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remaining issues in mobilenavbar need pruning, need rewrite the class.
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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 $T$.
...
QED
## 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)
QED
## 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!

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# Math 4111 Exam 3 review
## Relations between series and topology (compactness, closure, etc.)
Limit points $E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\backslash\{x\}\cap E\neq\phi\}$
Closure $\overline{E}=E\cup E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\cap E\neq\phi\}$
$p_n\to p\implies \forall \epsilon>0, \exists N$ 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 $E$ ($S'\cap E\neq\phi$).
## 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.

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# Math 4111 Final Review
## Weierstrass M-test
Let $\sum_{n=1}^{\infty} f_n(x)$ be a series of functions.
The weierstrass M-test goes as follows:
1. $\exists M_n \geq 0$ such that $\forall x\in E, |f_n(x)| \leq M_n$.
2. $\sum M_n$ converges.
Then $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly.
Example:
### Ver.0
$\forall x\in [-1,1)$,
$$
\sum_{n=1}^{\infty} \frac{x^n}{n}
$$
converges. (point-wise convergence on $[-1,1)$)
$\forall x\in [-1,1)$,
$$
\left| \frac{x^n}{n} \right| \leq \frac{1}{n}
$$
Since $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, we don't know if the series converges uniformly or not using the weierstrass M-test.
### Ver.1
However, if we consider the series on $[-1,1]$,
$$
\sum_{n=1}^{\infty} \frac{x^n}{n^2}
$$
converges uniformly. Let $M_n = \frac{1}{n^2}$. This satisfies the weierstrass M-test. And this series converges uniformly on $[-1,1]$.
### Ver.2
$\forall x\in [-\frac{1}{2},\frac{1}{2}]$,
$$
\sum_{n=1}^{\infty} \frac{x^n}{n}
$$
converges uniformly. Since $\left| \frac{x^n}{n} \right|=\frac{|x|^n}{n}\leq \frac{(1/2)^n}{n}\leq \frac{1}{2^n}=M_n$, by geometric series test, $\sum_{n=1}^{\infty} M_n$ converges.
M-test still not applicable here.
$$
\sum_{n=1}^{\infty} \frac{x^n}{n}
$$
converges uniformly on $[-\frac{1}{2},\frac{1}{2}]$.
> Comparison test:
>
> For a series $\sum_{n=1}^{\infty} a_n$, if
>
> 1. $\exists M_n$ such that $|a_n|\leq M_n$
> 2. $\sum_{n=1}^{\infty} M_n$ converges
>
> Then $\sum_{n=1}^{\infty} a_n$ converges.
## Proving continuity of a function
If $f:E\to Y$ is continuous at $p\in E$, then for any $\epsilon>0$, there exists $\delta>0$ such that for any $x\in E$, if $|x-p|<\delta$, then $|f(x)-f(p)|<\epsilon$.
Example:
Let $f(x)=2x+1$. For $p=1$, prove that $f$ is continuous at $p$.
Let $\epsilon>0$ be given. Let $\delta=\frac{\epsilon}{2}$. Then for any $x\in \mathbb{R}$, if $|x-1|<\delta$, then
$$
|f(x)-f(1)|=|2x+1-3|=|2x-2|=2|x-1|<2\delta=\epsilon.
$$
Therefore, $f$ is continuous at $p=1$.
_You can also use smaller $\delta$ and we don't need to find the "optimal" $\delta$._
## Play of open covers
Example of non compact set:
$\mathbb{Q}$ is not compact, we can construct an open cover $G_n=(-\infty,\sqrt{2})\cup (\sqrt{2}+\frac{1}{n},\infty)$.
Every unbounded set is not compact, we construct an open cover $G_n=(-n,n)$.
Every k-cell is compact.
Every finite set is compact.
Let $p\in A$ and $A$ is compact. Then $A\backslash \{p\}$ is not compact, we can construct an open cover $G_n=(\inf(A)-1,p)\cup (p+\frac{1}{n},\sup(A)+1)$.
If $K$ is closed in $X$ and $X$ is compact, then $K$ is compact.
Proof:
Let $\{G_\alpha\}_{\alpha\in A}$ be an open cover of $K$.
> $A$ is open in $X$, if and only if $X\backslash A$ is closed in $X$.
Since $X\backslash K$ is opened in $X$, $\{G_\alpha\}_{\alpha\in A}\cup \{X\backslash K\}$ is an open cover of $X$.
Since $X$ is compact, there exists a finite subcover $\{G_{\alpha_1},\cdots,G_{\alpha_n},X\backslash K\}$ of $X$.
Since $X\backslash K$ is not in the subcover, $\{G_{\alpha_1},\cdots,G_{\alpha_n}\}$ is a finite subcover of $K$.
Therefore, $K$ is compact.
## Cauchy criterion
### In sequences
Def: A sequence $\{a_n\}$ is Cauchy if for any $\epsilon>0$, there exists $N$ such that for any $m,n\geq N$, $|a_m-a_n|<\epsilon$.
Theorem: In $\mathbb{R}$, every sequence is Cauchy if and only if it is convergent.
### In series
Let $s_n=\sum_{k=1}^{n} a_k$.
Def: A series $\sum_{n=1}^{\infty} a_n$ converges if the sequence of partial sums $\{s_n\}$ converges.
$\forall \epsilon>0$, there exists $N$ such that for any $m>n\geq N$,
$$
|s_m-s_n|=\left|\sum_{k=n+1}^{m} a_k\right|<\epsilon.
$$
## Comparison test
If $|a_n|\leq b_n$ and $\sum_{n=1}^{\infty} b_n$ converges, then $\sum_{n=1}^{\infty} a_n$ converges.
Proof:
Since $\sum_{n=1}^{\infty} b_n$ converges, $\forall \epsilon>0$, there exists $N$ such that for any $m>n\geq N$,
$$
\left|\sum_{k=n+1}^{m} b_k\right|<\epsilon.
$$
By triangle inequality,
$$
\left|\sum_{k=n}^{m}a_k\right|\leq \sum_{k=n+1}^{m} |a_k|\leq \sum_{k=n+1}^{m} b_k<\epsilon.
$$
Therefore, $\forall \epsilon>0$, there exists $N$ such that for any $m>n\geq N$,
$$
|s_m-s_n|=\left|\sum_{k=n+1}^{m} a_k\right|<\epsilon.
$$
Therefore, $\{s_n\}$ is Cauchy, and $\sum_{n=1}^{\infty} a_n$ converges.

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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.
$\equiv\cancel{\exist} p\in \mathbb{Q}, p^2=2$
$\equiv p\in \mathbb{Q},p^2\neq 2$
<details>
<summary>Proof</summary>
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.
</details>
### 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 $A\neq B$

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# Math 4111
This is a course about real analysis.
Topics include:
1. Number Theory
2. Basic topology and set theory
3. Sequences and Series
4. Convergence of Series and Sequences
5. Limits and Continuity
The course is taught by [Alan Chang](https://math.wustl.edu/pQEDle/alan-chang).
<!--
It is easy in my semester perhaps, it is the first course I got 3 perfect scores in exams. (Unfortunately, I did not get the extra credit for the third midterm exam.)
## Midterms stats
Our semester is way more easier than the previous ones. The previous ones got median scores of 25.
Stats for first midterm:
| |out of|avg|stddev|25th|50th|75th|
|--|--|--|--|--|--|--|
|total|50|40.7|9.9|37.0|45.0|48.0|
|1a|10|9.3|2.0|10.0|10.0|10.0|
|1b|10|8.3|2.4|6.8|9.0|10.0|
|2a|10|7.5|3.1|6.0|8.0|10.0|
|2b|10|6.9|3.4|4.0|8.0|10.0|
|3|10|8.8|2.4|9.0|10.0|10.0|
I skipped the last half hour, still get 50. I can't believe how easy it is compared to homework assignments.
Stats for second midterm:
| |out of|avg|stddev|25th|50th|75th|
|--|--|--|--|--|--|--|
|total|50.0|37.3|12.4|30.5|41.0|47.5|
|1|10|7.6|3.2|4.0|10.0|10.0|
|2|10|8.9|2.7|10.0|10.0|10.0|
|3|10|6.6|3.4|4.0|8.0|10.0|
|4a|10|6.6|4.1|2.0|9.0|10.0|
|4b|10|7.7|3.5|8.0|10.0|10.0|
I skipped the last half hour again, still get 50. But I felt this time is more challenging than the first one. Much like what Math is.
Stats for third midterm:
| |out of|avg|stddev|25th|50th|75th|
|--|--|--|--|--|--|--|
|total|50.0|37.2|10.1|31.0|40.0|44.8|
|1|10|8.2|3.1|7.0|10.0|10.0|
|2|10|8.2|2.5|7.0|10.0|10.0|
|3|10|7.1|3.0|6.0|6.0|10.0|
|4a|10|5.6|4.2|0.0|6.5|9.8|
|4b|10|8.1|3.4|6.5|10.0|10.0|
|5|0|0.0|0.1|0.0|0.0|0.0|
I got 50 and just barely made it. The third midterm is the hardest one. The extra credit question is way too hard and I only got few minutes to solve it.-->