|=\frac{|s-s_n|}{|s||s_n|}<\frac{\frac{\epsilon|s|^2}{2}}{|s|^2}=\epsilon
+$$
+
+EOP
+
+### Subsequences
+
+#### Definition 3.5
+
+Given a sequence $(p_n)_{n=1}^\infty$, a sequence of $(n_i)_{i=1}^\infty$ is strictly increasing sequence in $\mathbb{N}$. i.e. $n_1<n_2<n_3<\cdots$.
+
+The sequence $(p_{n_i})_{i=1}^\infty$ is called a **subsequence** of $(p_n)_{n=1}^\infty$.
+
+Example:
+
+$p_n=\frac{1}{n},n_i=i^2$, then the subsequence is $(p_{n_i})_{i=1}^\infty=\left(\frac{1}{i^2}\right)_{i=1}^\infty$. i.e. $(\frac{1}{4},\frac{1}{9},\frac{1}{16},\cdots)$
+
+$(p_n)_{n=1}^\infty$ converges to $p$ if and only if every subsequence of $(p_n)_{n=1}^\infty$ converges to $p$.
+
+Proof:
+
+$\impliedby$:
+
+$(p_{n_i})_{i=1}^\infty$ is a subsequence of $(p_n)_{n=1}^\infty$.
+
+$\implies$:
+
+Thought process: show what if the sequence does not converge to $p$, then there exists a subsequence that does not converge to $p$.
+
+EOP
+
+#### Theorem 3.6
+
+(a) If $(p_n)$ is a sequence in a compact metric space $X$, then $(p_n)$ has a convergent subsequence converges to a point of $X$.
+
+(b) If $(p_n)$ is a bounded sequence in $\mathbb{R}^k$, then $(p_n)$ has a convergent subsequence in $\mathbb{R}^k$.
+
+Proof:
+
+(a) Let $E=\{p_n:n\in \mathbb{N}\}$. Note that $E$ is a set, not a sequence.
+
+Case 1: $E$ is finite.
+
+Then some term appears infinitely many times. i.e $\exists p\in E$ and subsequence $(p_{n_i})$ such that for all $i$, $p_{n_i}=p$.
+
+Then $(p_{n_i})$ converges to $p$.
+
+Case 2: $E$ is infinite.
+
+By **Theorem 2.37**, if $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
+
+$p\in E'\implies \forall r>0, B_r(p)\cap E\backslash \{p\}\neq \phi$
+
+- Choose $n_i$ such that $p_{n_i}\in B_i(p)$
+- If $n_1,\dots, n_{i-1}$ have bee chosen, choose $n_i$ such that $n_i>n_{i-1}$ and $p_{n_i}\in B_{\frac{1}{i}}(p)$. Then $p_{n_i}\to p$
+
+(b) Since $(p_n)$ is bounded , $\exists M$ such that $\forall n\in N$, $p_n\in \overline{B_M(0)}=\{y\in\mathbb{R}^k:|y|\leq M\}$
+
+$\overline{B_M(0)}$ is a closed and bounded set in $\mathbb{R}^k$.
+
+Then by Theorem 2.41, $\overline{B_M(0)}$ is compact.
+
+By part (a), $(p_n)$ has a subsequence $(p_{n_i})$ has a subsequence that converges to $B_M(0)$.
+
+#### Theorem 3.37
+
+Let $X$ be a metric space, $(p_n)$ is a sequence in $X$.
+
+Let $E^*=\{p\in X:\exists\textup{ subsequence }(p_{n_i})\textup{ such that }p_{n_i}\to p\}$.
+
+Then $E^*$ is closed in $X$.
+
+Example:
+
+$X=\mathbb{R}$
+
+1. $p_n=\frac{1}{n}$, $E^*=\{0\}$. (Specifically, if $p_n\to p$, then $E^*\to \{p\}$)
+2. $p_n=\begin{cases}1,n\textup{ is odd}\\ 0,n\textup{ is even}\end{cases}$, $E^*=\{0,1\}$
+3. $p_n=n$, $E^*=\phi$
+4. $p_n=\sin nx$, $E^*=\{0,1\}$
+5. $p_n=\sin n$, $E^*=[0,1]$

pages/Math4111/Math4111_L16.md

@@ -0,0 +1,152 @@
+# Lecture 16
+
+## Review
+
+Let $(s_n)$ be a sequence in $\mathbb{R}$ satisfying the following properties:
+
+1. It is bounded ($\exists M>0$ such that $\forall n\in \mathbb{N}, |s_n|\leq M$)
+2. It is monotonic increasing ($\forall n\in \mathbb{N}, s_n\leq s_{n+1}$)
+
+Let $E=\{s_n:n\in \mathbb{N}\}$ and $t=sup E$. Prove that $s_n\to t$. [Hint: The proof begins with "Let $\epsilon>0$ be arbitrary." What do we know about $t-\epsilon$?]
+
+Proof:
+
+Let $\epsilon>0$ be arbitrary. Then since $t-\epsilon$ is not an upper bound of $E$, $\exists N$ such that $t-\epsilon<s_N$.
+
+Let $n\geq N$. Since $(s_n)$ is monotonic increasing, $s_n\geq s_N>t-\epsilon$. Since $t$ is an upper bound of $E$, $s_n<t$. Therefore, $|s_n-t|<\epsilon$.
+
+So $(s_n)$ converges to $t$.
+
+EOP
+
+## New materials
+
+### Subsequences
+
+#### Theorem 3.7
+
+Let $X$ be a metric space. If $(p_n)$ is a sequence in $X$. $E^*=\{p\in X:\exists \textup{ subsequence } (p_{n_i}) \textup{ such that } p_{n_i}\to p\}$. Then $E^*$ is closed.
+
+Proof:
+
+Let $q\in (E^*)'$. We will show that $q\in E^*$.
+
+Step 1: Since $q\in (E^*)'$, $\exists x_1\in B_1(q)\cap E^*$.
+
+Since $x_1\in E^*$, $\exists n_1\in \mathbb{N}$ such that $p_{n_1}\in B_1(x_1)$.
+
+By triangle inequality, $d(p_{n_1},q)\leq d(p_{n_1},x_1)+d(x_1,q)<1+1=2$.
+
+Step 2: Since $q\in (E^*)'$, $\exists x_2\in B_{1/2}(q)\cap E^*$ and $n_2>n_1$ (by definition of $E^*$. If $x_2\in E^*$, then there are infinitely many $p\in \mathbb{N}$ such that $p_n\in B_{1/2}(x_2)$).
+
+Since $x_2\in E^*$, $\exists n_2\in \mathbb{N}$ such that $p_{n_2}\in B_{1/2}(x_2)$.
+
+By triangle inequality, $d(p_{n_2},q)\leq d(p_{n_2},x_2)+d(x_2,q)<\frac{1}{2}+\frac{1}{2}=1$.
+
+Step 3: By induction, we can get a sequence $n_1,n_2,\cdots$ such that $\forall i\in \mathbb{N}, d(p_{n_i},q)<\frac{2}{i}$.
+
+Then $(p_{n_i})$ is a subsequence of $(p_n)$ and $p_{n_i}\to q$.
+
+EOP
+
+### Cauchy Sequences
+
+#### Definition 3.8
+
+A sequence $(p_n)$ in a metric space $X$ is called a Cauchy sequence if for every $\epsilon>0$, there exists $N\in \mathbb{N}$ such that $\forall m,n\geq N$, $d(p_m,p_n)<\epsilon$.
+
+*The terms are getting closer to each other.*
+
+Example:
+
+$X=\mathbb{Q}$ with the usual metric. Let $(p_n)$ be a sequence
+
+$$
+3,3.1,3.14,3.141,3.1415,\cdots
+$$
+
+If $m\leq n$, $|p_m-p_n|<\frac{1}{10^{m}}$.
+
+Then $(p_n)$ is a Cauchy sequence. Let $\epsilon>0$ be arbitrary. Choose $N$ such that $\frac{1}{10^{N}}>\epsilon$. Then if $m,n\geq N$, then $|p_m-p_n|\leq \frac{1}{10^{m}}<\epsilon$.
+
+This sequence does not converge in $\mathbb{Q}$.
+
+$X=\mathbb{R}$ with the usual metric. Let $(p_n)$ be a sequence
+
+$$
+p_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}
+$$
+
+This sequence is not bounded above. (by Theorem 3.28), so (as we will prove) it is not a Cauchy sequence.
+
+The fact that $p_{n+1}-p_n=\frac{1}{n+1}\to 0$ is not relevant to determining whether $(p_n)$ is a Cauchy sequence.
+
+#### Theorem 3.11 (a)
+
+$(p_n)$ converges $\implies$ $(p_n)$ is a Cauchy sequence.
+
+Proof:
+
+Since $(p_n)$ converges, $\exists p\in X$ such that $p_n\to p$. Let $\epsilon>0$ be arbitrary. Then $\exists N\in \mathbb{N}$ such that $\forall n\geq N$, $d(p_n,p)<\epsilon$.
+
+If $m,n\geq N$, then $d(p_m,p_n)\leq d(p_m,p)+d(p,p_n)<\epsilon+\epsilon=2\epsilon$.
+
+*You can also use $\frac{\epsilon}{2}$ instead of $\epsilon$ in the above proof, just for fun.*
+
+EOP
+
+#### Lemma 3.11 (b)
+
+If $(p_n)$ is a Cauchy sequence, then $(p_n)$ is bounded above.
+
+Proof:
+
+Since $(p_n)$ is a Cauchy sequence, $\exists N\in \mathbb{N}$ such that $\forall m,n\geq N$, $d(p_m,p_n)<1$.
+
+Let $r=max\{d(p_i,p_j);1\leq i,j\leq N\}+1$.
+
+Then $\forall n\in \mathbb{N}$, $p_n\in B_r(p_N)$.
+
+EOP
+
+> Note: This proof is nearly identical to the proof of convergent sequences implies bounded.
+
+#### Definition 3.9
+
+Let $E$ be a nonempty subset of a metric space $X$, and let $S$ be the set of all real numbers of the form $d(p,q)$ for $p,q\in E$. The diameter of $E$, denoted by $diam E$, is defined to be the supremum of $S$.
+
+Exercise:
+
+Prove that $(p_n)$ is a Cauchy sequence if and only if $\lim_{N\to \infty}diam\{(p_n):n\geq N\}=0$.
+
+#### Theorem 3.10
+
+(a) $diam E=diam(\overline{E})$  
+(b) If $K_n$ is a sequence of nonempty compact sets and $K_1\supset K_2\supset \cdots$, then $\bigcap_{n=1}^{\infty}K_n$ has exactly one point.
+
+Proof:
+
+(a) The idea is still, triangle inequality.
+
+Since $E\subset \overline{E}$, $diam E\leq diam(\overline{E})$.
+
+Now we want to show that $diam(\overline{E})\leq diam E$.
+
+Claim: $\forall \epsilon>0$, $2\epsilon+diam E$ is an upper bound of $\{d(p,q):p,q\in \overline{E}\}$.
+
+Let $p,q\in \overline{E}$.
+
+Since $p\in \overline{E}$, $\exists p'\in E\cap B_\epsilon(p)$.
+
+Since $q\in \overline{E}$, $\exists q'\in E\cap B_\epsilon(q)$.
+
+Then $d(p,q)\leq d(p,p')+d(p',q')+d(q',q)<\epsilon+diam E+\epsilon=diam E+2\epsilon$.
+
+This proves the claim.
+
+By definition of supremum, the claim implies that $\forall \epsilon>0$, $diam(\overline{E})\leq 2\epsilon+diam E$. So $diam(\overline{E})\leq diam E$.
+
+(b) By **Theorem 2.36**, $\bigcap_{n=1}^{\infty}K_n\neq \emptyset$. Suppose for contradiction that there are at least two distinct points $p,q\in \bigcap_{n=1}^{\infty}K_n$. Then for all $n\in \mathbb{N}$, $x,y\in K_n$ so $diam K_n\geq d(p,q)>0$. Then diameter of $K_n$ does not converge to 0.
+
+EOP
+

pages/Math4111/Math4111_L17.md

@@ -0,0 +1,123 @@
+# Lecture 17
+
+## Review
+
+Given a sequence $(a_n)$ in $\mathbb{R}$, let $E_n=\{a_k:k\geq n\}$. Calculate $diam E_1$, $diam E_2$, $diam E_3$... for the following sequences:
+
+1. $a_n=0$: $E_n=\{0\}$, $$diam E_1=0, diam E_2=0, diam E_3=0, \ldots$$
+2. $a_n=n$: $E_n=\{n\}$, $$diam E_1=\infty, diam E_2=\infty, diam E_3=\infty, \ldots$$
+3. $a_n=(-1)^n$: $E_n=\{-1,1\}$, $$diam E_1=2, diam E_2=2, diam E_3=2, \ldots$$
+4. $a_n=1/n$: $E_n=\{1/n,1/(n+1),\dots\}$, $$diam E_1=\frac{1}{2}, diam E_2=\frac{1}{3}, diam E_3=\frac{1}{4}, \ldots$$
+5. $a_n=\frac{(-1)^n}{n}$: $E_n=\{-1/n,1/n,\dots\}$, $$diam E_1=\frac{1}{1}+\frac{1}{2}, diam E_2=\frac{1}{2}+\frac{1}{3}, diam E_3=\frac{1}{3}+\frac{1}{4}, \ldots$$
+
+## New materials
+
+### Cauchy sequence
+
+#### Theorem 3.11
+
+(b) If $X$ is a compact metric space, then every Cauchy sequence $(p_n)$ in $X$ converges.
+
+(c) In $\mathbb{R}^k$, every Cauchy sequence $(p_n)$ converges.
+
+Proof:
+
+(b) Let $E_N=\{p_n:n\geq N\}$. Since $(p_n)$ is Cauchy, $\lim_{N\to\infty} diam E_N=0$. By **Theorem 3.10 (a)**, $\lim_{N\to\infty} diam \overline{E_N}=0$.
+
+Since $X$ is compact, and $\overline{E_N}$ is closed, by **Theorem 2.35**, $\overline{E_N}$ is compact.
+
+Since $E_1\supset E_2\supset E_3\supset\cdots$, $\overline{E_1}\supset \overline{E_2}\supset \overline{E_3}\supset\cdots$. By **Theorem 3.10(b)**, $\exists p\in X$ such that $p\in\bigcap_{N=1}^{\infty}\overline{E_N}$.
+
+We claim that $(p_n)$ converges to $p$. Let  $\epsilon>0$, there exists $N_0$ such that $\forall N\geq N_0$, $diam \overline{E_N}<\epsilon$.
+
+For any $n\geq N_0$, $p_n\in \overline{E_{N_0}}$.
+
+So $d(p_n,p)\leq diam \overline{E_{N_0}}<\epsilon$, by definition of diameter.
+
+Therefore, $(p_n)$ converges to $p$.
+
+(c) Let $(p_n)$ be a Cauchy sequence in $\mathbb{R}^k$.
+
+By **Theorem 3.9**, $(p_n)$ is bounded. So $\exists R>0$ such that $p_n\in B(0,R)$ for all $n$. Moreover $p_n\in \overline{B(0,R)}$. and $\overline{B(0,R)}$ is closed and bounded. Thus by **Theorem 2.41**, $\overline{B(0,R)}$ is compact.
+
+Note that **Theorem 2.41** only works for $\mathbb{R}^k$.
+
+So by (b), $(p_n)$ converges to some $p\in \overline{B(0,R)}$.
+
+EOP
+
+#### Definition 3.12
+
+Let $X$ be a metric space. We say $X$ is **complete** if every Cauchy sequence in $X$ converges.
+
+**Theorem 3.11(b)** can also be rephrased as:
+
+$X$ is a compact metric space $\implies$ $X$ is complete.
+
+**Theorem 3.11(c)** can also be rephrased as:
+
+$\mathbb{R}^k$ is complete.
+
+> Note: completeness is a property of the "universe" $X$, not a property of any particular sequence in $X$.
+
+$\mathbb{Q}$ is not complete. $\{3,3.1,3.14,3.141,3.1415,\dots\}$ is a Cauchy sequence in $\mathbb{Q}$ but it does not converge in $\mathbb{Q}$.
+
+Fact: If $X$ is complete and $E$ is a closed subset of $X$, then $E$ is complete.
+
+#### Definition 3.13
+
+A sequence $(s_n)$ of real numbers is said to be
+
+- **monotone increasing** if $s_n\leq s_{n+1}$ for all $n$.
+- **monotone decreasing** if $s_n\geq s_{n+1}$ for all $n$.
+- **strictly monotone increasing** if $s_n<s_{n+1}$ for all $n$.
+- **strictly monotone decreasing** if $s_n>s_{n+1}$ for all $n$.
+- **monotone** if it is either monotone increasing or monotone decreasing.
+
+Example:
+
+1. $s_n=1/n$ is strictly monotone decreasing.
+2. $s_n=(-1)^n$ is neither monotone increasing nor monotone decreasing.
+
+#### Theorem 3.14
+
+Suppose $(s_n)$ is monotonic. Then $(s_n)$ converges $\iff$ $(s_n)$ is bounded.
+
+Proof:
+
+If $(s_n)$ is monotonic and bounded, then by previous result, $(s_n)$ converges.
+
+If $(s_n)$ is monotonic and converges, then by **Theorem 3.2(c)**, $(s_n)$ is bounded.
+
+EOP
+
+### Upper and lower limits
+
+#### Definition 3.15 (Divergence to $\infty$ or $-\infty$)
+
+Let $(s_n)$ be a sequence of real numbers with the following properties:
+
+For every real number $M$ there is an integer $N$ such that $n\geq N$ implies $s_n>M$. We then write $s_n\to\infty$
+
+For every real number $M$ there is an integer $N$ such that $n\geq N$ implies $s_n<M$. We then write $s_n\to-\infty$.
+
+_for every real number, we can find a element in the sequence that is greater than or less than it_
+
+#### Definition 3.16
+
+Let $(s_n)$ be a sequence of real numbers.
+
+Let $E_n=\{x\in[-\infty,\infty]:\exists \textup{ subsequence } (s_{n_k})\textup{ of } (s_n)\textup{ such that } s_{n_k}\to x\}$.
+
+Let $S^*=\sup E_n$, $S_*=\inf E_n$.
+
+We define $$\limsup_{n\to\infty} s_n=S^*$$ and $$\liminf_{n\to\infty} s_n=S_*$$
+
+Informally, $S^*$ is the largest possible value that a subsequence of $(s_n)$ can converge to.
+
+Example:
+
+$s_n=(-1)^n$, $E_n=\{-1,1\}$, $S^*=\sup E_n=1$, $S_*=\inf E_n=-1$. and $\lim_{n\to\infty} s_n$ does not exist.
+
+One advantage of $\limsup$ and $\liminf$ is that they always exist (they may be $\infty$ or $-\infty$), even if the sequence does not converge.
+

pages/Math4111/Math4111_L18.md

@@ -0,0 +1,157 @@
+# Lecture 18
+
+## Review
+
+Let $(s_n)$ be a sequence in $\mathbb{R}$, and suppose $\limsup_{n\to\infty} s_n=1$. Consider the following four sets:
+
+1. $\{n\in\mathbb{N}:s_n>2\}$
+2. $\{n\in\mathbb{N}:s_n<2\}$
+3. $\{n\in\mathbb{N}:s_n>0\}$
+4. $\{n\in\mathbb{N}:s_n<0\}$
+
+For each set, determine if the set $(1)$ must be infinite, or $(2)$ must be finite, or $(3)$ could be either finite or infinite, depending on the sequence $(s_n)$.
+
+If $\liminf_{n\to\infty} s_n=1$, then $\lim_{n\to\infty} \sup\{s_n,s_{n+1},s_{n+2},\dots\}=1$.
+
+So 1 must be finite, since if it is infinite, then $\limsup_{n\to\infty} s_n\geq 2$, which contradicts the given $\limsup_{n\to\infty} s_n=1$.
+
+2 and 3 are infinite.
+
+since $\liminf_{n\to\infty} s_n=1$, there exists infinitely many $n$ such that $2>s_n>0$.
+
+4 could be either finite or infinite.
+
+- $s_n=(-1)^n$ is example for 4 being infinite.
+- $s_n=1$ is example for 4 being finite.
+
+## Continue on Limit Superior and Limit Inferior
+
+### Limit Superior
+
+#### Definition 3.16
+
+Let $(s_n)$ **be a sequence of real numbers**.
+
+$S^*$ is the largest possible value that a subsequence of $(s_n)$ can converge to.
+
+(Normally, we need to be careful about the definition of "largest possible value", but in this case it does exist by **Theorem 3.7**.)
+
+Abbott's definition:
+
+$S^*=\limsup_{n\to\infty}\{s_k:k\geq n\}$.
+
+#### Theorem 3.17
+
+Let $(s_n)$ **be a sequence of real numbers**.
+
+$S^*$ is the unique number satisfying the following:
+
+1. $\forall x<S^*, \{n\in\mathbb{N}:s_n\geq x\}$ is infinite. (same as saying $\forall N\in\mathbb{N},\exists n\geq N$ such that $s_n\geq x$)
+2. $\forall x>S^*$, $\{n\in\mathbb{N}:s_n\geq x\}$ is finite. (same as saying $\exists N\in\mathbb{N}$ such that $n\geq N\implies s_n<x$)
+
+_In other words, $S^*$ is the boundary between when $\{n\in\mathbb{N}:s_n\geq x\}$ is infinite and when it is finite._
+
+Proof:
+
+(a) Case 1: $S^*=\infty$.
+
+Then $\sup E=\infty$, so $(s_n)$ is not bounded above.
+
+So $\forall x\in\mathbb{R}$, $\{n\in\mathbb{N}:s_n\geq x\}$ is infinite.
+
+Case 2: $S^*\in\mathbb{R}$.
+
+Then $S^*=\sup E\in \overline{E}=E$, by **Theorem 3.7**.
+
+So $\exists (s_{n_k})$ such that $s_{n_k}\to S^*$.
+
+So $\forall x<S^*$, $\{n\in\mathbb{N}:s_n\geq x\}$ is infinite.
+
+Case 3: $S^*=-\infty$: The statement is vacuously true. ($\nexists x<-\infty$)
+
+(b) We'll prove the contrapositive: If $\{n\in\mathbb{N}:s_n\geq x\}$ is infinite, then $S^*\leq x$.
+
+Case 1: $(s_n)$ is not bounded above.
+
+Then $\exists$ subsequence $(s_{n_k})$ such that $s_{n_k}\to\infty$. Thus $S^*=\infty\leq x$.
+
+Case 2: $(s_n)$ is bounded above.
+
+Let $M$ be an upper bound for $(s_n)$. Then $\{n\in\mathbb{N}:s_n\in[M,x]\}$ is infinite, by **Theorem 3.6 (b)** ($\exists$ subsequence $(s_{n_k})$ in $[x,M)$ and $\exists t\in[x,M]$ such that $s_{n_k}\to t$. This implies $t\in E$, so $x\leq t\leq S^*$).
+
+EOP
+
+#### Theorem 3.19 ("one-sided squeeze theorem")
+
+Let $(s_n)$ and $(t_n)$ be two sequences such that $s_n\leq t_n$ for all $n\in\mathbb{N}$, then
+
+$$
+\limsup_{n\to\infty} s_n\leq \limsup_{n\to\infty} t_n
+$$
+$$
+\liminf_{n\to\infty} s_n\leq \liminf_{n\to\infty} t_n
+$$
+
+Proof:
+
+By transitivity of $\leq$, for all $x\in\mathbb{R}$,
+
+$$
+|\{n\in\mathbb{N}:s_n\geq x\}|\leq |\{n\in\mathbb{N}:t_n\geq x\}|
+$$
+
+By **Theorem 3.17**, $\{n\in\mathbb{N}:t_n\geq x\}$ is finite $\implies$ $\{n\in\mathbb{N}:s_n\geq x\}$ is finite.
+
+Thus $\limsup_{n\to\infty} s_n\leq \limsup_{n\to\infty} t_n$.
+
+EOP
+
+> Normal squeeze theorem: If $s_n\leq t_n\leq u_n$ for all $n\in\mathbb{N}$, and $\lim_{n\to\infty} s_n=\lim_{n\to\infty} u_n=L$, then $\lim_{n\to\infty} t_n=L$.
+>
+> Proof: Exercise, hint: $u_n\to L\implies \limsup_{n\to\infty} u_n=\liminf_{n\to\infty} u_n=L$.
+
+#### Theorem 3.20
+
+> Binomial theorem: $(1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k$.
+
+Special sequences:
+
+(a) If $p>0$, then $\lim_{n\to\infty}\frac{1}{n^p}=0$.
+
+We want to find $\frac{1}{n^p}<\epsilon\iff n\geq\frac{1}{\epsilon^{1/p}}$.
+
+(b) If $p>0$, then $\lim_{n\to\infty}\sqrt[n]{p}=1$.
+
+We want to find $\sqrt[n]{p}-1<\epsilon\iff p<(1+\epsilon)^n$.
+
+> Bernoulli's inequality: for $\epsilon>0,n\in\mathbb{N}$, $(1+\epsilon)^n\geq 1+n\epsilon$.
+
+So it's enough to have $p<1+n\epsilon$
+
+So we can choose $N>\frac{p-1}{\epsilon}$.
+
+Another way of writing this: Let $x_n=\sqrt[n]{p}-1$.
+
+Then $p=(1+x_n)^n\geq 1+nx_n$.
+
+So $0\leq x_n\leq\frac{p-1}{n}$.
+
+By the squeeze theorem, $x_n\to 0$.s
+
+(c) $\lim_{n\to\infty}\sqrt[n]{n}=1$.
+
+We want to find $\sqrt[n]{n}-1<\epsilon\iff n<(1+\epsilon)^n$. (this will not work for bernoulli's inequality)
+
+So it's enough to have $n<\frac{n(n-1)}{2}\epsilon^2\iff n>1+\frac{2}{\epsilon^2}$. So choose $N>1+\frac{2}{\epsilon^2}$.
+
+(d) If $p>0$ and $\alpha$ is real, then $\lim_{n\to\infty}\frac{n^\alpha}{(1+p)^n}=0$.
+
+With binomial theorem, $(1+p)^n\geq \binom{n}{k}p^k(k\leq n)$.
+
+$\binom{n}{k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$.
+
+If $n\geq 2k$, then $n-k+1\geq n-\frac{n}{2}+1\geq\frac{n}{2}$.
+
+So $\binom{n}{k}\geq\frac{(n/2)^k}{k!}$.
+
+Continue on next class.

pages/Math4111/Math4111_L19.md

@@ -0,0 +1,194 @@
+# Lecture 19
+
+## Review
+
+> Binomial theorem: For $n\in\mathbb{N}$,
+> $$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$$
+
+1. Show that $2^n\geq \binom{n}{4}$ for all $n\geq 4$. (Hint: Expand $(1+1)^n$ using the binomial theorem)
+    Proof: 
+    $$
+    \begin{aligned}
+    (1+1)^n&=\sum_{k=0}^{n}\binom{n}{k}1^{n-k}1^k\\
+    &=\sum_{k=0}^{n}\binom{n}{k}\\
+    &=\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}\\
+    &\geq\binom{n}{4}
+    \end{aligned}
+    $$
+    EOP
+2. Using part 1, show that $\lim_{n\to\infty}\frac{n^3}{2^n}=0$.  
+   Proof:
+   $$
+   \frac{n^3}{2^n}\leq\frac{n^3}{\binom{n}{4}}
+   $$
+   The value of $\frac{n^3}{\binom{n}{4}}$ is decreasing when $n\geq 4$.
+   EOP
+
+## New materials
+
+### Series
+
+#### Definition 3.21
+
+Let $(a_n)_{n=1}^{\infty}$ be a sequence in $\mathbb{C}$. Let $s_n=\sum_{k=1}^{n}a_k$ denotes the sequence of partial sums.
+
+1. We say the series $\sum_{n=1}^{\infty}a_n$ converges if the sequence of partial sums $(s_n)_{n=1}^{\infty}$ converges.
+2. We define the sum of the series $\sum_{n=1}^{\infty}a_n$ to be the limit of the sequence of partial sums, i.e., $$\sum_{n=1}^{\infty}a_n=\lim_{n\to\infty}s_n=\lim_{n\to\infty}\sum_{k=1}^{n}a_k.$$
+
+#### Theorem 3.22 (Cauchy criterion for series)
+
+The series $\sum_{n=1}^{\infty}a_n$ converges if and only if for every $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $m,n\in\mathbb{N}$ with $m\geq n\geq N$,
+$$
+\left|\sum_{k=n}^{m}a_k\right|<\epsilon.
+$$
+
+Proof:
+
+$\sum_{n=1}^{\infty}a_n$ converges if and only if $(s_n)_{n=1}^{\infty}$ converges.
+
+Since $\mathbb{C}$ is complete, $(s_n)_{n=1}^{\infty}$ converges if and only if $(s_n)_{n=1}^{\infty}$ is Cauchy.
+
+Since $(s_n)_{n=1}^{\infty}$ is Cauchy, for every $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $m,n\in\mathbb{N}$ with $m\geq n\geq N$,
+$$
+|s_m-s_n|=\left|\sum_{k=n}^{m}a_k\right|<\epsilon.
+$$
+
+EOP
+
+Special case of this theorem.
+
+#### Corollary 3.23
+
+If $\sum_{n=1}^{\infty}a_n$ converges, then $\lim_{n\to\infty}a_n=0$.
+
+Note: the converse is not true. Example: $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges.
+
+The contrapositive of this corollary is: If $\lim_{n\to\infty}a_n\neq 0$, then $\sum_{n=1}^{\infty}a_n$ diverges. It is useful naming as ``n-th term test for divergence''.
+
+Observe:
+
+$\forall n,a_n\geq 0$
+
+$(a_n)$ is a non-negative sequence if and only if $(s_n)_{n=1}^{\infty}$ is increasing sequence.
+
+So if $(a_n)$ is a non-negative sequence, then $\sum_{n=1}^{\infty}a_n$ converges if and only if $(s_n)_{n=1}^{\infty}$ is bounded above.
+
+#### Theorem 3.25 (Comparison test)
+
+Let $(a_n)$ be a sequence in $\mathbb{C}$ and $(c_n)$ be a non-negative sequence in $\mathbb{R}$. Suppose $\forall n, |a_n|\leq c_n$.
+
+(a) If the series $\sum_{n=1}^{\infty}c_n$ converges, then the series $\sum_{n=1}^{\infty}a_n$ converges.  
+(b) If the series $\sum_{n=1}^{\infty}a_n$ diverges, then the series $\sum_{n=1}^{\infty}c_n$ diverges.
+
+Proof:
+
+(a) By **Theorem 3.22**, it's enough to show that for every $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $m,n\in\mathbb{N}$ with $m\geq n\geq N$,
+$$
+\left|\sum_{k=n}^{m}a_k\right|<\epsilon.
+$$
+
+Let $\epsilon>0$ be arbitrary.
+
+Since $\sum_{n=1}^{\infty}c_n$ converges, by **Theorem 3.22**, for the above $\epsilon$, there exists $N\in\mathbb{N}$ such that for all $m,n\in\mathbb{N}$ with $m\geq n\geq N$,
+$$
+\left|\sum_{k=n}^{m}c_k\right|\leq \sum_{k=n}^{m}c_k<\epsilon.
+$$
+
+EOP
+
+#### Theorem 3.26 (Geometric series)
+
+Let $x\in\mathbb{C}$.
+
+(a) If $|x|<1$, then the series $\sum_{n=0}^{\infty}x^n$ converges and $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$.  
+(b) If $|x|\geq 1$, then the series $\sum_{n=0}^{\infty}x^n$ diverges.
+
+Proof:
+
+(b) If $|x|\geq 1$, then $x^n$ does not converge to 0. So the series $\sum_{n=0}^{\infty}x^n$ diverges.
+
+(a) Let $s_n=\sum_{k=0}^{n}x^k=1+x+x^2+\cdots+x^n$.
+
+$xs_n=x+x^2+x^3+\cdots+x^n+x^{n+1}=s_n+x^{n+1}$.
+
+So $s_n=\frac{1-x^{n+1}}{1-x}$.
+
+Since $|x|<1$, $x^{n+1}$ converges to 0. So $\lim_{n\to\infty}s_n=\frac{1}{1-x}$.
+
+EOP
+
+#### Lemma 3.28 
+
+(a) $\sum_{n=0}^{\infty}\frac{1}{n}$ diverges.  
+(b) $\sum_{n=0}^{\infty}\frac{1}{n^2}$ converges.
+
+Proof:
+
+(a)
+$$
+\begin{aligned}
+\sum_{n=0}^{\infty}\frac{1}{n}&=\frac{1}{1}+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots\\
+&>\frac{1}{2}+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\cdots\\
+&=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots\\
+&=\infty
+\end{aligned}
+$$
+
+(b)
+$$
+\begin{aligned}
+\sum_{n=0}^{\infty}\frac{1}{n^2}&=\frac{1}{1}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots\\
+&<\frac{1}{1}+\left(\frac{1}{2^2}+\frac{1}{2^2}\right)+\left(\frac{1}{4^2}+\cdots+\frac{1}{4^2}\right)+\left(\frac{1}{8^2}+\cdots+\frac{1}{8^2}\right)+\cdots\\
+&=\frac{1}{1}+\frac{2}{2^2}+\frac{4}{4^2}+\frac{8}{8^2}+\cdots\\
+&=\frac{1}{1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots\\
+&=\frac{1}{1-\frac{1}{2}}\\
+&=\frac{1}{\frac{1}{2}}\\
+&=2
+\end{aligned}
+$$
+
+> Fun fact: $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$.
+
+EOP
+
+#### Theorem 3.27 (Cauchy condensation test)
+
+Suppose $(a_n)$ is a non-negative sequence. The series $\sum_{n=1}^{\infty}a_n$ converges if and only if the series $\sum_{k=0}^{\infty}2^ka_{2^k}$ converges.
+
+Proof:
+
+Let $s_n=\sum_{k=1}^{n}a_k$ and $t_k=\sum_{k=0}^{k}2^ka_{2^k}$.
+
+If $n\leq 2^k$, then 
+
+$$
+\begin{aligned}
+s_n&=a_1+a_2+\cdots+a_n\\
+&\leq a_1+(a_2+a_3)+(a_4+a_5+\cdots+a_7)+\cdots+(a_{2^k}+a_{2^k+1}+\cdots+a_{2^{k+1}-1})\\
+&\leq a_1+2a_2+4a_4+\cdots+2^ka_{2^k}\\
+&=t_k.
+\end{aligned}
+$$
+
+If $n\geq 2^{k+1}$, then
+
+$$
+\begin{aligned}
+s_n&=a_1+a_2+\cdots+a_n\\
+&\geq a_1+a_2+(a_3+a_4)+(a_5+a_6+\cdots+a_7)+\cdots+(a_{2^k}+a_{2^k+1}+\cdots+a_{2^{k+1}-1})\\
+&\geq a_1+a_2+2a_4+\cdots+2^{k-1}a_{2^k}\\
+&\geq \frac{1}{2}\left(a_1+2a_2+4a_4+\cdots+2^ka_{2^k}\right)\\
+&=\frac{1}{2}t_k.
+\end{aligned}
+$$
+
+We have shown that
+
+- If $n\leq 2^k$, then $s_n\leq t_k$.
+- If $n\geq 2^{k+1}$, then $s_n\geq \frac{1}{2}t_k$.
+
+So $(s_n)_{n=1}^{\infty}$ is a bounded above.
+
+By **Theorem 3.14**, $(s_n)_{n=1}^{\infty}$ converges if and only if $(t_k)_{k=0}^{\infty}$ converges.
+
+EOP

pages/Math4111/Math4111_L2.md

@@ -0,0 +1,106 @@
+# Lecture 2
+
+Ordered sets, least upper bounds and fields.
+
+## Warm up
+
+(a) The statements says: $\forall a\in A, \exists s\in a$ such that $s\geq 7$.
+
+The negation is $\exist a\in A,\forall s\in a$, such that $s<7$.
+
+## Ordered sets
+
+Let $S$ be a set. An order on $S$ is a relation satisfying:
+
+1. "trichotomy". If $x,y\in S$, then exactly on eof the these statements are hold: $x<y,x=y,x>y$.
+2. "transitivity". If $x,y,z\in S$, then $x<y,y\implies x<z$.
+
+An ordered set is a set with order.
+
+### Definition 1.7
+
+Let $S$ be an ordered set and let $E\subset S$ ($E$ is the "universe", all the element you can ever think of...)
+
+1. $\beta\in S$ is an upper bound of $E$ if $\forall x\in E,x\leq \beta$
+2. $E$ is bounded above if $\exist \beta \in S$ such that $\beta$ is the upper bound of $E$.
+
+#### Example
+
+1. $S=\mathbb{Q}, E=\{1,2,3\}$ ($E$ is bounded above)
+    * 3,4,3.5 are all upper bounds of $E$.
+    * 2,2.5 is not upper bounds of $E$.
+    * $\pi$ is not upper bound of $S$ since $\pi\notin S$
+2. $S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x<1\}$  ($E$ is bounded above)
+    * The upper bound is $1$.
+3. $S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x\}$ ($E$ is not bounded above)
+    * Sad
+4. $S=\mathbb{Q}, E=\phi$.
+    * $\beta\in S$ is an upper bound of $E$ if $\forall x\in E,x\leq \beta\equiv\beta\in S$ is not an upper bound of $E$ if $\exists x\in E,x> \beta$
+    * So this statement is true for any rational numbers since $\cancel{\exist} a\in E$ such that $x>\beta$.
+
+### Definition 1.8
+
+Least upper bound, LUB, supremum, SUP
+
+Let $S$ be an ordered set and $E\subset S$. We say $\alpha\in S$ is the LUB of $E$ if 
+
+1. $\alpha$ is the UB of $E$. ($\forall x\in E,x\leq \alpha$)
+2. if $\gamma<\alpha$, then $\gamma$ is not UB of $E$. ($\forall \gamma <\alpha, \exist x\in E$ such that $x>\gamma$ )
+
+#### Lemma 
+
+Uniqueness of upper bounds.
+
+If $\alpha$ and $\beta$ are LUBs of $E$, then $\alpha=\beta$.
+
+Proof:
+
+Suppose for contradiction $\alpha$ and $\beta$ are both LUB of $E$, then $\alpha\neq\beta$
+
+WLOG $\alpha>\beta$ and $\beta>\alpha$.
+
+EOP
+
+We write $SupE$ to denote the LUB of $E$.
+
+This also applies to $GLB$ (greatest lower bound) and infinum of $E$
+
+#### Example 
+
+1. $S=\mathbb{Q}, E=\{1,2,3\}$ ($E$ is bounded above)
+    * $SupE=3$, $Inf E=1$
+2. $S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x<1\}$  ($E$ is bounded above)
+    * $SupE=3$, $Inf E=1$
+
+$SupE$ and $Inf E=1$ don't have to $\in E$
+
+3. $S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x\}$ ($E$ is not bounded above)
+   * $SupE=\infty$ or not defined, $Inf E=0$
+4. $S=\mathbb{Q}, E=\phi$.
+   * $SupE=-\infty$ or not defined, $Inf E=\infty$ or not defined, we don't put $\infty$ in $\mathbb{Q}$
+
+Important example
+
+5. $S=\mathbb{Q}, A=\{p\leq \mathbb{Q}:p>0, p^2<2\}$.
+    * $A$ is not empty and bounded above. However, $Sup A$ des not exists.
+
+If $S=\mathbb{R}, A=\{p\leq \mathbb{Q}:p>0, p^2<2\}$.
+    * $A$ is not empty and bounded above. However, $Sup A=\sqrt{2}$.
+
+#### Least upper bound property (LUBP)
+
+if $\forall E\subset S$ that tis non-empty and bounded above, $\exist Sup E\in S$.
+
+#### Greatest upper bound property (GLBP)
+
+S has greatest lower bound property (GLBP) if $\exist E\subset S$ that is non-empty and bounded below, $\exists Inf E\in S$
+
+$\mathbb{Q}$ does not have LUBP and GLBP.
+
+#### Theorem 1.11
+
+Let $S$ be an ordered set. Then $S$ has the LUBP $\iff$ $S$ has the GLBP
+
+Proof:
+
+Let $S$ be a set with LUBP. (we want to show $S$ has GLBP)

pages/Math4111/Math4111_L20.md

@@ -0,0 +1,270 @@
+# Lecture 20
+
+## Review
+
+Using the binomial theorem, prove that 
+
+$$
+\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{n!}\geq \left(1+\frac{1}{n}\right)^n
+$$
+
+> Binomial theorem: $$(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$$ $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$
+
+Proof:
+$$
+\begin{aligned}
+\left(1+\frac{1}{n}\right)^n &= \sum_{k=0}^{n} \binom{n}{k} \left(1\right)^{n-k} \left(\frac{1}{n}\right)^k \\
+&= \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} \\
+&= \sum_{k=0}^{n} \frac{1}{k!} \prod_{j=1}^{k} \frac{n-j+1}{n} \\
+\end{aligned}
+$$
+
+Since $j\geq 1$, $\frac{n-j+1}{n} \leq1$.
+
+$$
+\begin{aligned}
+&= \sum_{k=0}^{n} \frac{1}{k!} \prod_{j=1}^{k} \frac{n-j+1}{n} \\
+&\geq \sum_{k=0}^{n} \frac{1}{k!} \\
+\end{aligned}
+$$
+
+## New material
+
+### Series
+
+#### Definition 3.30
+
+$$
+e=\sum_{n=0}^{\infty} \frac{1}{n!}
+$$
+
+#### Lemma 3.30
+
+$\sum_{n=0}^{\infty} \frac{1}{n!}$ converges.
+
+Proof:
+
+If $n\geq 2$,
+
+$$
+\begin{aligned}
+\frac{1}{n!} &= \frac{1}{n} \cdot \frac{1}{(n-1)!} \\
+&\leq \frac{1}{2} \cdot \frac{1}{2} \cdot \dots \cdot \frac{1}{2} \\
+&= \frac{1}{2^{n-1}}
+\end{aligned}
+$$
+
+$$
+\frac{1}{n!} \leq \frac{1}{2^{n-1}}
+$$
+
+So $\sum_{n=0}^{\infty} \frac{1}{n!}$ converges.
+
+#### Theorem 3.31
+
+$$
+\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e
+$$
+
+Proof:
+
+Let $s_n = \sum_{k=0}^{n} \frac{1}{k!}$, let $t_n = \left(1+\frac{1}{n}\right)^n$.
+
+Goal: $\lim_{n\to\infty} s_n = \lim_{n\to\infty} t_n$. we already proved $\lim_{n\to\infty} s_n$ exists. But we don't know yet if $\lim_{n\to\infty} t_n$ exists.
+
+By warmup exercise, $\forall n\geq 0, t_n \leq s_n$.
+
+So if $\limsup_{n\to\infty} t_n \leq \limsup_{n\to\infty} s_n$, then $\lim_{n\to\infty} t_n$ exists and $\lim_{n\to\infty} t_n = \lim_{n\to\infty} s_n$.
+
+Now we will show $\limsup_{n\to\infty} t_n \geq e$.
+
+Idea: (special case of the argument)
+
+If $n\geq 2$, then
+
+$$
+\begin{aligned}
+t_n &= \sum_{k=0}^{n} \binom{n}{k} \left(\frac{1}{n}\right)^k \\
+&\geq \binom{n}{0} + \binom{n}{1}\frac{1}{n} + \binom{n}{2}\left(\frac{1}{n}\right)^2 + \cdots + \binom{n}{n}\left(\frac{1}{n}\right)^n \\
+&= 1 + \frac{n}{n} + \frac{n(n-1)}{2n^2} + \cdots + \frac{1}{n^n} \\
+\end{aligned}
+$$
+
+Let $n\to\infty$, then
+
+$$
+\liminf_{n\to\infty} t_n \geq 1 + 1 + \frac{1}{2} + \frac{1}{3} + \cdots
+$$
+
+Fix $m\geq 2$, for any $n\geq m$,
+
+$$
+t_n \geq \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!}\frac{n}{n}\frac{n-1}{n}\cdots+\frac{1}{m!}\frac{n}{n}\frac{n-1}{n}\cdots\frac{n-m+1}{n}
+$$
+
+Let $n\to\infty$, then
+
+$$
+\liminf_{n\to\infty} t_n \geq \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{m!}=s_m
+$$
+
+So $\liminf_{n\to\infty} t_n \geq \lim_{n\to\infty} s_n = e$.
+
+Therefore, $e\leq \liminf_{n\to\infty} t_n\leq \limsup_{n\to\infty} t_n\leq e$.
+
+So $\lim_{n\to\infty} t_n$ exists and $\lim_{n\to\infty} t_n = e$.
+
+EOP
+
+#### Theorem 3.32
+
+$e$ is irrational.
+
+Q: How good is the approximation is $s_n$ to $e$?
+
+A: Very good actually.
+$$
+\begin{aligned}
+e-s_n &= \sum_{k=n+1}^{\infty} \frac{1}{k!} \\
+&<\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\cdots\right) \\
+&=\frac{1}{(n+1)!}\sum_{k=0}^{\infty}\left(\frac{1}{n+1}\right)^k \\
+&=\frac{1}{(n+1)!}\frac{1}{1-\frac{1}{n+1}} \\
+&=\frac{1}{n!}\cdot\frac{1}{n} \\
+&<\frac{1}{n!n}
+\end{aligned}
+$$
+
+Proof:
+
+Suppose $e=\frac{p}{q}$ for some $p,q\in\mathbb{N}$.
+
+Observe that:
+
+$$
+s_q=1+1+\frac{1}{2}+\cdots+\frac{1}{q!}
+$$
+
+So $q! s_q$ is an integer.
+
+Since $e=\frac{p}{q}$, $q!e$ is an integer, $q!(e-s_q)$ is an integer.
+
+However,
+
+$$
+0<q!(e-s_q)<\frac{q!}{q!q}<\frac{1}{q}
+$$
+
+Contradiction.
+
+EOP
+
+### The root and ratio tests
+
+This is a fancy way of using comparison test with geometric series.
+
+#### Theorem 3.33 (Root test)
+
+> $$ \sqrt[n]{|a_n|} \leq \alpha \implies |a_n|\leq \alpha^n$$
+
+Given a series $\sum_{n=0}^{\infty} a_n$, put $\alpha = \limsup_{n\to\infty} \sqrt[n]{|a_n|}$.
+
+Then
+
+(a) If $\alpha < 1$, then $\sum_{n=0}^{\infty} a_n$ converges.  
+(b) If $\alpha > 1$, then $\sum_{n=0}^{\infty} a_n$ diverges.  
+(c) If $\alpha = 1$, the test gives no information
+
+Proof:
+
+(a) Suppose $\alpha < 1$. Then $\exists \beta$ such that $\alpha < \beta < 1$.
+
+By **Theorem 3.17(b)**, $\forall n\geq N, \sqrt[n]{|a_n|} < \beta$.
+
+So $\forall n\geq N, |a_n| < \beta^n$.
+
+By comparison test, $\sum_{n=0}^{\infty} a_n$ converges.
+
+(b) Suppose $\alpha > 1$. By **Theorem 3.17(a)**, $\{n\in \mathbb{N}: \sqrt[n]{|a_n|} > 1\}$ is infinite.
+
+Thus $a_n\not\to 0$, $\sum_{n=0}^{\infty} a_n$ diverges.
+
+(c) $\sum_{n=0}^{\infty} \frac{1}{n}$ and $\sum_{n=0}^{\infty} \frac{1}{n^2}$ both have $\alpha = 1$. but the first diverges and the second converges.
+
+EOP
+
+#### Theorem 3.34 (Ratio test)
+
+> $$ \left|\frac{a_{n+1}}{a_n}\right| \leq \alpha \implies |a_n|\leq \alpha^n$$
+
+Given a series $\sum_{n=0}^{\infty} a_n$, $a_n\in\mathbb{C}\backslash\{0\}$.
+
+Then
+
+(a) If $\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$, then $\sum_{n=0}^{\infty} a_n$ converges.  
+(b) If $\left|\frac{a_{n+1}}{a_n}\right| \geq 1$ for all $n\geq n_0$ for some $n_0\in\mathbb{N}$, then $\sum_{n=0}^{\infty} a_n$ diverges.
+
+Remark:
+
+1. If $\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$, the test gives no information.
+2. If $\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| > 1$, the test gives no information.
+
+Proof:
+
+(b) $\forall n\geq n_0, \left|\frac{a_{n+1}}{a_n}\right| \geq 1$.
+
+So $a_{n_0}\not\to 0$, $\sum_{n=0}^{\infty} a_n$ diverges.
+
+(a) $\beta \in(\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|, 1)$.
+
+By **Theorem 3.17(b)**, $\exists N$ such that $\forall n\geq N, \left|\frac{a_{n+1}}{a_n}\right| < \beta < 1$.
+
+So,
+
+$$
+\begin{aligned}
+|a_N| &< \beta|a_N|\\
+|a_{N+1}| &< \beta|a_{N+1}|\\
+|a_{N+2}| &< \beta|a_{N+2}|\\
+\end{aligned}
+$$
+
+i.e. $\forall n\geq N, |a_n| < \beta^{n-N}|a_N|=\beta^n(\beta^{-N}|a_N|)$.
+
+Since $\sum_{n=N}^{\infty} \beta^n$ converges, by comparison test, $\sum_{n=0}^{\infty} a_n$ converges.
+
+EOP
+
+We will skip **Theorem 3.37**. One implication is that if ratio test can be applied, then root test can be applied.
+
+### Power series
+
+#### Definition 3.38
+
+Let $(c_n)$ be a sequence of complex numbers. A power series is a series of the form
+
+$$
+\sum_{n=0}^{\infty} c_n z^n
+$$
+
+#### Theorem 3.39
+
+Given a power series $\sum_{n=0}^{\infty} c_n z^n$, let $R=\frac{1}{\limsup_{n\to\infty} \sqrt[n]{|c_n|}}$.
+
+Then
+
+(a) The series converges absolutely for all $z\in\mathbb{C}$ with $|z| < R$.  
+(b) The series diverges for all $z\in\mathbb{C}$ with $|z| > R$.  
+(c) If $0\leq r < R$, then the series converges uniformly on the closed disk $\{z\in\mathbb{C}: |z|\leq r\}$.
+
+Proof:
+
+$$
+\begin{aligned}
+\limsup_{n\to\infty} \sqrt[n]{|c_n z^n|} &= \limsup_{n\to\infty} \sqrt[n]{|c_n|} \cdot |z| \\
+&= \frac{|z|}{R}
+\end{aligned}
+$$
+
+By root test, the series converges absolutely for all $z\in\mathbb{C}$ with $|z| < R$.
+
+EOP

pages/Math4111/Math4111_L21.md

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+# Lecture 21
+
+## Review
+
+Recall the alternating series test from calculus: "Suppose $(a_n)^\infty_{n=1}$ is a sequence satisfies the following conditions:
+
+1. The sequence is nonnegative. (For all $n\in \mathbb{N}$, $a_n\geq 0$.)
+2. The sequence is decreasing. ($a_1\geq a_2\geq a_3\geq \cdots$)
+3. $\lim_{n\to\infty}a_n=0$.
+
+Then $\sum_{n=1}^\infty (-1)^{n+1}a_n$ converges."
+
+Exercise: Show that the statement above is false if we remove the second condition.
+
+[Hint: Use the fact that $\sum_{n=1}^\infty \frac{1}{n}$ diverges.]
+
+Let the sequence $a_n$ be defined as $a_n=\frac{1}{n},a_{n+1}=0$ for all $n\in \mathbb{N}$. This sequence satisfies the 1,3 but not the 2. 
+
+And the harmonic series is not convergent.
+
+## New Material
+
+### Other tests for convergence of series
+
+Recall the integration by parts formula:
+
+Let $A(t),a(t),b(t)$ be functions of $t$ and $A'(t)=a(t)$.
+
+Then 
+$$
+\begin{aligned}
+\int_p^q a(t)b(t)\,dt&=\int_p^q b(t)A'(t)\,dt\\
+&=\left.b(t)A(t)\right|_p^q-\int_p^q A(t)b'(t)\,dt
+\end{aligned}
+$$
+
+#### Theorem 3.41 Summation by parts
+
+Let $a_n,b_n$ be sequences.
+
+Let $A(n)=\sum_{k=1}^n a_k$. ($A_{-1}=0$). If $0\leq p\leq q$, then
+
+$$
+\sum_{n=p}^q a_nb_n=A_q b_q-A_{p-1}b_p-\sum_{n=p}^{q-1}A_n (b_n-b_{n+1})
+$$
+
+Proof:
+
+$$
+\begin{aligned}
+\sum_{n=p}^q a_nb_n&=\sum_{n=p}^q (A_n-A_{n-1})b_n\\
+&=\sum_{n=p}^q A_nb_n-\sum_{n=p}^q A_{n-1}b_n\\
+&=\sum_{n=p}^q A_nb_n-\sum_{n=p-1}^{q-1}A_n b_{n+1}\\
+&=A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1} A_nb_n-\sum_{n=p}^{q-1} A_n b_{n+1}\\
+&=A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1} A_n (b_n-b_{n+1})
+\end{aligned}
+$$
+
+EOP
+
+#### Theorem 3.42 (Dirichlet's test)
+
+Suppose
+
+(a) the partial sum $A_n$ of $\sum a_n$ form a bounded sequence.  
+(b) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)  
+(c) $\lim_{n\to\infty}b_n=0$.
+
+Then $\sum a_nb_n$ converges.
+
+Proof:
+
+By Cauchy criterion, it's enough to prove
+
+$\forall \epsilon >0, \exists N\in \mathbb{N}$ such that for all $p\geq q\geq N$,
+
+$$
+\left|\sum_{n=p}^q a_nb_n\right|<\epsilon
+$$
+
+By the partial sum $A_n$ of $\sum a_n$ form a bounded sequence. Let $\left|A_n\right|\leq M$ for all $n\in \mathbb{N}$.
+
+$$
+\begin{aligned}
+\left|\sum_{n=p}^q a_nb_n\right|&=\left|A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1}A_n (b_n-b_{n+1})\right|\\
+&\leq |A_qb_q|+|A_{p-1}b_p|+\sum_{n=p}^{q-1}|A_n (b_n-b_{n+1})|\\
+&\leq M|b_q|+M|b_p|+\sum_{n=p}^{q-1}M(b_n-b_{n+1})\\
+&=M|b_q|+M|b_p|+M\sum_{n=p}^{q-1}(b_n-b_{n+1})\\
+&=M|b_q|+M|b_p|+M(b_p-b_q)\\
+&=2M|b_p|
+\end{aligned}
+$$
+
+Then we let $\epsilon >0$ be given. Since $b_n\to 0$, there exists $N\in \mathbb{N}$ such that for all $n\geq N$, $|b_n|<\frac{\epsilon}{2M}$.
+
+If $q\geq p\geq N$, then
+
+$$
+\left|\sum_{n=p}^q a_nb_n\right|\leq 2M|b_p|<\epsilon
+$$
+
+So $\sum a_nb_n$ converges.
+
+EOP
+
+#### Theorem 3.43 (Alternating series test)
+
+Let $(b_n)^\infty_{n=1}$ be a sequence such that:
+
+(a) $b_1\geq b_2\geq b_3\geq \cdots$ (non-increasing)
+(b) $\lim_{n\to\infty}b_n=0$
+
+Then $\sum_{n=1}^\infty (-1)^{n+1}b_n$ converges.
+
+Proof:
+
+Let $a_n=(-1)^{n+1}$
+
+$A_n=\sum_{k=1}^n a_k=1$ if $n$ is odd, $0$ if $n$ is even.
+
+So $|A_n|\leq 1$ for all $n\in \mathbb{N}$.
+
+By Theorem 3.42, $\sum_{n=1}^\infty a_n b_n$ converges.
+
+EOP
+
+Example:
+
+Consider the power series $\sum_{n=0}^\infty \frac{z^n}{n}$.
+
+The radius of convergence is $1$.
+
+We claim that the series converges for all $z\in \mathbb{C}$ with $|z|=1$ and $z\neq 1$.
+
+#### Theorem 3.44 Abel's test
+
+Let $(b_n)^\infty_{n=0}$ be a sequence such that:
+
+(a) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)
+(b) $\lim_{n\to\infty}b_n=0$
+
+Then if $|z|=1$ and $z\neq 1$, $\sum_{n=0}^\infty b_nz^n$ converges.
+
+Proof:
+
+Fix $z\in \mathbb{C}$ with $|z|=1$ and $z\neq 1$.
+
+Let $a_n=z^n$.
+
+Then $A_n=\sum_{k=0}^n z^k=\frac{1-z^{n+1}}{1-z}$._
+
+$|A_n|\leq \frac{|1-z^{n+1}|}{|1-z|}$ for all $n\in \mathbb{N}$.
+
+By triangle inequality, $|1-z^{n+1}|\leq |1|+|z^{n+1}|=1+|z^{n+1}|$.
+
+And since $|z|=1$, $|z^{n+1}|=|z|^{n+1}=1$.
+
+So $|1-z^{n+1}|\leq 2$.
+
+So $|A_n|\leq \frac{2}{|1-z|}$ for all $n\in \mathbb{N}$.
+
+By Dirichlet's test, $\sum_{n=0}^\infty b_nz^n$ 
+
+EOP
+
+### Absolute convergence
+
+The series $\sum_{n=0}^\infty a_n$ is said to **converge absolutely** if $\sum_{n=0}^\infty |a_n|$ converges.
+
+If $\sum_{n=0}^\infty a_n$ converges but does not converge absolutely, then $\sum_{n=0}^\infty a_n$ is said to **converge conditionally**.
+
+_Absolute convergence are nice but conditionally convergent series are not._
+
+#### Theorem 3.45 (Absolute convergence)
+
+If $\sum_{n=0}^\infty a_n$ converges absolutely, then $\sum_{n=0}^\infty a_n$ converges.
+
+Proof:
+
+Use comparison test.
+
+$$
+\sum_{n=0}^\infty |a_n|\geq \sum_{n=0}^\infty a_n
+$$
+
+EOP
+
+Rearrangement of series:
+
+Let $f:\mathbb{N}\to \mathbb{N}$ be a bijection.
+
+If $\sum_{n=0}^\infty a_n$ is a sequence and $b_n=a_{f(n)}$, then $(b_n)^\infty_{n=0}$ is a rearrangement of $(a_n)^\infty_{n=0}$.
+
+If $\sum_{n=0}^\infty a_n$ converges absolutely, then any rearrangement of $\sum_{n=0}^\infty a_n$ converges to the same sum.
+
+Example:
+
+$a_n=\frac{(-1)^{n+1}}{n}$. $b_n=a_{f(n)}$.
+
+|n|1|2|3|4|5|6|7|8|9|
+|:-:|:-:|:-:|:-:|:-:|:-:|:-:|:-:|:-:|:-:|
+|$f(n)$|1|2|4|3|6|8|5|10|12|
+|$b_n$|1|-1/2|-1/4|1/3|-1/6|-1/8|1/5|-1/10|-1/12|
+
+$$
+\sum_{n=1}^\infty a_n=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\log 2
+$$
+
+$$
+\begin{aligned}
+\sum_{n=1}^\infty b_n&=1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10}-\frac{1}{12}+\cdots\\
+&=\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac{1}{5}-\frac{1}{10}\right)-\frac{1}{12}+\cdots\\
+&=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots\\
+&=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)\\
+&=\frac{1}{2}\log 2
+\end{aligned}
+$$
+
+You cannot always rearrange series.
+
+But, if $\sum_{n=0}^\infty a_n$ converges absolutely, then you can rearrange the series.
+
+#### Theorem 3.55
+
+Let $(a_n)^\infty_{n=0}$ be a sequence in $\mathbb{C}$ such that $\sum_{n=0}^\infty |a_n|$ converges absolutely.
+
+Then any rearrangement of $\sum_{n=0}^\infty a_n$ converges absolutely to the same sum.
+
+$$
+\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty a_{f(n)}
+$$
+
+Idea of proof:
+
+Let $f:\mathbb{N}\to \mathbb{N}$ be a bijection.
+
+and let $b_n=a_{f(n)}$.
+
+Let $s_n=\sum_{k=0}^n a_k,t_n=\sum_{k=0}^n b_k=\sum_{k=0}^n a_{f(k)}$.
+
+$I_n=\{1,2,\cdots,n\}$.
+
+$J_n=\{f(1),f(2),\cdots,f(n)\}$.
+
+$$
+\begin{aligned}
+s_n-t_n&=\sum_{k=0}^n a_k-\sum_{k=0}^n a_{f(k)}\\
+&=\sum_{k\in I_n} a_k-\sum_{k\in J_n} a_k\\
+&= \sum_{k\in I_n\setminus J_n} a_k+\sum_{k\in J_n\setminus I_n} a_k\\
+&\leq \sum_{k\in I_n\setminus J_n} |a_k|+\sum_{k\in J_n\setminus I_n} |a_k|
+\end{aligned}
+$$
+
+Key observation:
+
+For every $n\in \mathbb{N}$, there exists a $p$ such that $\{1,2,\cdots,n\}\subset I_n\cap J_n$.
+
+Then $|s_n-t_n|\leq \sum_{k=N+1}^\infty |a_k|$.
+
+EOP

pages/Math4111/Math4111_L22.md

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+# Lecture 22

pages/Math4111/Math4111_L23.md

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+# Lecture 23

pages/Math4111/Math4111_L24.md

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+# Lecture 24

pages/Math4111/Math4111_L25.md

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+# Lecture 25

pages/Math4111/Math4111_L26.md

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+# Lecture 26

pages/Math4111/Math4111_L27.md

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+# Lecture 27

pages/Math4111/Math4111_L28.md

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+# Lecture 28

pages/Math4111/Math4111_L3.md

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+# Lecture 3
+
+## Review
+
+Let $S=\mathbb{Z}$.
+
+1. Let $E=\{x\in S:x>0,x^2<5\}$. What are $sup\ E$ and $inf\ E$?
+  
+    $sup\ E=2,inf\ E=1$
+
+2. Can you find a subset $E\subset S$ which is bounded above but not bounded below?
+
+    $E=\{x\in S:x<0\}$
+
+3. Does $S$ have the least upper bound property?
+
+    Yes, $\forall E\subset S$ that tis non-empty and bounded above, $\exist Sup E\in S$.
+
+4. Does $S$ have the greatest lower bound property?
+
+    Yes, $\forall E\subset S$ that tis non-empty and bounded below, $\exist Inf E\in S$.
+
+## Continue
+
+### LUBP
+
+Proof that $LUBP\implies GLBP$.
+
+Let $S$ be an ordered set with LUBP. Let B<S be non-empty and bounded below.
+
+Let $L=y\in S:y$ is a lower bound of B$\}$. From the picture, we expect $sup\ L=inf\ B$ First we'll show $sup\ L$ exists.
+
+1. To show $L\neq \phi$.
+
+    $B$ is bounded below $\implies L\neq\phi$
+2. To show $L$ id bounded above.
+
+    $B$ is not empty $\implies \exists x\in B\implies x$ is a upper bound of $L$.
+
+3. Since $S$ has the least upper bound property, $sup L$ exists (in $S$).
+
+Let's say $\alpha=sup\ L$. We claim that $\alpha=inf\ B$. We need to show $2$ things.
+
+1. To show $\alpha$ is a lower bound of $B$, $\forall \gamma\in B,\alpha\leq \gamma$.
+
+    Let $\gamma\in B$, then $\gamma$ is an upper bound of $L$.
+
+    Since $\alpha$ is the least upper bound of $L$, $\alpha\leq \gamma$.
+
+2. To show $\alpha$ is the greatest lower bound of $B$, $\forall \beta>\alpha,\beta$ is not a lower bound of $B$.
+
+    Let $\beta>\alpha$. Since $\alpha$ is an upper bound of $L$, $\beta\notin L$.
+
+    By definition of $L$, $\beta$ is not a lower bound of $B$.
+
+Thus $\alpha=inf\ B$
+
+### Field
+
+|                | addition                                                    | multiplication                                                 |
+| -------------- | ----------------------------------------------------------- | -------------------------------------------------------------- |
+| closure        | $\checkmark$                                                | $\checkmark$                                                   |
+| commutativity  | $\checkmark$                                                | $\checkmark$                                                   |
+| associativity  | $\checkmark$                                                | $\checkmark$                                                   |
+| identity       | $\checkmark$ (denoted $0$)                                  | $\checkmark$ (denoted $1$)                                     |
+| inverses       | $\checkmark$ (denoted $-x$)                                 | $\checkmark$ (exists when $x\neq 0$ denoted $1/x$ or $x^{-1}$) |
+| distributivity | $\checkmark$ (distributive of multiplication over addition) ||
+
+Examples: $\mathbb{Q},\mathbb{R},\mathbb{C}$
+
+Non-examples: $\mathbb{N}$ fails A4,A5,M5, $\mathbb{Z}$ fails M5
+
+Another example of field: $\mathbb{Z}/5\mathbb{Z}=\{1,2,3,4,5\}$, $\forall a,b\in \mathbb{Z}/5\mathbb{Z}$, $a+b=(a+b)\mod 5$, $a\cdot b=(a\cdot b)\mod 5$
+
+Some properties of fields: see Proposition 1.14,1.15,1.16
+
+Remark:
+
+1. It's more helpful if you try to prove these yourselves. The proofs are "straightforward".
+2. For this course, it's not important to remember which properties are axioms, etc.
+
+Example of proof: 
+
+#### 1.14(a) $x+y=x+z\implies y=z$
+
+Proof: 
+
+$x+y=x+z$,
+
+$(-x)+(x+y)=(-x)+(x+z)$,
+
+by A3, $(-x+x)+(y)=(-x+x)+(z)$,
+
+$0+y=0+z$,
+
+$y=z$.
+
+Chain of equalities.
+
+#### 1.16(a) $\forall x\in \mathbb{F}, 0x=0$
+
+1. A4, where 0 is defined.
+2. Since $0$ is defined in the addition, identity. The proposition says something about multiplication by 0. The only proposition that relates the addition and multiplication is Distributive law.
+
+$0x=(0+0)x=0x+0x$, cancel $0x$ on both side we have $0x=0$.
+
+### Ordered Field (1.17)
+
+An _ordered field_ is a _field_ $F$ which is also an _ordered set_, such that
+
+1. $x+y<x+z$ if $x,y,\in F$ and $y<z$,
+2. $xy>0$ if $x\in F,y\in F,x>0$ and $y>0$.
+
+#### Prop 1.18
+
+If $x>0$ and $y<z$, then $xy<yz$.
+
+Proof: $y<z\implies 0<z-y$, $x(z-y)>0\implies xz>xy$
+
+We define $\mathbb{R}$ to be the unique ordered field with $LUBP$. (The existence and uniqueness are discussed in the appendix of this chapter).
+
+#### Theorem 1.20
+
+1. (Archimedean property) If $x,y\in \mathbb{R}$ and $x>0$, then $\exists n\in \mathbb{N}$ such that $nx>y$.
+2. ($\mathbb{Q}$ is dense in $\mathbb{R}$) If $x,y\in \mathbb{R}$ and $x<y$, then $\exists p\in \mathbb{Q}$$ such that $x<p<y$.

pages/Math4111/Math4111_L4.md

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+# Lecture 4
+
+## Review
+
+1. Let $F$ be a field. Let $a,b,c,...,z\in F$ . Using he field axioms, simplify 
+
+    $$
+    (x-a)(x-b)(x-c)...(x-z)
+    $$
+
+    $x\in F$, it must be at least one $0$ in the product...
+
+2. Suppose $A,B\subset\mathbb{R}$. Suppose $A$ and $B$ are nonempty and bounded above,$A\subset B$. WHat can you say about $sup\ A$ and $sup\ B$? Please justify.
+
+    $$
+    \forall x\in A, x\in B.  sup\ A\leq sup\ B
+    $$
+
+    Any UB of $B$ is also an UB of $A$.
+
+    $sup\ B$ is an UB of $B$ by def
+    $sup\ B$ is an UB of $A$
+
+## Continue
+
+### Archimedean property
+
+(Archimedean property) If $x,y\in \mathbb{R}$ and $x>0$, then $\exists n\in \mathbb{N}$ such that $nx>y$.
+
+Proof
+
+Suppose the property is false, then $\exist x,y\in \mathbb{R}$ with $x>0$ such that $\forall v\in \mathbb{N}$, nx\leq y$
+
+Let $A=\{nx:n\in\mathbb{N}\}$. Then $A\neq\phi$ (Since $x\in A$) and $A$ is bounded above by $y$. Since $\mathbb{R}$ has LUBP, $sup\ A$ exists. Let $\alpha=sup\ A$.
+
+$x>0\implies \alpha-x<\alpha$, $\alpha-x$ is not an upper bound of $A$. (Since $\alpha$ is the LUB of $A$) $\implies \exist m\in \mathbb{N}$ such that $mx>\alpha-x$ by definition of $A$.
+
+This implies $(m+1)x>\alpha$
+
+Since $(m+1)x\in \alpha$, this contradicts the fact that $\alpha$ is an upper bound of $A$.
+
+EOP
+
+### $\mathbb{Q}$ is dense in $\mathbb{R}$
+
+$\mathbb{Q}$ is dense in $\mathbb{R}$) If $x,y\in \mathbb{R}$ and $x<y$, then $\exists p\in \mathbb{Q}$$ such that $x<p<y$.
+
+Some thoughts:
+
+$$
+x<\frac{m}{n}<y\iff nx<m<ny
+$$
+
+Proof:
+
+Let $x,y\in\mathbb{R}$, with $x<y$. We'll find $n\in \mathbb{N},\mathbb{m}\in \mathbb{Z}$ such that $nx<m<ny$.
+
+By Archimedean property, $\exist n\in \mathbb{N}$ such that $n(y-x)>1$, and $\exist m_1\in \mathbb{N}$ such that $m_1\cdot 1>nx$, $\exist m_2\in \mathbb{N}$ such that $m_2\cdot 1>-nx$.
+
+So $-m_2<nx<m_1$. Thus $\exist m\in \mathbb{Z}$ such that $m-1\leq nx<m$ (Here we use a property of $\mathbb{Z}$) We have $ny>1+nx\geq 1+(m-1)=m$
+
+EOP
+
+### $\sqrt{2}\in \mathbb{R}$, $(\sqrt[n]{x}\in\mathbb{R})$
+
+Notation $\mathbb{R}_{>0}$= the set of positive numbers.
+
+#### Theorem 1.21
+
+$\forall x\in \mathbb{R}_{>0},\forall n\in \mathbb{N},\exist$ unique $y\in \mathbb{R}_{>0}$ such that $y^n=x$.
+
+(Because of this Theorem we can define $x^{1/x}=y$ and $\sqrt{x}=y$)
+
+Proof:
+
+We cna assume $n\geq 2$ (For $n=1,y=x$)
+
+Step 1 (uniqueness): If $0<y_1<y_2$, then $y_1^n<y_2^n$ (by properties of ordered field)
+
+Step 2 (existence): Let $E=\{t\in \mathbb{R}_{>0}: t^n<x\}$ We want to let $y=sup\ E$, but to do this we need to check 2 things.
+
+1. To show $E\neq \phi$:
+
+    If $x\geq 1$, then $1/2\in E$.
+
+    If $x<1$, then $x\in E$.
+
+2. To show $E$ is bounded above. We need to find an upper bound.
+
+    If $x\geq 1$, then $x\in E$
+
+    If $x<1$, then $1 \in E$.
+
+So we can let $y=sup\ E$
+
+Step 2b ($y^n\geq x$) Suppose for contradiction $y^n<x$.
+
+Thoughts: If we can find $h>0$ such that $(y+h)^n<x$, then $y+h\in E$. This would contradict the facts $y$ is an upper bound of $E$.
+
+$$
+(y+h)^n=y^n+ny^{n-1}h+{more\ terms}
+$$
+
+We want $ny^{n-1}h+{more\ terms}<x-y^n$
+
+Observe: If $0<a<b$, then
+
+$$
+\frac{b^n-a^n}{b-a}=b^{n-1}+b^{n-2}a+...+a^{n-1}\leq b^{n-1}+b^{n-2}b+...+b^{n-1}=nb^{n-1}
+$$

pages/Math4111/Math4111_L5.md

@@ -0,0 +1,174 @@
+# Lecture 5
+
+## Review
+
+In each case, determine (with justification) whether the claim or its negation is the true statement.
+
+(a) For all real numbers satisfying $a<b$, there exists an $n\in \mathbb{N}$ such that $a+1/n<b$.
+
+negation: $\exists a<b$, $\forall n\in \mathbb{N}$ such that $a+1/n\geq b$.
+
+By Archimedean property, the statement is true.
+
+(b) There exists a real number $x>0$ such that $x<1/n$ for all $n\in \mathbb{N}$.
+
+The statement is ambiguous because we can arrange the statement in two ways.
+
+$\exists x\in \mathbb{R}_{>0}$ such that $\forall n\in \mathbb{N},x\leq \frac{1}{n}$
+
+negation: $\forall x\in \mathbb{R}_{>0}$, $\exists n\in \mathbb{N}$, such that $x\leq \frac{1}{n}$.
+
+The statement is true, let $x=\frac{1}{n+1}$.
+
+## New Materials
+
+### Continue on the theorem
+
+#### Theorem 1.21
+
+$\forall x\in \mathbb{R}_{>0},\forall n\in \mathbb{N},\exist$ unique $y\in \mathbb{R}_{>0}$ such that $y^n=x$.
+
+(Because of this Theorem we can define $x^{1/x}=y$ and $\sqrt{x}=y$)
+
+Proof:
+
+We cna assume $n\geq 2$ (For $n=1,y=x$)
+
+Step 1 (uniqueness): If $0<y_1<y_2$, then $y_1^n<y_2^n$ (by properties of ordered field)
+
+Step 2 (existence): Let $E=\{t\in \mathbb{R}_{>0}: t^n<x\}$ We want to let $y=sup\ E$, but to do this we need to check 2 things.
+
+1. To show $E\neq \phi$:
+
+    If $x\geq 1$, then $1/2\in E$.
+
+    If $x<1$, then $x\in E$.
+
+2. To show $E$ is bounded above. We need to find an upper bound.
+
+    If $x\geq 1$, then $x\in E$
+
+    If $x<1$, then $1 \in E$.
+
+So we can let $y=sup\ E$
+
+Step 2b ($y^n\geq x$) Suppose for contradiction $y^n<x$.
+
+Thoughts: If we can find $h>0$ such that $(y+h)^n<x$, then $y+h\in E$. This would contradict the facts $y$ is an upper bound of $E$.
+
+$$
+(y+h)^n=y^n+ny^{n-1}h+{more\ terms}
+$$
+
+We want $ny^{n-1}h+{more\ terms}<x-y^n$
+
+Observe: If $0<a<b$, then
+
+$$
+\frac{b^n-a^n}{b-a}=b^{n-1}+b^{n-2}a+...+a^{n-1}\leq b^{n-1}+b^{n-2}b+...+b^{n-1}=nb^{n-1}
+$$
+
+The fact tells us $(y+h)^n-y^n\leq n(y+h)^{n-1}h$.
+
+And we want $n(y+h)^{n-1} h+y^n<x$.
+
+So want $h<\frac{x-y^h}{n(y+h)^{n-1}}$.
+
+To achieve this, choosey any $h>0$ satisfying $h<1$ and $h<\frac{x-y^h}{n(y+h)^{n-1}}$
+
+[For actual proof, see the text.]
+
+Step 2c showing ($y^n\leq x$)
+
+Suppose for contradiction $y^n>x$
+
+Thoughts: Find $k>0$ such that $(y-k)^n>x$.
+
+Then $y-k$ is an upper bound for $E$, which contradicts the fact that $y$ is the least upper bound of $E$.
+
+$y^n-(y-k)^n\leq ny^{n-1}k$.
+
+We want $y^n-ny^{n-1}k\geq x$.
+
+So want $k\leq \frac{y^n-x}{ny^{n-1}}$
+
+[For actual proof, see the text.]
+
+### Complex numbers
+
+1. $=\{a+bi:a,b\in \mathbb{R}\}$.
+
+Conjugate: $z=a+bi,\bar{z}=a-bi$.
+
+#### Theorem 1.31 (see text)
+
+Pure computational proof: boring...
+
+$z\bar{z}=a^2-(bi)^2=a^2+b^2$
+
+You can also use vector sum for representing operation in complex numbers.
+
+#### Theorem 1.33 (see text)
+
+More computation and still, boring...
+
+some fun theorems:
+
+- $|Re\ z|\leq |z|$ (equal when no imaginary parts)
+- $|z+w|\leq |z|+|w|$ (equal when both $z,w$ have no imaginary parts) (Triangle inequality)
+
+Proof for $|z+w|\leq |z|+|w|$:
+
+$$
+|z+w|^2=(z+w)(\overline{z+w})=(z+w)(\bar{z}+\bar{w})=|z|^2+|w|^2+z\bar{w}+\bar{z}w
+$$
+
+Since 
+
+$$
+z\bar{w}+\bar{z}w\leq 2Re(z\bar{w})
+$$
+
+$$
+(z+w)(\bar{z}+\bar{w})=|z|^2+|w|^2+z\bar{w}+\bar{z}w\leq |z|^2+|w|^2+2|z||w|\leq |z|+|w|
+$$
+
+#### Theorem 1.35 Cauchy-Schwarz inequality
+
+If $\vec{a},\vec{b}\in \mathbb{C}^n$, then
+
+$$
+|\vec{a}\vec{b}|^2\leq (\vec{a}\vec{a})(\vec{b}\vec{b})
+$$
+
+> Remark: The proof is very tricky.
+
+To help us motivate the proof in text, let's consider the special case of real numbers.
+
+$$
+(\sum a_j b_j)^2=(\sum a_j^2)(\sum b_j^2)
+$$
+
+Proof for real numbers:
+
+Let $A=\sum a_j^2,B=\sum b_j^2, C=\sum a_j b_j$, want to show $C^2\leq AB$
+
+Note: if $B=0$, then $b_1=b_2=...=0$, so $C=0$ and we are done, so we may assume $B\neq 0$ so $B>0$.
+
+Clever step: For any $t\in \mathbb{R}$,
+
+$$
+0\leq \sum (a_j-t b_j)^2=\sum (a_j^2-2ta_jb_j+t^2b_i^2)=A-2tC+t^2B
+$$
+
+let $t=C/B$ to get $0\leq A-2(C/B)C+(C/B)^2B=A-\frac{C^2}{B}$
+
+to generalize this to $\mathbb{C}$, $A=\sum |a_j|^2,B=\sum |b_j|^2,C=\sum |a_j \bar{b_j}|$.
+
+### Euclidean spaces
+
+Nothing much to say. lol.
+
+Normal dot product as inner product.
+
+read text... Theorem 1.37

pages/Math4111/Math4111_L6.md

@@ -0,0 +1,149 @@
+# Lecture 6
+
+## Review
+
+Let $A$ and $B$ be subset of $\mathbb{R}$, and consider the function $f:A \to B$ defined by $f(x)=cos(x)$. Find choices for the domain $A$ and co-domain $B$ which make $f$...
+
+(a) neither injective nor surjective.
+
+$A=\mathbb{R},B=\mathbb{R}$
+
+(b) surjective but not injective.
+
+$A=[0,\frac{3\pi}{2}],B=[-1,1]$
+
+(c) injective but not surjective.
+
+$A=[0,\pi],B=[-1,1.1]$
+
+(d) bijective.
+
+$A=(0,\pi),B=(-1,1)$
+
+> injective: y don't repeat, surjective: there exists x for each y
+
+## New Materials (Chapter 2: Basic topology (4171). Finite, countable, and uncountable sets)
+
+### Functions
+
+#### Definition 2.1/2.2
+
+"$f:A\to B$" means "$f$ is a function from $A$ to $B$"
+
+$A$:"domain", and $B$: "co-domain".
+
+If $S\subset A$, the **range** of $S$ under $f$ is $f(S)=\{f(x):s\in S\}$
+
+The "**image**" of $f$ is $f(A)$.
+
+If $T\subset B$, **the inverse image** (pre-image) of $T$ under $f$ is 
+
+$$
+f^{-1}(T)=\{x\in A: f(x)\in T\}
+$$
+
+* $f$ is injective or one-to-one if $\forall x_1,x_2\in A$ such that $f(x_1)=f(x_2)$, then $x_1=x_2$. ($f(x_1)=f(x_2)\implies x_1=x_2 \equiv x_1\neq x_2\implies f(x_1)\neq f(x_2)$)
+
+* $f$ is surjective or onto if $\forall y\in B, \exists x\in A$ such that $f(x)=y$. ($f(A)=B$)
+
+* $f$ is bijective if it's both injective and surjective.
+
+#### Definition 2.3
+
+If $\exists$ bijection $f:A\to B$, we say:
+
+* $A$ and $B$ can be put into 1-1 correspondence
+* $A$ and $B$ b oth have the same cardinality
+* $A$ and $B$ are equivalent $A\sim B$
+
+### Cardinality
+
+#### Definition 2.4
+
+(a) $A$ is finite if $A\neq \phi$ or $\exists n\in \mathbb{N}$ such that $A\sim \{1,2,...,n\}$
+
+(b) $A$ is **infinite** if it's not finite
+
+(c) $A$ is **countable** if $A\sim \mathbb{N}$
+
+(d) $A$ is **uncountable** if $A$ is neither finite nor countable
+
+(e) $A$ is **at most countable** if it's finite or countable
+
+> Note in some other books call (c) **countable infinite**, and (e) for **countable**
+
+#### Definition 2.7
+
+A **sequence** in $A$ is a function $f:\mathbb{N}\to\mathbb{A}$
+
+Note: By conversion, instead of $f(1),...f(n)$, we usually write $x_1,x_2,...,x_3$ and we say $\{x_n\}_{n=1}^{\infty}$ is a sequence.
+
+#### Theorem 2.8
+
+Every infinite subset of countable set $A$ is countable.
+
+Ideas of proof: if $A$ is countable, so we can list its element in a sequence. and we iterate $E\subset A$ to create a new order function by deleting element $\notin E$
+
+#### Definition 2.9 (arbitrary unions and intersections)
+
+Let $A$ be a set (called the "index set"). For each $\alpha\in A$, let $E_{\alpha}$ be a set.
+
+Union: $\bigcup_{\alpha \in A}E_{\alpha}=\{x:\exists \alpha \in A$ such that $x\in E_{\alpha}\}$
+
+Intersection: $\bigcap_{\alpha \in A}E_{\alpha}=\{x:\exists \alpha \in A$ such that $x\in E_{\alpha}\}$
+
+Special notation for special cases:
+
+$\bigcup^{n}_{m=1}E_m$ and $E_1\cup E_2\cup ...\cup E_n$ are by definition $\bigcup_{\alpha \in \{1,..,n\}}E_{\alpha}$
+
+and $\bigcup^{\infty}_{m=1}E_m$ and $E_1\cup E_2\cup ...\cup E_n$ are by definition $\bigcup_{\alpha \in \mathbb{N}}E_{\alpha}$
+
+> Note: Despite the $\infty$ symbol this def makes no references to limits, different from infinite sums.
+
+### Countability
+
+#### Theorem 2.12
+
+"A countable union of countable sets is countable".
+
+Let $\{E_n\},n=1,2,3,...$ be a sequence of countable sets, and put 
+
+$$
+S=\bigcup^{\infty}_{n=1} E_n
+$$
+
+The $S$ is countable.
+
+Proof by infinite grid, and form a new sequence and remove duplicates.
+
+#### Corollary
+
+An at most countable union of at most countable sets is at most countable.
+
+#### Theorem 2.13
+
+$A$ is countable, $n\in \mathbb{N}$,
+
+$\implies A^n=\{(a_{1},...,a_{n}):a_1\in A, a_n\in A\}$, is countable.
+
+Proof: Induct on $n$, 
+
+Base case $n=1$,
+
+True by assumptions
+
+Induction step: suppose $A^{n-1}$ is countable. Note $A^n=\{(b,a):b\in A^{n-1},a\in A\}=\bigcup_{b\in A^{n-1}\{(b,a),a\in A\}}$.
+
+Since $b$ is fixed, so this is in 1-1 correspondence with $A$, so it's countable by Theorem 2.12.
+
+#### Theorem 2.14
+
+Let $A$ be the set of all sequences for 0s and 1s. Then $A$ is uncountable.
+
+Proof: Let $E\subset A$ be a countable subset. We'll show $A\backslash E\neq \phi$ (i.e.$\exists t\in A$ such that $t\notin E$)
+
+$E$ is countable so we can list it's elements $S_1,S_2,S_3,...$.
+
+Then we define a new sequence $t$ which differs from $S_1$'s first bit and $S_2$'s second bit,...
+
+This is called Cantor's diagonal argument.

pages/Math4111/Math4111_L7.md

@@ -0,0 +1,103 @@
+# Lecture 7
+
+## Review
+
+Let $S=\{(x,y,z)\in \mathbb{R}^3:x=1,y=4\}=\{(1,4,z):z\in\mathbb{R}\}$
+
+1. How can we describe the set $S$ geometrically in three-dimensional space?
+
+   Just a line
+2. Show that $S$ and $\mathbb{R}$ are in one-to-one correspondence.
+
+    We can find a bijective function $f:S\to \mathbb{R}$
+3. Show that for any $(a,b)\in\mathbb{Z}^2$, the set $\{(a,b,z):z\in\mathbb{Z}\}$ is in one-to-one correspondence with $\mathbb{Z}$
+
+    Use **Theorem 2.13** $A$ is countable, $n\in \mathbb{N} \implies A^n=\{(a_{1},...,a_{n}):a_1\in A, a_n\in A\}$, is countable.
+
+## New materials
+
+### Metric spaces
+
+#### Definition 2.15
+
+Let $X$ be a set. A function $d:X\times X\to \mathbb{R}$ is called a distance function or a metric if it satisfies:
+
+1. Positivity: $\forall p,q\in X,p\neq q\implies d(p,q)>0$, and $\forall p\in X,d(p,p)=0$.
+2. Symmetry: $\forall p,q\in X, d(p,q)=d(q,p)$.
+3. Triangle inequality: $\forall p,q,r\in X$, $d(p,q)\leq d(p,r)+d(r,q)$
+
+We say **$(X,d)$ is a metric space**. If $d$ is understood, $X$ is a metric space.
+
+Examples:
+
+The most important example:
+
+$X\subset \mathbb{R}^k(k\geq 1)$
+
+$d(x,y)=|x-y|$
+
+And other examples: function spaces...
+
+An example from graph theory (not needed for this class):
+
+$d(p,q)$ can be defined by the shortest path fro $p$ to $q$.
+
+#### Definition 2.17
+
+By the *segment* $(a,b)$ we mean the set of all real numbers $x$ such that $a<x<b$.
+
+*segment excludes the bound* $(a,b)$
+
+By the *interval* $[a,b]$ we mean the set of all real numbers $x$ such that $a\leq x\leq b$
+
+* interval include the bound* $[a,b]$
+
+Convex: $E\subset \mathbb{R}^k$ is convex if $\forall x,y\in E,\{\lambda x+(1-\lambda)y:\lambda\in (0,1)\}\subset E$
+
+### Open sets
+
+#### Definition 2.18
+
+Let $(X,d)$ be a metric space.
+
+1. $p\in X,r>0$. The r-neighborhood of $p$ is $B_r(p)=N_r(o)=\{q\in X: d(p,q)<r\}$ (a ball in metric space)
+2. $E\subset X$, $p\in X$. We say $p$ is an interior point of $E$ if $\exists r>0$ such that $B_r(p)\subset E$. Notation $E^{\circ}=$set of interior points of $E$
+3. $E\subset X$, we say $E$ is **open** if $E\subset E^{\circ}$, i.e. $\forall p\in E, \exists r>0$ such that $B_r(p)\subset E$.
+
+*Note: is follows from definitions that $E^{\circ}\subset E$ is always true.*
+
+Example:
+
+$X=\mathbb{R}^2$($d$ be the euclidean distance) $E=[0,1)\times [0,1)$.
+
+$E^{\circ}=(0,1)\times (0,1)$
+
+So $E=(0,1)\times (0,1)$ is a open set.
+
+#### Theorem 2.19
+
+Let $(X,d)$ be  a metric space, $\forall p\in X,\forall r>0$, $B_r(p)$ is an open set.
+
+*every ball is an open set*
+
+Proof: Let $q\in B_r(p)$.
+
+Let $h=r-d(p,q)$.
+
+Since $q\in B_r(p),h>0$. We claim that $B_h(q)$. Then $d(q,s)<h$, so $d(p,s)\leq d(p,q)+d(q,s)<d(p,q)+h=r$. (using triangle inequality) So $S\in B_r(p)$.
+
+### Closed sets
+
+1. $E\subset X,p\in X$. We say $p$ is a limit point of $E$ if $\forall r>0, (B_r(p)\cap E)\backslash {p}\neq \phi$.
+
+    Let $E'$ be the set of limit points of $E$.
+
+2. $E$ is closed if $E'\subset E$
+
+Example: $X=\mathbb{R}^2$, $E=[0,1)\times [0,1)$.
+
+$(1,1)$ is a limit point.
+
+$X=\mathbb{R},E=\{\frac{1}{n},n\in \mathbb{N}\}$
+
+$0$ is the only limit point. $E'=\{0\}$

pages/Math4111/Math4111_L8.md

@@ -0,0 +1,150 @@
+# Lecture 8
+
+## Review
+
+Let $(X,d)$ be a metric space. Recall that $B_r(x)=\{z\in X:d(x,z)<r\}$.
+
+Let $x,y\in X$ and let $r=\frac{1}{2}d(x,y)$. What do you think is true about $B_r(x)\cap B_r(y)$? Can you prove it?
+
+It should be empty. Proof any point cannot be in two balls at the same time. (By triangle inequality or contradiction)
+
+### Metric space defs
+
+1. $p\in X,r>0$, $B_r(p)=\{q\in X:d(p,q)<0\}$, also called **neighborhood**.
+2. $p$ is a **limit point** of $E(p\in E')$ if $\forall r>0$, $(B_s(p)\cap E)\backslash \{p\}\neq \phi$
+3. If $p\in E$ and $p$ is not a limit point of $E$, then $p$ is called an **isolated point** of $E$.
+4. $E$ is **closed** if $E'\subset E$
+5. $p$ is a **interior point** of $E(p\in E^{\circ})$ if $\exists r>0$ such that $B_r(p)\subset E$.
+
+## New materials
+
+### Metric space
+
+#### Theorem 2.20
+
+$p\in E'\implies \forall r>0,B_r(p)\cap E$ is infinite.
+
+Proof:
+
+We will prove the contrapositive.
+
+want to prove $\exists r>0$ such that $B_r(p)\cap E$ is finite $\implies p\notin E'$ ($\exists s>0$ such that $(B_s(p)\cap E)\backslash \{p\}=\phi$)
+
+Suppose $\exists r>0$ such that $B_r(p)\cap E$ is finite
+
+let $B_s(p)\cap E)\backslash \{p\}={q_1,...,q_n}$ 
+
+- If $n=0$, then $B_s(p)\cap E)\backslash \{p\}=\phi$, so $p\in E'$
+- If $n\geq 1$, then let $s=min\{d(p,q_m):1\leq m\leq n\}$
+
+    Each $d(p,q_m)$ is positive and the set is finite, so $s>0$.
+
+Then $(B_s(p)\cap E)\backslash \{p\}=\phi$, so $p\notin E$
+
+EOP
+
+#### Theorem 2.22 De Morgan's law
+
+$$
+\left(\bigcup_a E_a\right)^c=\bigcap_a(E^c_a)
+$$
+
+$E^c=X\backslash E$
+
+Proof:
+
+$x\in \cup_{a\in A} E_x\iff \exists a\in A$ such that $x\in E_a$
+
+So $x\in \left(\bigcup_a E_a\right)^c\iff \forall a\in A, x\notin  E_a\iff \forall a\in A,x\in E_a^c\iff \bigcap_a(E^c_a)$
+
+#### Theorem 2.23
+
+$E$ is open $\iff$ $E^c$ is closed.
+
+> Warning: $E$ is open $\cancel{\iff}$ $E$ is closed.
+> $E$ is closed $\cancel{\iff}$ $E$ is open.
+>
+> Example:  
+>$\phi$, $\R$ is both open and closed. "clopen set"  
+>$[0,1)$ is not open and not closed. bad...
+
+Proof:
+
+$\impliedby$ Suppose $E^c$ is closed. Let $x\in E$, so $x\notin E^c$
+
+$E^c$ is closed and $x\notin E^c\implies x\notin (E^c)'\implies \exists r >0$ such that $(B_r(x)\cap E^c)\backslash \{x\}=\phi$
+
+So $\phi=(B_r(x)\cap E^c)\backslash \{x\}=B_r(x)\cap E^c$
+
+So $B_r(x)\in E$
+
+$\implies$
+
+Suppose $E$ is open
+
+$$
+\begin{aligned}
+    x\in (E^c)'&\implies \forall r>0, (B_r(x)\cap E^c)\backslash \{x\}\neq \phi\\
+    &\implies \forall r>0, (B_r(x)\cap E^c)\neq \phi\\
+    &\implies \forall r>0, B-r(x)\notin E\\
+    &\implies x\notin  E^{\circ}\\
+    &\implies x\notin E\\
+    &\implies x\in E^c
+\end{aligned}
+$$
+
+So $(E^c)'\subset E^c$
+
+EOP
+
+#### Theorem 2.24
+
+##### An arbitrary union of open sets is open
+
+Proof:
+
+Suppose $\forall \alpha, G_\alpha$ is open. Let $x\in \bigcup _{\alpha} G_\alpha$. Then $\exists \alpha_0$ such that $x\in G_{\alpha_0}$. Since $G_{\alpha_0}$ is open, $\exists r>0$ such that $B_r(x)\subset G_{\alpha_0}$ Then $B_r(x)\subset G_{\alpha_0}\subset \bigcup_{\alpha} G_\alpha$
+
+EOP
+
+##### A finite intersection of open set is open
+
+Proof:
+
+Suppose $\forall i\in \{1,...,n\}$, $G_i$ is open.
+
+Let $x\in \bigcap^n_{i=1}G_i$, then $\forall i\in \{1,..,n\}$ and $G_i$ is open, so $\exists r_i>0$, such that $B_{r_i}(x)\subset G_i$
+
+Let $r=min\{r_1,...,r_n\}$. Then $\forall i\in \{1,...,n\}$. $B_r(x)\subset B_{r_i}(x)\subset G_i$. So $B_r(x)\subset \bigcup_{i=1}^n G_i$
+
+EOP
+
+The other two can be proved by **Theorem 2.22,2.23**
+
+#### Definition 2.26
+
+The closure $\bar{E}=E\cup E'$
+
+Remark: Using the definition of $E'$, we have, $\bar{E}=\{p\in X,\forall r>0,B_r(p)\cap E\neq \phi\}$
+
+#### Definition 2.27
+
+$\bar {E}$ is closed.
+
+Proof:
+
+We will show $\bar{E}^c$ is open.
+
+Suppose $p\in \bar{E}^c$. Then by remark, $\exists r>0$ such that $B_r(p)\cap E=\phi$ (a)
+
+Furthermore,, we claim $B_r(p)\cap E'=\phi$ (b)
+
+Suppose for contradiction that $\exists q\in B_r(p)\cap E'$ By **Theorem 2.19**, $\exists s>0$ such that $B_s(q)\subset B_r(p)$
+
+Since $q\in E',(B_s(q)\cap E)\backslash \{q\}\neq \phi$. This implies $B_r(p)\cap E=\phi$, which contradicts with (a)
+
+This proves (b)
+
+So $\bar{E}^c$ is open
+
+EOP

pages/Math4111/Math4111_L9.md

@@ -0,0 +1,104 @@
+# Lecture 9
+
+## Review
+
+1. Let $X=\mathbb{R}$ (and as usual, let $d(x,y)=|x-y|$). What is the set $B_1(0)$
+    $B_1(0)=(-1,1)$
+2. Let $X=[0,5]$ (and let $d$ be as usual). What is the set $B_1(0)$.
+    $B_1(0)=[0,1)$
+3. Let $X=\mathbb{R}$ and let $E=[0,2)$. Is $E$ open?
+    No, $0$ is not a interior point.
+4. Let $X=[0,5]$ and let $E=[0,2)$. Is $E$ open?
+    Yes, $0$ is a interior point, we can set radius to $1$ and all the points of $B_1(0)\subset E$
+
+## Continue on new materials
+
+### Metric space
+
+#### Theorem 2.27
+
+If $X$ is a metric spae and $E\subset X$, then.
+
+1. $\bar{E}$ is closed.
+2. $E=\bar{E}$ if and only if $E$ is closed.
+    Proof: $E=\bar{E}\iff E=E\cup E'\iff E'\subset E\iff E$ is closed.
+3. $\bar{E}\subset F$ for every closed set $F\subset X$ such that $E\subset F$.
+    $F\subset X$ closed, and $E'\subset F'\subset F$, so $\bar{E}=E\cup E'\subset F$
+
+#### Theorem 2.28
+
+$E\subset \mathbb{R}$ non-$\phi$ and bounded above, $\sup E\in \bar{E}$
+
+Proof:  
+
+Let $y\sup E$, To show $y\in \bar{E}$, we need to show $\forall h>0$, $B_h(y)\cap E\neq \phi$
+
+Let $h>0$. Since $y-h$ is not an upper bound of $E$, $\exists x\in E$ such that $x>y-h$.
+
+Since $y$ is an upper bound of $E$, $x\leq y$. So $x\in B_n(y)\cap E$, so $B_h(y)\cap E\neq \phi$.
+
+EOP
+
+#### Remark 2.29
+
+Let $(X,d)$ be a metric space, $E\subset X$. "$E$ is open" is short for "$E$ is open in $X$"/$E$ is open relative to $X$.
+
+This means $\forall p\in E$, $\exists r>0\{q\in X:d(p,q)<r\}\subset E$.
+
+Suppose $E\subset Y\subset X$. $\exists r>0\{q\in Y:d(p,q)<r\}\subset E$.
+
+Example: $X=\mathbb{R}$, $Y=[0,5]$, $E=[0,2)$, $E$ is open in $Y$, $E$ is not open on $X$.
+
+#### Theorem 2.30
+
+Let $E\subset Y\subset X$,then $E$ is open in $Y\iff \exists G$ open in $X$ such that $E=Y\cap G$
+
+Proof:
+
+Observe. If $p\in Y, r>0$, then $\{q\in Y:d(p,q)<r\}=\{q\in X:d(p,q)<r\}$ (ball in $X$ and ball in $Y$).
+
+$\impliedby$
+
+Suppose $\exists G$ open in $X$ such that $E=Y\cap G$. Let $p\in E$. Then $p\in G$ so $\exists r>0$ such that $\{q\in X:d(p,q)<r\}\subset G$ intersect both sides with $Y$ to get $\{q\in Y:d(p,q)<r\}\subset G\cap Y=E$
+
+$\implies$ 
+
+Suppose $E$ is open in $Y$. Then $\forall p\in E,\exists r_p>0$ such that $\{q\in Y:d(p,q)<r_p\}\subset E$
+
+Let $V_p=\{q\in X:d(p,q)<r_p\}$, $G=\bigcup_{p\in E}V_p$.
+
+Each $V_p$ is open in $X$ so $G$ is open in $X$. We need to show $E=G\cap Y$.
+
+To show $E\subset G\cap Y$. $p\in E\implies p\in V_p\implies p\in G$. So $E\subset G$. Also, by assumption, $E\subset Y$.
+
+To show $G\cap Y\subset E$.
+
+$G\cap Y=\left(\bigcup_{p\in E}V_p\right)\cap Y=\bigcup_{p\in E}(V_p\cap Y)\subset E$
+
+EOP
+
+### Compact sets
+
+#### Definition 2.31/2.32
+
+Let $(X,d)$ be a metric space $K\subset X$.
+
+Let $A$ be an index set. Suppose $\forall \alpha\in A$, $G_\alpha$ is an open set (in $X$) and suppose $K\subset \bigcup_{\alpha\in E}G_\alpha$. Then we say $\{G_\alpha:\alpha\in A\}$ is an **open cover** of $K$.
+
+Let $\{G_\alpha\}_{\alpha \in A}$ be an open cover of $K$. Suppose $\exists \alpha_1,...,\alpha_n\in A$ such that $K\subset\bigcup_{i=1}^n G_{\alpha_i}$. Then we say $\{G_{\alpha_i}\}_{i=1}^n$ is a **finite subcover** of $\{G_\alpha\}_{\alpha\in A}$ of $K$.
+
+#### Definition
+
+We say $K$ is **compact** if every open cover of $K$ contains a finite subcover.
+
+i.e. $\forall$ (This is important) open cover $\{G_\alpha\}_{\alpha\in A}$ of $K$, $\exists$ finite subcover $\{G_{\alpha_i}\}_{i=1}^n$ of $\{G_\alpha\}_{\alpha\in A}$ of $K$.
+
+Examples: $X=\mathbb{R}$.
+
+1.$K=\mathbb{R}$. $\mathbb{R}$ is not compact.
+
+as we can build an infinite open cover $\bigcup_{i\in Z} (i,i+2)$ and it does not have a finite subcover because:
+
+Suppose we consider the sub-collection $\{n_i,n_i+2:i=1,..,k\}$, Then $N+3$ is not in the union, where $N=max\{n_1,...,n_k\}$.
+
+EOP

pages/_app.jsx

@@ -0,0 +1,3 @@
+export default function App({ Component, pageProps }) {
+    return <Component {...pageProps} />
+  }

pages/_meta.js

@@ -0,0 +1,5 @@
+export default {
+    index: 'My Homepage',
+    contact: 'Contact Us',
+    about: 'About Us'
+  }

pages/index.mdx

@@ -0,0 +1,3 @@
+# Welcome to Nextra
+ 
+Hello, world!

theme.config.jsx

@@ -0,0 +1,21 @@
+export default {
+    footer: <p>MIT 2023 © Nextra.</p>,
+    head: ({ title, meta }) => (
+      <>
+        {meta.description && (
+          <meta name="description" content={meta.description} />
+        )}
+        {meta.tag && <meta name="keywords" content={meta.tag} />}
+        {meta.author && <meta name="author" content={meta.author} />}
+      </>
+    ),
+    readMore: 'Read More →',
+    postFooter: null,
+    darkMode: false,
+    navs: [
+      {
+        url: 'https://github.com/shuding/nextra',
+        name: 'Nextra'
+      }
+    ]
+  }