====== Martin Zhai's Review Note ======
===== Content Summary =====
==== Week 1 ====

=== Lecture 1 (Jan 19) - Covered Ross Section 1.1, 1.2, 1.3 ===
  * __Natural Numbers__($\natnums$) $\{ 1, 2, 3, ... \}$\\
  * __Integers__($\Z$) $\{..., -2, -1, 0, 1, 2, ... \}$ (an example of Ring structure)\\
  * __Rational Numbers__($\mathbb{Q}$) $\{\frac{n}{m}, n,m\isin \Z, m\neq0\}$\\
      * //Proposition//: if $r\isin\mathbb{Q}$ with gcd(c,d)=1, and is a root of $C\scriptstyle{n} \normalsize {x^n}+ C\scriptstyle{n-1} \normalsize {x^{n-1}}+ ... +C\scriptstyle{0}$, $C\scriptstyle{0} \normalsize \, \mathrlap{\,/}{=} \, 0, C\scriptstyle{n} \normalsize \, \mathrlap{\,/}{=}\,0$, then d divides $C\scriptstyle{n}$, c divides $C\scriptstyle{0}$
      * //Corollary//: if $r\,=\,\frac{c}{d}\,\mathrlap{\,/}{=}\,0$ is a root of a "monic polynomial", i.e. leading term has coefficient 1, then r is an integer. 


=== Lecture 2 (Jan 21) - Covered Ross Section 1.4 ===
  * __Maximum and Minimum__: let $S\, \subset\, \Reals$ and $S\, \neq\, \text{\o}$, we say $\alpha\, \isin\, S$ is a maximum if $\forall\beta\, \isin\, S,\, \alpha\, \geq\, \beta$. Similarly, $\alpha\, \isin\, S$ is a minimum if $\forall\beta\, \isin\, S,\, \alpha\, \leq\, \beta$.
    * Remark: both maximum and minimum are elements in a set, and they are not guaranteed to exist.
    * Examples: $S\, =\, [\sqrt{2},100]\, \subset\, \Reals$, $\max(S)\, =\, 100,\, \min(S)\, =\, \sqrt{2}$
    * Non-Examples: $S\, =\, [\sqrt{2},100]\, \subset\, \mathbb{Q}$, $\max(S)\, =\, 100,\, \min(S)$ does not exist (since $\sqrt{2}\, \mathrlap{\,/}{\isin}\, \mathbb{Q}$ and $\mathbb{Q}$ is dense)
  * __Upper and Lower Bounds__: let $\text{\o}\, \neq\, S\, \subset\, \Reals$
    - $\alpha\, \isin\, \Reals$, we say $\alpha$ is an upper bound of S, if $\forall\, \beta\, \isin\, S,\, \beta\, \leq\, \alpha$
       *Examples: Define $S={x^2-10x+24,\, x \isin \Reals}$, $10$ would be an upper bound for $S$, since for $x\geq100$, then ${x^2}-10x+24 \geq 0$, hence $x \mathrlap{\,/}{\isin}S$.
       *Non-Examples: With the same set S defined as above, 5 is not an upper bound, since consider $6\, \geq\, 5$ and ${x^2}-10x+24$ evaluated at $x\, =\, 6$ is 0, i.e. $6\, \isin\, S$.
    - $\alpha\, \isin\, \Reals$, we say $\alpha$ is a lower bound of S, if $\forall\, \beta\, \isin\, S,\, \beta\, \geq\, \alpha$
       *Examples: Define $S\, =\, [5,10]\cup\Z$, both 1 and 4 would be a lower bound for S, since for $x\, \isin\, S,\, x\, \geq\, 4$ and $x\, \geq\, 1$.
       *Non-Examples: Consider $S\, =\, \Reals$, there is no such lower bound for $\Reals$, since for any number $s$, we could always find an element $\alpha\, \isin\, S$ and $\alpha\, \leq\, s$.
    *Remark: upper bounds and lower bounds are not unique, and may not exist.
  *__Supremum and Infimum__: let $\text{\o}\, \neq\, S\, \subset\, \Reals$, i.e. $S$ be a non-empty subset of $\Reals$
    - if $S$ is bounded above (i.e. there exists an upper bound for $S$), and $S$ has a least upper bound, then it is called supremum, denoted as $\sup(S)$
      * Examples: $S\, =\, [0,\sqrt{5}]\cup\Z$, then $\sup(S)\, =\, \sqrt{5}$ regardless of $\sqrt{5}\, \mathrlap{\,/}{\isin}\, S$.
      * Non-Examples: $S\, =\, \Z$, then $S$ does not have a supremum since $S$ is not bounded above, hence no such least upper bound exist.
    -  if $S$ is bounded below (i.e. there exists a lower bound for $S$), and $S$ has a greatest lower bound, then it is called infimum, denoted as $\inf(S)$
      * Examples: $S\, =\, \natnums$, then $\inf(S)\, =\, 1$.
      * Non-Examples: $S\, =\, \mathbb{Q}$, then there is no infimum for $S$.
  *__Completeness Axiom__: Every non-empty subset $S\, \subset\, \Reals$ that is bounded above has a least upper bound, i.e. a supremum.
    * Corollary: if $S$ is bounded from below, then $\inf(S)$ exists.
  * __Archimedean Property__: If $a\, >\, 0$, $b\, >\, 0$ are real numbers, then for some $n\, \isin\, \natnums$, we have $n*a\, >\, b$.

=== Homework 1 ===
  * if $S\, \subset\, T$, then $\inf(T)\, \leq\, \inf(S)\, \leq\, \sup(S)\, \leq\, \sup(T)$
  * $\sup({S}\cup{T})\, =\, \max(\sup(S),\, \sup(T))$
  * $\sup(A+B)\, =\, \sup(A)\, +\, \sup(B)$
  * $\inf(A+B)\, =\, \inf(A)\, +\, \inf(B)$
  * if $a\, \leq\, b+\frac{1}{n},\, \forall\, n\, \isin\, \natnums$, then $a\, \leq\, b$


==== Week 2 ====

=== Lecture 3 (Jan 26) - Covered Ross Section 2.7, 2.9 ===
  *__Sequence__: $a\scriptstyle{1}\normalsize,\, a\scriptstyle{2}\normalsize,\, a\scriptstyle{3}\normalsize,\, ...,\, a\scriptstyle{n}\normalsize,\,\isin\, \Reals$, could be denoted as $(a\scriptstyle{n}\normalsize)\scriptstyle{n\isin\natnums}$
    *Remark: sequence is not a set where sequence considers the order of all its elements while set only record what element is in it.
  *__Limit of Sequence__: we say a sequence $(a\scriptstyle{n}\normalsize)\scriptstyle{n}$ has limit $\alpha\, \isin\, \Reals$, if $\forall\, \epsilon>0$, there exists a real number $N>0$ such that for all integer $n>N$, we have $\lvert\, a\scriptstyle{n}\normalsize\, -\, \alpha\, \rvert\, <\, \epsilon$. We denote this by $\lim_{n\to\infty}a_{n}=\alpha$
    *Example: Prove $\lim_{n\to\infty}\frac{1}{n}=0$
      *Fix a $\epsilon>0$, we take $N=\frac{1}{\epsilon}$, then $n>N \iff n>\frac{1}{\epsilon} \iff \frac{1}{n}<\epsilon \iff \lvert \frac{1}{n}-0 \rvert\, <\epsilon$, completing the proof.
    *Non-Examples: Determine if $\lim_{n\to\infty}n=5$
      *Consider for all $\epsilon>0$. In order to have $\lvert a_{n}-5 \rvert <\epsilon$, then $a_{n} < 5 + \epsilon$, but since $n\rightarrow\infty$, it is impossible to find a $N$ satisfying $\forall n>N,\, a_{n} <5+\epsilon$, hence the claim is incorrect.
  ***Properties and Tools to find Limit**:
    * __Bounded sequence__: $(a_{n})_{n}$ is a bounded sequence if $\exists M>0$ such that $-M\leq a_{n} \leq M,\, \forall n \isin \natnums$
      *Examples: Consider $a_{n} = \frac{1}{n},\, \forall n \isin \natnums$, then it is a bounded sequence where we could take $M=1$ such that $a_{n} \leq M=1\, \forall n\isin \natnums$.
      *Non-Examples: Define $a_{n} = n,\, \forall n \isin \natnums$. For this sequence, there is no such $M$ to bound the elements in the sequence since for arbitrary $M$, it is always possible to find a $a_{n}$ such that $a_{n} > M$ ($a_{n}$ tends to $\infty$ as $n \rightarrow \infty$).
    - __Theorem 9.1__: All convergent sequences are bounded.
    - __Theorem 9.2__: If $\lim a_{n} = \alpha$, and if $k\isin \Reals$, then $\lim (k*a_{n})= k* \alpha$.
    - __Theorem 9.3, 9.4, 9.6__: Let $a_{n},\, b_{n}$ be convergent sequences with $\lim a_{n} =\alpha$ and $\lim b_{n} =\beta$, then:
      - $\lim (a_{n}+b_{n}) = \alpha + \beta$
      - $\lim (a_{n}*b_{n}) = \alpha * \beta$
      - if $a_{n} \neq 0\,\, \forall n$ and $\alpha \neq 0$, then $\lim (\frac{b \scriptscriptstyle n \normalsize}{a \scriptscriptstyle n \normalsize}) = \frac{\beta}{\alpha}$

=== Lecture 4 (Jan 28) - Covered Ross Section 2.9, 2.10 ===
  ***Properties and Tools to find Limit**:
    - __Theorem 9.7__:
      - $\lim_{n\to\infty} \frac{1}{n^p} = 0,\, \forall p > 0$
      - $\lim_{n\to\infty} {a^n} = 0,$ if $\lvert a \rvert < 1$
      - $\lim_{n\to\infty} {n^\frac{1}{n}} = 1$
      - $\lim_{n\to\infty} {a^\frac{1}{n}} = 1$ for $a>0$
    *__Limit of Infinity__: we say $\lim s_n = +\infty$ provided for each $M>0$, $\exists N$ such that $n>N \implies s_n >M$; similarly $\lim s_n = -\infty$ provided for each $M>0$, $\exists N$ such that $n>N \implies s_n <M$
      *Example: $\lim (\sqrt{n} + 7)= +\infty$
        *Fix a $M>0$, define $N = (M-7)^2$. Then $n>N \implies n>(M-7)^2 \implies \sqrt{n} > M-7 \implies \sqrt{n} +7>M$, completing the proof.
      *Non-Example: Determine if $\lim \frac{1}{n}= +\infty$
        *No. Consider $M = 1$, in order to have $\frac{1}{n} > 1$, $n$ must be smaller than $1$. Thus it is not possible to define a $N$ such that $\forall n>N \implies \frac{1}{n} > 1$.
      - __Theorem 9.9__: Let $(s_n),(t_n)$ be sequences such that $\lim s_n = +\infty,\, \lim t_n > 0$, then $\lim (s_n*t_n) = +\infty$.
      - __Theorem 9.10__: Let a sequence $(s_n)$ of positive real numbers, then $\lim s_n = +\infty \iff \lim \frac{1}{s_n} = 0$.
  ***Monotone Sequence and Cauchy Sequence**:
    * __Cauchy Sequence__: $(a_n)$ is a Cauchy sequence if $\forall \epsilon >0,\, \exists N>0$ such that $\forall n_1,n_2>N$, we have $\lvert a_{n_1} - a_{n_2} \rvert < \epsilon$ i.e. oscillation amplitude gets smaller and smaller.
      * Example: $a_n = \frac{1}{n^2}\, \forall n \isin \natnums$ is a Cauchy sequence. 
        * Fix a $\epsilon >0$. Since we already know from Theorem 9.7a), $\lim \frac{1}{n^2} = 0$, then we could find a $N>0$ such that $n>N \implies \frac{1}{n^2} < \epsilon$. Thus take the same $N$, we get $\forall a,b >N$, $\lvert \frac{1}{a^2} - \frac{1}{b^2} \rvert \leq \lvert \frac{1}{a^2} \rvert < \epsilon$, completing the proof.
      * Non-Example: Consider $a_n = n$, then if we fix $\epsilon = 0.5$, there is no such $N>0$ such that $n_1, n_2>N \implies  \lvert a_{n_1} - a_{n_2} \rvert < 0.5$ since the smallest difference between any $a_n$ is 1, thus not a Cauchy sequence.
    * __Monotone Sequence__:
      - A monotone increasing sequence is such that $\forall n > m,\, a_{n} \geq a_m$.
      - A monotone decreasing sequence is such that $\forall n > m,\, a_{n} \leq a_m$.
        * Example: $a_n = 1 - \frac{1}{n}$ is a monotone increasing sequence, since $\forall n > m,\, \frac{1}{n} < \frac{1}{m}$, hence $1 - \frac{1}{n} > 1 - \frac{1}{m}$.
        * Non-Example: $a_n = (n-2)^2$ is not a monotone sequence. 
          * Consider $n = 2,\, m = 1,\, n>m$, but $a_2 = 0 < a_1 = 1$, hence not monotone increasing. 
          * Then consider $n = 2,\, m = 3,\, n<m$, but $a_2 = 0 < a_3 = 1$, hence not monotone decreasing. 
    - __Theorem 10.2__: All bounded monotone sequences are convergent.
    - __Theorem 10.11__: Let $(a_n)$ be a sequence. Then $(a_n)$ is Cauchy $\iff (a_n)$ converges.

=== Homework 2 ===
  * 9.9c): If there exists $N_0$ such that $s_n \leq t_n\, \forall n > N_0$,and $\lim s_n$ and $\lim t_n$ exists, then $\lim s_n \leq \lim t_n$.

==== Week 3 ====
=== Lecture 5 (Feb 2) - Covered Ross Section 2.10 ===
  ***Monotone and Cauchy Sequences**:
    * __$\limsup$__: Let $(a_n)$ be a sequence in $\Reals$, $\limsup a_n = \lim_{N\to\infty}(\sup_{n>N} {a_n})$
    * __$\liminf$__: Let $(a_n)$ be a sequence in $\Reals$, $\liminf a_n = \lim_{N\to\infty}(\inf_{n>N} {a_n})$
    * Example:$(a_n) = 0$ if $\frac{n}{2} = 0$, $(a_n) = \frac{1}{n}$ if $\frac{n}{2} = 1$. 
      * $\limsup a_n = 0$ since the supremum of the subsequence $(a_n)_{n>N}$ for all $N$ is $\frac{1}{N+1}$ if $N$ is even and $\frac{1}{N}$ if $N$ is odd, which converges to $0$ as $N \rightarrow \infty$.
      * $\liminf a_n = 0$ since the infimum of the subsequence $(a_n)_{n>N}$ for all $N$ is $0$ (there is always an even number $\geq N$)
      * __Property of Monotone Sequences__: If it is bounded, then its limit exists; if it is unbounded, then $\lim a_n = +/- \infty$.
      * __Lemma__: If $(a_n)$ is a bounded sequence and $\alpha_+ = \limsup a_n$, then for any $\epsilon >0,\, \exists N$ such that $\forall n>N$, we have $a_n \leq \alpha _+ + \epsilon$
      * __Theorem 10.7__: Let $(a_n)$ be a bounded sequence, then $\lim a_n$ exists $\iff \limsup a_n = \liminf a_n$.

=== Lecture 6 (Feb 4) - Covered Ross Section 2.11 ===
  ***Subsequences**:
    *__Subsequence__: Let $(s_n)_{n\isin \natnums}$ be a sequence of real numbers. Given a list of indices, $n_1 < n_2 < ... <n_k< ...$. Take $t_k = s_{n_k}$, then $(t_m)$ is called a subsequence of $(s_n)$, we write $(s_{n_k})_k$ for the subsequence. 
      *Example: Let $(s_n)$ denote some sequence with real numbers, define $(t_m)$ as $t_m = s_{2*m}$. Then $(t_m)$ is a subsequence of $(s_n)$ (could be denoted as $(s_{2n})_n$)
    *__Theorem 11.2__: Let $(s_n)$ be a sequence.$\\$ If $t\isin \Reals$, then there is a subsequence of $(s_n)$ converging to $t \iff$ the set $\{n\isin \natnums :\, \lvert s_n - t\rvert < \epsilon\}$ is infinite for all $\epsilon >0$. 
    *__Theorem 11.3__: If $(s_n)$ is convergent, then any subsequence converges to the same point.
    *__Theorem 11.4__: Every sequence has a monotonic subsequence.

=== Homework 3 ===
  * $(s_n)$ a bounded sequence, then $\liminf s_n \leq \limsup s_n$ and $\limsup s_n = \inf \{\sup_{n\geq N} s_n:\, n\isin \natnums\}$.
  * If $(a_n), (b_n)$ are two bounded sequences, then $\limsup (a_n+b_n) \leq \limsup (a_n) + \limsup (b_n)$.

==== Week 4 ====
=== Lecture 7 (Feb 9) - Covered Ross Section 2.11 ===
  ***Subsequences**
    *__Theorem 11.5 (Bolzano-Weiestrass Theorem)__: Every bounded sequence has a convergent subsequence.
    *__Subsequential Limit__: Let $(s_n)$ be a sequence in $\Reals$, a subsequential limit is any real number or $+/- \infty$ that is the limit of a subsequence of $(s_n)$.
      *Example: Let $(r_n)$ be the enumeration of $\mathbb{Q}$, then $\forall r \isin \Reals$ is a subsequential limit of $(r_n)$. This is because the denseness of $\mathbb{Q}$ in $\Reals$, which means for any $r\isin \Reals,\, \forall \epsilon >0,\, (r - \epsilon, r+\epsilon)$ is an infinite set. Hence by Theorem 11.2, there is some subsequence of $(s_n)$ that converges to such $r$, i.e. a subsequential limit.
    *__Theorem 11.7__: Let $(s_n)$ be any sequence. Then there exists a monotonic subsequence that converges to $\limsup s_n$ and a monotonic subsequence that converges to $\liminf s_n$.
    *__Theorem 11.8__: Let $(s_n)$ be a sequence, $S$ be the set of subsequential limits of $(s_n)$, then 
      *S is non-empty
      *$\sup S = \limsup s_n$, and $\inf S = \liminf s_n$
      *$S=\{ \alpha \} \iff \lim s_n = \alpha$
    *__Theorem 11.9__: Let $S$ be the set of subsequential limits of $(s_n)$. Suppose $(t_n)$ is a sequence in $S\cap \Reals$ and $t = \lim t_n$. Then $t \isin S$.

=== Lecture 8 (Feb 11) - Covered Ross Section 2.12 ===
  ***lim sup's and lim inf's**:
    * Recall that in general, $\limsup s_n \geq \liminf s_n$, and if they are equivalent, then $\lim s_n$ exists.
    * __Theorem 12.1__: Let $(s_n)$ be a sequence that converges to a positive real number $s$, and $(t_n)$ be any sequence. Then $\limsup (s_n*t_n) = s*\limsup t_n$. (Here we allow notation of $s*(+\infty) = +\infty$ and $s*(-\infty) = -\infty$ for $s>0$).
    * __Theorem 12.2__: Let $(s_n)$ be a sequence of positive real numbers, then we have$\\$ $\enspace \liminf (\frac{s_{n+1}}{s_n}) \leq \liminf (s_n)^{\frac{1}{n}} \leq \limsup (s_n)^{\frac{1}{n}} \leq \limsup (\frac{s_{n+1}}{s_n})$.

=== Homework 4 ===
  * $(s_n)$ bounded $\iff \limsup \lvert s_n \rvert < +\infty$
  * Let $(s_n)$ be a bounded sequence in $\Reals$. If $A = \{ a \isin \Reals:$ only finitely many $s_n <a\}$ and $B = \{ b \isin \Reals:$ only finitely many $s_n >b\}$, then $\sup A = \liminf s_n$ and $inf B = \limsup s_n$.

==== Week 5 ====
=== Lecture 9 (Feb 16) - Covered Ross Section 2.13 ===
  ***Metric Space & Topology**:
    *__Metric Space__: A metric space is a set $S$ together with a distance function $d: S \times S \rightarrow \Reals$ such that 
      - $d(x,y) \geq 0,$ and $d(x,y) = 0 \iff x=y$
      - $d(x,y) = d(y,x)$
      - $d(x,y) + d(y,z) \geq d(x,z)$ (Triangle Inequality)
      * Example: Let $d(x,y) = \sqrt{\lvert x-y \rvert}$, then with the set $\Reals$, it forms a metric space. The first two criteria is obvious. For the triangle inequality, $(\sqrt{\lvert x-y \rvert} + \sqrt{\lvert y-z \rvert})^2 = \lvert x-y \rvert + \lvert y-z \rvert + 2\sqrt{\lvert x-y \rvert}  \sqrt{\lvert y-z \rvert}$. By Euclidean distance metric, we know $\lvert x-y \rvert + \lvert y-z \rvert \geq \lvert x-z \rvert$. Also $2 \sqrt{\lvert x-y \rvert} \sqrt{\lvert y-z \rvert} \geq 0$. Therefore $d(x,y) = \sqrt{\lvert x-y \rvert}$ is a metric, hence with $\Reals$ is a metric space.
      * Non-Example: Let $d(x,y) = (x-y)^2$. It is not a metric since consider $x=2, y = 1, z=0$, then $d(x,y) = 1$ and $d(y,z)=1$ but $d(x,z) = 4$ which means $d(x,y) + d(y,z) < d(x,z)$, thus not a metric.
    * __Cauchy Sequence (in metric space (S,d))__: A sequence $(s_n)$ in $S$ is Cauchy if $\forall \epsilon >0,\, \exists N >0$ such that $\forall n,m >N,\, d(s_n,s_m) < \epsilon$.
    * __Convergence (in metric space (S,d))__: A sequence $(s_n)$ converge to a point $s\isin S$ if $\forall \epsilon >0,\exists N>0$ such that $d(s_n,s)<\epsilon \enspace \forall n>N$.
    * __Completeness__: A metric space $(S,d)$ is complete if every Cauchy sequence is convergent.
      * Example: The Euclidean k-space $\Reals^k$ is complete by Theorem 13.4.
      * Non-Example: $\mathbb{Q}$ with the Euclidean distance function is not complete. Consider a sequence of rational numbers converging to $\sqrt{2}$ (possible by example in Lecture 7). This sequence is clearly Cauchy but not convergent since $\sqrt{2} \mathrlap{\,/}{\isin} \mathbb{Q}$. Thus $\mathbb{Q}$ with Euclidean distance formula is not a complete metric space.
    * __Induced Distance Function__: If $(S,d)$ is a metric space and $A \subset S$ is any subset, then $(A, d|_{A\times A})$ is a metric space.
    * __Theorem 13.5 (Bolzano-Weiestrass Theorem for $\Reals^n$)__: Every bounded sequence $(s_m)_m \isin \Reals^n$ has a convergent subsequence.
    * __Topology__: Let $S$ be a set. A topological structure on $S$ is the data of a collection of subsets in S. This collection needs to satisfy:
      - $S$ and $\text{\o}$ are open
      - arbitrary union of open subsets is still open
      - finite intersections of open sets are open

=== Lecture 10 (Feb 18) - Midterm 1 ===

=== Homework 5 ===
  * Every open subset of $\Reals$ is the disjoint union of finite or countably infinite sequence of open intervals.

==== Week 6 ====
=== Lecture 11 (Feb 23) - Covered Ross Section 2.13, Rudin Chapter 2 ===
  ***Topology of Metric Space**:
    *__Basic Notions__: Let $E \subset S$
      * //Interior point//: $p\isin E$ is an interior point of E if $\exist \delta>0$ such that $B_{\delta}(p) = \{q \isin S| d(p,q)< \delta\} \subset E$.
        * Example: Consider $E = [0,1] \cap \Reals \subset \Reals$, then $p=0.5$ is an interior point of $E$. If we take $\delta = 0.3$, then $B_{0.3}(0.5) = [0.2, 0.8]\cup \Reals \subset E$.
        * Non-Example: Take the same E as above, consider $p=0$. P is not an interior point since for any $\delta \neq 0$, the points in $\Reals$ on the left side of $0$ is not in $E$, hence not an interior point.
      * //Interior//: The set of all interior points of E, denote as $E^o$.
    * __Open__: $E\subset S$ is an open subset of $S$ if $E=E^o$, i.e. $\forall p \isin E,\, \exists \delta >0$ such that $B_{\delta}(p) \subset E$.
      * Example: Take $E = (0,1) \cap \Reals \subset \Reals$, $E$ is a subset of $\Reals$. For any point $p \isin E$, let $\delta_o = \max (d(p,0), d(p,1))$. Then we could choose $\delta = \frac{\delta_o}{2}$, then $B_{\delta}(p) \subset E$ for arbitrary point $p \isin E$, hence $E$ is an open subset of $\Reals$.
      * Non-Example: Take $E = [0,1] \cap \Reals \subset \Reals$. As the non-example for interior points shown, $0$ is not an interior point in $E$, hence $E$ is not an open subset of $\Reals$.
    * __Close__: $E\subset S$ is a closed subset of S if the complement $E^c = S \setminus E$ is open.
      * Propositions: 
        - $S$ and $\text{\o}$ are closed
        - An arbitrary collection of closed sets is closed
        - A finite intersection of closed sets is closed
    * __Limit Point__: Let $E\subset S$, $p\isin S$ is a limit point of $E \iff \forall \delta >0,\, B_{\delta}(p)$ intersects $E$ non-empty, i.e. $\exists q\isin E,\, q \neq p,\, d(p,q)<\delta$. $E' =$ set of limit points of $E$.
      *Example: Let $S=\Reals,\, E = \{\frac{1}{n}:n\isin \natnums\}$. Then $0$ is a limit point since for any $\delta$, we could always find a $N>0$ such that $\frac{1}{n} < \delta, \forall n>N$, hence $B_{\delta}(0) \cap E \neq \text{\o}$.
      *Non-Example: Consider the same scenario as the previous example. $\frac{1}{2}$ is not a limit point if we take $\delta = \frac{1}{10}$, then $B_{\frac{1}{10}}(\frac{1}{2}) = [\frac{4}{10}, \frac{6}{10}] \cap E = \text{\o}$.
    * __Closure__: $E\subset S$, the closure of $E$ is the intersection of all closed subsets containing $E$. Denote as $E^-$.
      * Proposition: $E^- = E \cup E'$
      * Example: Let $S=\Reals,\, E = \{\frac{1}{n}:n\isin \natnums\}$. Then $E^- = \{\frac{1}{n}:n\isin \natnums\} \cup \{0\}$ by proposition above ($\{0\} = E'$) 
    * __Boundary__: The boundary points of $E\subset S$ is the set $E^-\setminus E^o$.
    * __Proposition 13.9__: Let $E$ be a subset of a metric space $(S,d)$
      - The set $E$ is closed if and only if $E=E^-$
      - The set $E$ is closed if and only if it contains the limit of every convergent sequence of points in $E$
      - An element is in $E^-$ if and only if it is the limit of some sequence of points in $E$
      - A point is in the boundary of $E$ if and only if it belongs to the closure of both $E$ and its complement
    *__Isolated Point__: If $p \isin E$ and $p$ is not a limit point of $E$, then $p$ is called an isolated point.
    *__Perfect__: $E$ is perfect if $E$ is closed and every point of $E$ is a limit point of $E$.
    *__Dense__: $E$ is dense in $S$ if every point of $S$ is a limit point of $E$ or a point of $E$ or both.
    *__Rudin 2.30__: Suppose $Y\subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y\cap G$ for some open subset $G$ of $X$.
  ***Compact Set**:
    *__Open Cover__: Let $(S,d)$ be a metric space, $E\subset S$, $\{ G_{\alpha} \}$ is a collection of open sets. We say $\{ G_{\alpha} \}$ is an open cover of $E$ if $E \subset \cup_{\alpha} G_{\alpha}$.
    *__Compact Set__: $K\subset S$ is a compact subset, if for any open cover of $K$, there exists a finite subcover, i.e. if $\{ G_{\alpha} \}$ is an open cover, then $\alpha_1, ..., \alpha_n$ indices such that $K\subset G_{\alpha_1} \cup ... \cup G_{\alpha_n}$.
      *Example: $K=[0,1]$ is a compact subset of $\Reals$.
      *Non-Example: $K=(0,1]$ is not a compact subset of $\Reals$. Consider the open cover $G_n = B_{\frac{1}{2n}}(\frac{1}{n}), n\isin \natnums,$ then $K \subset \cup_{n\isin \natnums}$. In this case, there is no finite subcover. If we take indices $n_1 < n_2 < ... < n_m$, then $x < \frac{1}{2n_m}$ is not in the union.
    *__Sequentially Compact__: $E\subset S$ is sequentially compact if any sequence in $E$ has a convergent subsequence in $E$ (the limit point is also in $E$).
    *__Theorem__: For any metric space $(S,d)$, $E\subset S$, $E$ compact $\iff E$ sequentially compact.
    *__Theorem 13.12 (Heine-Borel Theorem)__: Consider $\Reals^n$ with Euclidean metric $d(x,y)= \lvert x-y \rvert$, $E\subset \Reals^n$ is compact $\iff E$ is closed and bounded.
    *__Rudin 2.33__: $K\subset Y\subset X$, then $K$ is compact relative to $Y$ if and only if $K$ is compact relative to $X$.
    *__Rudin 2.34__: Compact subsets of metric spaces are closed.
    *__Rudin 2.35__: Closed subsets of compact sets are compact.
    

=== Lecture 12 (Feb 25) - Covered Ross Section 2.14, 2.15 ===
  ***Series**:
    * __Infinite Sum__: An infinite sum of sequence $(a_n)$ is defined as $a_1 + a_2 + ... = \sum_{n=1}^{\infty} a_n$.
    * __Partial Sum__: Defined as $a_1 + a_2 + ... + a_n = \sum_{i=1}^{n} a_i$.
    * __Convergence__: A series converge to $\alpha$ if the corresponding partial sum converges to $\alpha$.
    * __Cauchy Condition for Series Convergence__: $\forall \epsilon>0,\, \exists N>0$ such that $\forall n,m>N,\, \lvert \sum_{i=n+1}^{m} a_i \rvert < \epsilon$.
    * __Absolute Convergence__: If $\sum \lvert a_n \rvert < \infty$, we say $\sum a_n$ converges absolutely.
    * Recall Geometric Series: $\sum_{n=0}^{\infty} ar^n$ converges to $a\frac{1}{1-r}$ if $\lvert r \rvert <1$.
  * **Tests for Series Convergence**:
    * __Comparison Test__: 
      * Suppose $\sum_{n} a_n <\infty,\, a_n >0$, and $b_n \isin \Reals < a_n$, then $\sum_{n} b_n < \infty$.
      *  Suppose $\sum_{n} a_n =\infty,\, a_n >0$, and $b_n \geq a_n$, then $\sum_{n} b_n = \infty$.
    * __Ratio Test (Test for Absolute Convergence)__: 
      *  If $\limsup \lvert \frac{a_{n+1}}{a_n} \rvert <1$, then $\sum_{n} \lvert a_n \rvert$ converges.
      *  If $\liminf \lvert \frac{a_{n+1}}{a_n} \rvert >1$, then $\sum_{n} \lvert a_n \rvert$ diverges.
      *  Otherwise the test does not give any information on its convergence.
    * __Root Test__: Let $\sum_{n} a_n$ be a series, $\alpha = \limsup (\lvert a_n \rvert)^{\frac{1}{n}}$, then $\sum_{n} a_n$:
      * converges absolutely if $\alpha <1$.
      * diverges if $\alpha >1$.
      * gives no information if $\alpha = 1$.
    * __Alternating Series Test__: Let $a_1 \geq a_2 \geq ...$ be a monotone decreasing series, $a_n \geq 0$. And assuming $\lim a_n = 0$. Then $\sum_{n=1}^{\infty} (-1)^{n+1}a_n = a_1 - a_2 + a_3 - ...$ converges. Moreover, the partial sums $s_n = \sum_{k=1}^{n} (-1)^{k+1}a_k$ satisfy $\lvert s- s_n \rvert \leq a_n$ for all $n$.
    * __Integral Test__: If the terms $a_n$ in $\sum_{n} a_n$ are non-negative and $f(n) = a_n$ is a decreasing function on $[1, \infty)$, then let $\alpha = \lim_{n\to\infty} \int_{1}^{n} f(x)dx$
      * If $\alpha = \infty$, then the series diverge
      * If $\alpha < \infty$, then the series converge

=== Homework 6 ===
  * Rudin 2.36: If $\{K_{\alpha}\}$ is a collection of compact subsets of a metric space such that the intersection of every finite subcollection of $\{K_{\alpha}\}$ is non-empty, then $\cap K_{\alpha}$ is non-empty. (This statement is incorrect if compact is replaced by closed or bounded).
  * If $a_n >0$ and $\sum a_n$ diverges, then $\sum \frac{a_n}{1+a_n}$ also diverges.
  * If $a_n >0$ and $\sum a_n$ converges, then $\sum \frac{\sqrt{a_n}}{n}$ also converges.

==== Week 7 ====
=== Lecture 13 (Mar 2) - Covered Rudin Chapter 4 ===
  ***Continuous Functions**:
    *__Function__: A function from set $A$ to set $B$ is an assignment for each element $\alpha \isin A$ an element $f(\alpha) \isin B$.
      -//Injective//: A function $f$ is injective/one-to-one if $\forall x,y\isin A,\, x \neq y$, then $f(x) \neq f(y)$
      -//Surjective//: A function $f$ is surjective/onto if $\beta \isin B,$ there exists at least one element $\alpha \isin A$ such that $f(\alpha)=\beta$
      -//Bijective//: A function $f$ is bijective if $f$ is both injective and surjective
      * Example: $f: \Reals \rightarrow \Reals,\, f(x)=x$ is a bijective function since if $f(x) = f(y)$, then $x = y$. And for any $y \isin \Reals$, we take $x=y \isin \Reals$ as the domain will prove its surjectivity. Hence a bijection.
      * Non-Example: 
        - $f: \Reals \rightarrow \Reals,\, f(x)=x^2$ is not injective since for $x=1$ and $y=-1$, $f(x) = f(y)$ even if $x\neq y$. 
        - $f: \Reals \rightarrow \Reals,\, f(x)=\frac{1}{x}$ is not surjective since for $y = 0 \isin \Reals$, there is no element in domain that get mapped to $0$ under such $f$.
    *__Pre-image__: If $f: A\rightarrow B$. Given a subset $E\subset B$, $f^{-1}(E) = \{\alpha \isin A|\, f(\alpha) \isin E\}$ is called the pre-image of $E$ under $f$, which is a subset of $A$.
      *Example: $f: \Reals \rightarrow \Reals,\, f(x)=\ln x$ and let $E=[0,\infty)$. Then pre-image of $E$ under $f$ is $[1,\infty)$.
    *__Limit of a Function__: Suppose $p\isin E'$(set of limit points of $E$), we write $f(x) \rightarrow q(\isin Y)$ as $x \rightarrow p$ or $\lim_{x\to p} f(x) = q$ if $\forall \epsilon >0,\, \exists \delta >0$ such that $\forall x \isin E,\, 0<d_X(x,p)<\delta  \implies d_Y(f(x),q)<\epsilon$.
      *Example: Consider $f: \Reals \rightarrow \Reals$ $f(x)=\frac{x^2-1}{x-1}$. Claim $\lim_{x\to 1} f(x) = 2. 
        *Fix a $\epsilon >0$, consider $\delta = \lvert \epsilon - 1\rvert$. Using the Euclidean distance formula, $\lvert x - 1\rvert < \delta = \lvert \epsilon - 1\rvert \implies -\epsilon + 1 < x - 1< \epsilon -1 \implies -\epsilon + 3 < x + 1< \epsilon + 1 \implies -\epsilon + 3 < \frac{(x + 1)(x-1)}{x-1}< \epsilon + 1 \implies -\epsilon + 2 < \frac{(x + 1)(x-1)}{x-1} - 1< \epsilon \implies \lvert \frac{(x + 1)(x-1)}{x-1} - 1 \rvert < \epsilon$, hence proving the claim.
    *__Rudin 4.2__: With the same notation as above, $\lim_{x\to p} f(x) = q$ if and only iff $\lim_{n\to\infty} f(p_n) = q$ for every sequence $(p_n)$ in $E$ such that $p_n \neq p, \lim_{n\to\infty} p_n = p$.
      *Example: $f:\Reals \rightarrow \Reals$ and $f(x)=x^2$. $\lim_{x\to 0} f(x) = 0$ and then consider sequence $p_n = \frac{1}{n}$ which converges to $0$, and $f(p_n) = \frac{1}{n^2}$ which also converges to $0$ when $n \rightarrow \infty$
    *__Corollary__: If $f$ has a limit at point $p$, then it is unique.
    *__Rudin 4.4__: Suppose $f,g: E \rightarrow \Reals$, suppose $p\isin E;$ and $\lim_{x\to p} f(x) = A, \lim_{x\to p} g(x) = B$, then
      *$\lim_{x\to p} f(x) +g(x)= A+B$
      *$\lim_{x\to p} f(x)g(x) = AB$
      * $\lim_{x\to p} \frac{f(x)}{g(x)} = \frac{A}{B}$ if $B\neq 0$ and $g(x)\neq 0 \, \forall x\isin E$
      * $\forall c\isin \Reals$, $\lim_{x\to p} c*f(x)=cA$
  * **Continuity of Functions**:
    * __Continuity at a Point__: Let $(X,d_X), (Y,d_Y)$ be metric spaces, $E\subset X$, $f:E\rightarrow Y$, $p \isin E$, $q=f(p)$. We say $f$ is continuous at $p$, if $\forall \epsilon>0, \exists \delta>0$ such taht $\forall x\isin E$ with $d_X(x,p) <\delta \implies d_Y(f(x),q)<\epsilon$.
    * __Rudin 4.6__: If $p\isin E$ is also a limit point of $E$, then $f$ is continuous at $p \iff \lim_{x\to p} f(x) = f(p)$.
    * __Continuity__: We say $f$ is continuous on $E$ if $f$ is continuous at every point in $E$.
    * __Rudin 4.7__: $(X,d_X), (Y,d_Y)$, $f:X\rightarrow Y$ as above. Then $f$ is continuous $\iff$ for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X$.
    * __Lemma__: If $f: A\rightarrow B$ is a function and $E\subset A, F\subset B$. The $f(E)=F \iff E\subset f^{-1}(F)$.
    * __Rudin 4.7__: Let $X,Y,Z$ be metric spaces and $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ continuous functions. We define $h:X\rightarrow Z$ by $h(x)=g(f(x)$. Then $h$ is also continuous ($h$ is called the composition of $f$ and $g$).
    * __Rudin 4.9___: If $f,g:X\rightarrow \Reals$ continuous, then $f+g,\, f-g,\, fg$ are continuous functions, and if $g(x)\neq 0$ for any $x\isin X$, then $\frac{f}{g}$ is also continuous.
      * Result: All polynomials are continuous since $f(x)=x$ is continuous.
    * __Rudin 4.10__: Let $f:X\rightarrow \Reals^n$ with $f(x) = (f_1(x), f_2(x), ..., f_n(x))$. Then $f$ is continuous $\iff$ each $f_i$ is continuous.

=== Lecture 14 (Mar 4) - Covered Rudin Chapter 4 ===
  * **Review of Compact Sets**:
    * __Compactness__: $K\subset X$ is compact if $\forall$ open cover of $K$, $\exists$ a finite subcover.
    * __Propositions__:
      * $K$ compact $\implies K$ bounded
      * $K$ compact $\implies K$ closed
      * $E\subset X$ is closed, $K$ is compact, $E\subset K \implies E$ is compact.
    * __Theorems__:
      * Compactness $\iff$ Sequential Compactness
      * Heine-Borel (Rudin 2.41): in $\Reals^n$, $K$ compact $\iff K$ closed and bounded.
    * Remark: The notion of "compact" is intrinsic, while open and closed depends on the ambient space.
  * **Continuous Maps and Compactness**:
    * __Three Definitions of Continuous Maps__:
      * $f$ is continuous if and only if $\forall p\isin X, \forall \epsilon >0, \exists \delta >0$ such that $f(B_{\delta}(p)) \subset B_{\epsilon}(f(p))$
      * $f$ is continuous if and only if $\forall V\subset Y$ open, $f^{-1}(V)$ is open
      * $f$ is continuous if and only if $\forall x_n \rightarrow x$ in $X$, we have $f(x_n) \rightarrow f(x)$ in $Y$
    * __Rudin 4.14__: Suppose $f$ is a continuous map from a compact metric space $X$ to another compact metric space $Y$, then $f(X) \subset Y$ is compact.
    * __Rudin 4.16__: Suppose $f$ is a continuous real function on a compact metric space $X$, and $M = \sup_{p\isin X} f(p)$, $m=\inf_{p\isin X} f(p)$. Then there exists point $p,q \isin X$ such that $f(p)=M$ and $f(q)=m$.
      * Recall: if $K\subset \Reals$, $K$ is compact, then $\sup(K) \isin K$ and $\inf(K) \isin K$.
      * Remark: If $f:X\rightarrow Y$ is continuous, $f$ sends compact set $X$ to compact set $Y$, but given $E\subset Y$ compact, $f^{-1}(E)$ is not guaranteed to be compact. 

=== Homework 7 ===
  * If $f:X\rightarrow Y$ is a continuous function from a metric space $X$ to a metric space $Y$, $f(E^-)\subset {f(E)}^-$ for any $E\subset X$.
  * Let $f,g$ be continuous maps from $X$ to $Y$. Suppose there is a dense set $E\subset X$ such that $f|_E=g|_E$, then $f=g$.

==== Week 8 ====
=== Lecture 15 (Mar 9) - Covered Ross Section 3.19 Rudin Chapter 2 and 4 ===
  ***Uniform Continuity**:
    *__Uniform Continuous Function__: $f:X\rightarrow Y$. Suppose for all $\epsilon >0$, $\exists \delta >0$ such that $\forall p,q\isin X$ with $d_X(p,q)<\delta$, we have $d_Y(f(p),f(q)) <\epsilon$. Then we say $f$ is a uniform continuous function.
      *Example: $f:[0,1] \rightarrow \Reals$ and $f(x)=x^2$. Then $f$ is uniformly continuous function. $\forall \epsilon >0$, we can take $\delta=\frac{\epsilon}{2}$, then $\forall p,q \isin [0,1]$, $\lvert p-q \rvert < \delta$ we have $\lvert p^2-q^2 \rvert = \lvert (p-q)(p+q) \rvert = \lvert (p-q) \rvert \lvert (p+q) \rvert < \delta * 2 = \epsilon$.
      *Non-Example 1: $f:\Reals \rightarrow \Reals$ and $f(x)=x^2$. Then $f$ is not uniformly continuous. $\delta > 0 $, we could always find $p,q\isin \Reals$ and $\lvert p-q \rvert < \delta$ such that $\lvert f(p) - f(q) \rvert = 1$. If we take $p-q = \frac{\delta}{2}$ and $p+q = \frac{2}{\delta}$, then $\lvert f(p) - f(q)\rvert = \lvert p^2-q^2 \rvert = \lvert (p-q)(p+q) \rvert = \lvert (p-q) \rvert \lvert (p+q) \rvert = \frac{\delta}{2} * \frac{2}{\delta} = 1$.
      *Non-Example 2: $f:(0, \infty) \rightarrow \Reals$, $f(x)=\frac{1}{x}$ is not uniformly continuous. Intuition: when $x\rightarrow 0$, then distance between two points $p,q$ may be close enough but $\lvert \frac{1}{p} - \frac{1}{q} \rvert$ may be large.
    *__Theorem__: Suppose $f:X\rightarrow Y$ is a continuous function between metric spaces. If $X$ is compact, then $f$ is uniformly continuous.
      * Theorem 19.2: If $f$ is continuous on a closed interval $[a,b]$, then $f$ is uniformly continuous on $[a,b]$.
    * __Proposition__: If $f:X\rightarrow Y$ is uniformly continuous and $S\subset X$ subset with induced metric, then the restriction $f|_S:S\rightarrow Y$ is uniformly continuous. 
  * **Connectedness**:
    * __Connected__: Let $X$ be a set. We say $X$ is connected if $\forall S\subset X$ and $S$ is both open and closed, then $S$ has to be either $X$ or $\text{\o}$.
      *Non-Example (From Midterm 2): $(0,2)\cap \mathbb{Q}$ is not connected. Consider the set $(0,\sqrt{2}) \cap \mathbb{Q}$ and $(\sqrt{2}, 2) \cap \mathbb{Q}$. Both open and closed but none of those two are $\text{\o}$.
    *__Proposition__: $X$ is connected $\iff$ if $X = U\sqcup V$ and $U\&V$ are both open, then one of $U,V$ is empty set.
    *__Rudin 4.22__: If $f:X\rightarrow Y$ is continuous, if $E\subset X$ is connected, then $f(E)$ is connected. (Continuity preserves connectedness)
    *__Proposition__: $[0,1] \subset \Reals$ is a connected subset.

=== Lecture 16 (Mar 11) - Covered Rudin Chapter 2 and 4 ===
  ***Review**:
    *__Induced Topology__: Given a topological space $X$, and $S\subset X$, we endow $S$ with the induced toplogy: $U\subset S$ is open in $S$ if and only if $\exists V \subset X$ open in $X$ such that $U = V\cap S$.
      *Example: $X= \Reals$ and $S= \Z \subset \Reals$, with induced topology on $S$, $\forall n \isin \Z$, $\{ n \}$ is open in $S$ since $\{ n \} = (n-\frac{1}{2}, n+\frac{1}{2}) \cap S$
    *__Corollary__:
      * If $S\subset X$ is open in $X$, then $U\subset S$ is open in $S$ if and only if $U$ is open in $X$.
      * If $S\subset X$ is closed in $X$, then $E\subset S$ is closed in $S$ if and only if $E$ is closed in $X$.
  * **Connectedness**:
    * __Lemma__: $E$ is connected if and only if $E$ cannot be written as $A\cup B$ when $A^- \cap B = \text{\o}$ and $A\cap B^- = \text{\o}$ (closure taken with respect to ambient space $X$).
    * __Rudin 4.27__: $E\subset \Reals$ is connected$\iff \forall x,y\isin E, x<y$, we have $[x,y]\subset E$
    * __Rudin 4.23__: Let $f$ be a continuous real function on the interval $[a,b]$. If $f(a)<f(b)$ and if $c$ is a number such that $f(a)<c<f(b)$, then there exists a point $x\isin [a,b]$ such that $f(x)=c$.
  * **Discontinuities**：
    * __Discontinuous__: $f:X\rightarrow Y$ is discontinuous at $x\isin X$ if and only if $x$ is a limit point of $X$ and $\lim_{x\to p} f(q)$ either does not exist or $\neq f(x)$.
    * __Right and Left Limit__: Let $f:(a,b)\rightarrow \Reals$ (not necessarily continuous) 
      * $\forall x\isin [a,b)$, we say $f(x+) = q$ if for all sequence $(t_n)$ in $(x,b)$ that converge to $x$, we have limit $\lim_n f(t_n) = q$.
        * Example: $f(x) = 1$ if $x\geq 0$, $f(x)=0$ if $x < 0$. Then $f(x+) =1 $.
      * $\forall x\isin (a,b]$, we say $f(x-) = q$ if for all sequence $(t_n)$ in $(a,x)$ that converge to $x$, we have limit $\lim_n f(t_n) = q$.
        * Example: $f(x) = 1$ if $x\geq 0$, $f(x)=0$ if $x < 0$. Then $f(x+) =1$.
    * __Discontinuity of First and Second Kind__: $f:(a,b)\rightarrow \Reals$, $x\isin (a,b)$. Suppose $f$ is discontinuous at $x$
      * We say $f$ has a simple discontinuity or discontinuity of the first kind at $x_o$ if both $f(x_o+)$ and $f(x_o-)$ exists.
        * Example: $f(x) = \frac{1}{n}$ if $x\isin \mathbb{Q}, x= \frac{m}{n}$ with $m,n$ coprime, $f(x) =0$ otherwise. $\forall x\isin \mathbb{Q}$, we have a simple discontinuity.
      * We say $f$ has a discontinuity of second kind, if it is not a simple discontinuity.
        *Example: $f(x) = \sin \frac{1}{x}, x>0$, $f(x) = x, x\leq 0$. Since $f(0+)$ does not exist, thus $f$ has a discontinuity of second kind at $x=0$.

=== Homework 8 ===
  * If $K\subset \Reals^n$ is compact and $C\subset \Reals^n$ is closed, then $K+C$ is closed.

==== Week 9 ====
=== Lecture 17 (Mar 16) - Covered Rudin Chapter 4 and 7 ===
  ***Monotonic Functions**:
    * __Monotonic Functions__: A function $f:(a,b)\rightarrow \Reals$ is monotone increasing if $\forall x>y$, we have $f(x) \geq f(y)$. Similarly one can define monotone decreasing functions.
      * Example: $f(x) = \log(x)$ is a monotone increasing function, $g(x) = 1$ is both a monotone increasing function and a monotone decreasing function.
      * Non-Example: $f(x) = x^2+2x+1$.
    * __Rudin 4.29__: Suppose $f:(a,b)\rightarrow \Reals$ is a monotone increasing function, then $\forall x\isin (a,b)$, the left limit $f(x-)$ and the right limit $f(x+)$ exists, satisfying $\sup\{f(t)|\, t<x\} = f(x-) \leq f(x+) = \inf\{f(t)|\, t>x\}$; and given $x<y$ in $(a,b)$, then $f(x+) \leq f(y-)$.
    * __Corollary__: If $f$ is monotone, then $f(x)$ only has discontinuity of the first kind/simple discontinuity.
    * __Rudin 4.30__: If $f$ is monotone, then there are at most countably many discontinuities.
  * **Sequence and Convergence of Functions**:
    * __Pointwise Convergence of Sequence of Sequences__: Let $(x_n)_n$ be a sequence of sequences, $x_n\isin \Reals^{\natnums}$, we say $(x_n)_n$ converges to $x\isin \Reals^{\natnums}$ pointwise if $\forall i\isin \natnums$, we have $\lim_{n\to\infty} x_{ni} = x_i$.
      * Example: $x_{ni} = \frac{i}{n+i}$, then this seq to $0$ pointwise, since for arbitrary fixed $i$, we have $\lim_{n\to\infty} x_{ni} = \lim_{n\to\infty} \frac{i}{n+i} = 0$.
    * __Uniform Convergence of Sequence of Sequences__: Let $(x_n)_n$ be a sequence of sequences, $x_n\isin \Reals^{\natnums}$, we say $x_n \rightarrow x$ uniformly if $\forall \epsilon >0$, $\exists N>0$ such that $\forall n>N$, $\sup\{\lvert x_{ni} - x_i \rvert :\, i\isin\natnums\} <\epsilon$ (also known as $d_{\infty}(x_n, x)$).
      * Non-Example: $x_{ni} = \frac{i}{n+i}$ failed to converge uniformly to $0$, since $d_{\infty}(x_n,0) = \sup \{\lvert x_{ni} \rvert :\, i\isin\natnums\} = \sup \{\frac{i}{n+i}:\, i\isin\natnums\} = 1$ (n is fixed).
    * __Pointwise Convergence of Sequence of Functions__: Given a sequence of functions $f_n \isin$ Map$(\Reals, \Reals)$, we say $f_n$ converge to $f$ pointwise if $\forall x\isin\Reals$, $\lim_{n\to\infty} f_n(x) = f(x) \iff \lim_{n\to\infty} \lvert f_n(x) - f(x) \rvert = 0$.
      * Examples：Running and Shrinking Bumps.
{{ math104-s21:s:img_d0796d4b49e0-1.jpeg?400 |}}

=== Lecture 18 (Mar 18) - Covered Rudin Chapter 7 ===
  * **Uniform Convergence**:
    * __Uniform Convergence of Sequence of Functions__: Given a sequence of functions $(f_n): X\rightarrow Y$, is said to converge uniformly to $f:X \rightarrow Y$, if $\forall \epsilon >0, \exists N>0$ such that $\forall n>N, \forall x\isin X$, we have $\lvert f_n(x) - f(x) \rvert <\epsilon$.
      * Remark: The integer $N$ depends only on $\epsilon$ in uniform convergence while $N$ could depend on both $\epsilon$ and $x$ in pointwise convergence.
      * Example: $f_n(x) = \frac{\sin(x)}{1+nx^2}$ converges uniformly. See Homework 9 Question 4 for the detailed proof.
    * __Rudin 7.8__: Suppose $f_n: X\rightarrow \Reals$ satisfies that $\forall \epsilon >0,\exists N>0$ such that $\forall x\isin X, \lvert f_n(x) - f_m(x) \rvert < \epsilon$, then $f_n$ converges uniformly (Uniform Cauchy $\iff$ Uniform Convergence).
    * __Rudin 7.9__: Suppose $f_n\rightarrow f$ pointwise, then $f_n \rightarrow f$ uniformly $\iff \lim_{n\to\infty} (\sup \lvert f_n(x) - f(x) \rvert) = 0$.
    * A sequence of functions $\{ f_n\}$ is uniformly convergent to $f:D\to\Reals\iff \lim_{n\to\infty} \sup \{\lvert f_n(x) - f(x) \rvert : x\isin D\}$.
    * __Rudin 7.10 (Weiestrass M-Test)__: Suppose $f(x) = \sum_{n=1}^{\infty} f_n(x)\, \forall x\isin X$. If $\exists M_n>0$ such that $\sup_x \lvert f_n(x) \rvert \leq M_n$ and $\sum_{n} M_n < \infty$, then the partial sum $F_N(x)=\sum_{n=1}^{N} f_n(x)$ converges to $f(x)$ uniformly.
      * Example: See last question on Midterm 2 version A.
  ***Uniform Convergence and Continuity**:
    * __Rudin 7.11__: Suppose $f_n \rightarrow f$ uniformly on set $E$ in a metric space.  of $E$, and suppose that $\lim_{t\to x} f_n(t) = A_n$. Then $\{ A_n\}$ converges and $\lim_{t\to x} f(t) = \lim_{n\to\infty} A_n$. In conclusion, $\lim_{t\to x} \lim_{n\to\infty} f_n(t) = \lim_{n\to\infty} \lim_{t\to x} f_n(t)$.
    * __Rudin 7.12__: If $\{f_n\}$ is a sequence of continuous functions on $E$, and if $f_n\rightarrow f$ uniformly on $E$, then $f$ is continuous on $E$.
    * __Rudin 7.13__: Suppose $K$ compact and
      -$\{f_n\}$ is a sequence of continuous functions on $K$
      -$\{f_n\}$ converges pointwise to a continuous function $f(x)$ on $K$
      -$f_n(x) \geq f_{n+1}(x) \forall x\isin K,\, \forall n = 1,2,...$
      * Then $f_n\rightarrow f$ uniformly on $K$.

=== Homework 9 === 
  *Let $f:X\rightarrow \Reals$ be a function on a metric space. We say $f$ is Lipschitz Continuous if there exists a $K>0$ such that for any $x,y\isin X$, we have $\lvert f(x)-f(y) \rvert \leq K*d(x,y)$. Such $K$ is called a Lipschitz constant for $f$. 

==== Week 10 ====
=== Spring Break (Mar 23, Mar 25) ===

==== Week 11 ====
=== Lecture 19 (Mar 30) - Review for Midterm 2 ===

=== Lecture 20 (Apr 1) - Midterm 2 ===

==== Week 12 ====
=== Lecture 21 (Apr 6) - Covered Rudin Chapter 5 ===
  ***Derivative**:
    *__Differentiability__: Let $f:[a,b]\rightarrow \Reals$ be a real valued function. Define $\forall x\isin [a.b]$, $f'(x) = \lim_{t\to x} \frac{f(t)-f(x)}{t-x}$. If $f'(x)$ exists, we say $f$ is differentiable at this point $x$.
      *Example: $f(x)=x^2$. For $x=3$, define $g(x) = \frac{f(x)-f(3)}{x-3} = \frac{x^2-9}{x-3} = x+3$. Then $f'(3) = \lim_{x\to 3} g(x) = \lim_{x\to 3} x+3 = 6$.
      *Non-Example: $f(x) = x\sin \frac{1}{x}$ if $x>0$ and $f(x)=0$ if $x\leq 0$, then f is not differentiable at $x=0$. For $x>0$, $g(x) = \frac{f(x) - f(0)}{x-0} = \frac{x\sin \frac{1}{x}}{x} = \sin \frac{1}{x}$, and $\lim_{x\to 0^+} g(x)$ does not exist, thus $f'(0)$ does not exist as well.
    *__Proposition__: If $f:[a,b]\rightarrow \Reals$ is differentiable at $x_o\isin [a,b]$, then $f$ is continuous at $x_o$, i.e. $\lim_{x\to x_o} f(x) = f(x_o)$.
    *__Rudin 5.3__: Let $f,g: [a,b]\rightarrow \Reals$. Assume $f,g$ are differentiable at point $x_o\isin [a,b]$, then 
      - $\forall c\isin \Reals$, $(cf)'(x_o)=cf'(x_o)$
      - $(f+g)'(x_o) = f'(x_o)+g'(x_o)$
      - $(fg)'(x_o)=f'(x_o)g(x_o)+f(x_o)g'(x_o)$ (Leibniz's Rule)
      - if $g(x_o)\neq 0$, then $(\frac{f}{g})'(x_o) = \frac{f'(x_o)g(x_o)-f(x_o)g'(x_o)}{(g(x_o))^2}$
    * __Rudin 5.5 (Chain Rule)__: Suppose $f:[a,b]\rightarrow I\subset \Reals$ and $g: I\rightarrow \Reals$. Suppose for some $x_o\isin [a,b]$, $f(x_o) = y_o$, $y_o\isin \Reals$, $f'(x_o)$ and $g'(y_o)$ exists. Then, the composition $h=g \circ f:[a,b]\rightarrow \Reals$. $h(x)=g(f(x))$ is differentiable at $x_o$, $h'(x_o) = g'(y_o)f'(x_o)$.
      * Example: $h(x) = \sin(x^2)$, then by Chain rule, $h'(x) = 2x\cos(x^2)$
  * **Mean Value Theorem**:
    * __Local Maximum and Minimum__: Let $f:[a,b]\rightarrow \Reals$. We say $f$ has a local maximum at point $p\isin [a,b]$, if $\exists \delta >0$ and $\forall x\isin[a,b]\cap B_{\delta}(p),\, f(x) \leq f(p)$. Local minimum is defined in the similar way.
      * Example: Say $f:[-\pi,\pi] \rightarrow \Reals$ and $f(x)=\sin(x)$. Then we can say $p= \frac{\pi}{2}$ is a local maximum with $\delta = \frac{\pi}{4}$. And similarly, we can say $q=0$ is a local minimum with $\delta = \frac{\pi}{2}$.
    * __Rudin 5.8__: Let $f:[a,b]\rightarrow \Reals$. If $f$ has a local maximum or minimum at $p\isin (a,b)$, and if $f$ is differentiable at $p$, then $f'(p)=0$. 
      * Remark: The point $p$ cannot be taken at the endpoints of the domain.
    * __Rolle's Theorem__: Suppose $f:[a,b]\rightarrow \Reals$ is a continuous function and $f$ is differentiable in $(a,b)$. If $f(a)=f(b)$, then there is some $d\isin (a,b)$ such that $f'(d) = 0$.

=== Lecture 22 (Apr 8) - Covered Rudin Chapter 5 ===
  ***Mean Value Theorem**:
    *__Rudin 5.9 (Mean Value Theorem)__: Let $f,g: [a,b]\rightarrow \Reals$ be continuous function differentiable on $(a,b)$. Then $\exists d\isin (a,b)$ such that $[f(a)-f(b)]g'(d) = [g(a)-g(b)]f'(d)$.
    *__Rudin 5.10__: Let $f: [a,b]\rightarrow \Reals$ be continuous function differentiable on $(a,b)$. Then $\exists d\isin (a,b)$ such that \\ $[f(b)-f(a)] = [b-a]f'(d)$.
      *Remark : Mean Value Theorem relates slope at a point to the difference of values of the functions.
    *__Corollary__: Suppose $f: [a,b]\rightarrow \Reals$ be continuous function, $f'(x)$ exists for all $x\isin (a,b)$, and $\lvert f'(x) \rvert \leq M$ for some constant $M$. Then $f$ is uniformly continuous.
    *__Rudin 5.11__: Suppose $f$ is differentiable in $(a,b)$, then
      - If $f'(x) \geq 0$ for all $x\isin (a,b)$, then $f$ is monotonically increasing.
      - If $f'(x) = 0$ for all $x\isin (a,b)$, then $f$ is constant.
      - If $f'(x) \leq 0$ for all $x\isin (a,b)$, then $f$ is monotonically decreasing.
  ***Intermediate Value Theorem**:
    *__Rudin 5.12__: Assume $f$ is differentiable over $[a,b]$ with $f'(a)<f'(b)$. Then from each $\lambda \isin (f'(a),f'(b))$, there exists a $d\isin (a,b)$ such that $f'(d)=\lambda$.
  ***L'Hospital's Rule**:
    *__Rudin 5.13__: Suppose $f$ and $g$ are real and differentiable in $(a,b)$, and $g'(x)\neq 0$ for all $x\isin (a,b)$, where $-\infty \leq a<b\leq \infty$. Suppose $\frac{f'(x)}{g'(x)} \to A$ as $x\to a$. Then if $f(x) \to 0$ and $g(x) \to 0$ as $x\to a$, or if $g(x) \to +\infty$ as $x\to a$, then $\frac{f(x)}{g(x)}\to A$ as $x\to a$.
      *Example: $f(x)=\frac{\sin(x)}{x}$. $\lim_{x\to 0} \sin(x) = 0$ and $\lim_{x\to 0} x= 0$, then by L'Hospital's Rule, $\lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{\cos(x)}{1} = 1$.
      *Another Example: $f(x)=\frac{\log(x)}{x}$. $\lim_{x\to\infty} x = +\infty$. Thus by L'Hospital's Rule, $\lim_{x\to\infty} f(x) = \lim_{x\to\infty} \frac{\frac{1}{x}}{1} = 0$.

=== Homework 10 ===
  * Let $f:[a,b]\rightarrow \Reals$ be differentiable, then $f'(x)$ cannot have any simple discontinuities.
  * Even if a sequence of differentiable functions converges uniformly to $f$, $f$ is not guaranteed to be differentiable.

==== Week 13 ====
=== Lecture 23 (Apr 13) - Covered Rudin Chapter 5 ===
  * **Higher Order Derivatives**:
    * __Definition__: If $f'(x)$ is differentiable at $x_o$, then we define $f"(x_o)=(f')'(x_o)$. Similarly, if the (n-1)th derivative $f^{(n-1)}$ exists and differentiable at $x_o$, we define $f^{(n)}(x_o) = (f^{(n-1)})'(x_o)$
      * Example: $f(x)=\sin(x)$. Then $f'(x)=\cos(x)$, $f"(x)=-\sin(x)$, $f^{(3)}(x)=-\cos(x)$, $f^{(4)}(x)=\sin(x)$, ...
    * __Smooth Function__: $f(x)$ is a smooth function on $(a,b)$ if $\forall x\isin (a,b)$, $\forall k\isin \{1,2,...\}$, $f^{(k)}(x)$ exists.
      *Example: $f(x)=\sin(x)$; $f(x)=e^{\frac{-1}{x^2}}$; smooth step function; smooth bump function.
  * **Taylor Theorem**:
    * __Rudin 5.15__: Suppose $f$ is a real function on $[a,b]$, $n$ is a positive integer, $f^{(n-1)}$ is continuous on $[a,b]$, $f^{(n)}(t)$ exists for every $t\isin (a,b)$. Let $\alpha, \beta$ be distinct points of $[a,b]$, and define $P(t) = \sum_{k=0}^{n-1} \frac{f^{(k)}(\alpha)}{k!} (t-\alpha)^k$. Then there exists a point $x$ between $\alpha$ and $\beta$ such that $f(\beta) = P(\beta) + \frac{f^{(n)}(x)}{n!}(\beta - \alpha)^n$.
    *__Taylor Series for a Smooth Function__: If $f$ is a smooth function on $(a,b)$, and $\alpha \isin (a,b)$, we can form the Taylor Series: \\ $P_{\alpha}(x)=\sum_{k=0}^{\infty} \frac{f^{(k)}(\alpha)}{k!} (x-\alpha)^k$.
      *Remark: RHS is not guaranteed to converge and even if it converges, it may not equal to $f(x)$.

=== Lecture 24 (Apr 15) - Covered Rudin Chapter 3 and 6 ===
  * **Taylor Series**:
    * Remark: Taylor expansion is finite term expansion with remainder while Taylor series is infinite sum with no remainder.
    * __Nth Order Taylor Expansion__: $P_{x_o,N}(x) = \sum_{n=0}^{N} f^{n)}(x_o) * \frac{1}{n!} (x-x_o)^n$.
    * __Definition__: Let $N\to\infty$, we write $P_{x_o}(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_o)}{n!} (x-x_o)^n$ (Taylor Series at $x_o$).
    * __Rudin 3.39__: Consider power series $\sum_{n} c_n z^n$, put $\alpha = \lim_{n\to\infty} \sup \lvert c_n \rvert^{\frac{1}{n}}$. Let $R=\frac{1}{\alpha}$ (if $\alpha = 0$ then $R=+\infty$; if $\alpha =+\infty$ then $R=0$), then the series is convergent if $\lvert z \rvert <R $ and the series is divergent if $\lvert z \rvert >R$. Such $R$ is called the radius of convergence.
      * Remark: If $\lvert z \rvert = R$, it depends.
    * Example: $f(x)=\frac{1}{1+x^2}$, find its Taylor series based at $x=0$.
      * Directly manipulate, $\forall \lvert x^2 \rvert < 1$, $\frac{1}{1+x^2} = \frac{1}{1-\alpha}= 1+\alpha +{\alpha}^2 + ... = 1 + (-x^2) + (-x^2)^2+...=1-x^2+x^4-... +(-x^2)^n+...$
      * Radius of convergence: $\lvert c_n \rvert = 0$ if odd and $\lvert c_n \rvert = 1$ if even. Hence $\lim_{n\to\infty} \sup \lvert c_n \rvert^{\frac{1}{n}} = 1 =\alpha$, then $R=\frac{1}{\alpha}=1$.
    * Remark: Taylor expansion is a way to approximate a smooth function near a given point, but the approximation is not uniform over the entire domain of $f$.
  * **Riemann Integral**:
    * __Partition__: Let $[a,b]\subset \Reals$ be a closed interval. A partition $P$ of $[a,b]$ is finite set of number in $[a,b]$: $a=x_0 \leq x_1 \leq ... \leq x_n=b$. Define $\Delta x_i=x_i-x_{i-1}$.
      * Example: $[10, 20]\subset \Reals$, then a partition would be $P = \{10, 15, 18, 19, 20\}$.
    * __U(P,f) and L(P,f)__: Given $f:[a,b]\to \Reals$ bounded, and partion $p = \{x_0 \leq x_1 \leq ... \leq x_n\}$, we define $U(P,f) = \sum_{i=1}^{n} \Delta x_i M_i$ where $M_i= \sup \{f(x), x\isin [x_{i-1},x_i]\}$; $L(P,f) = \sum_{i=1}^{n} \Delta x_i m_i$ where $m_i= \inf \{f(x), x\isin [x_{i-1},x_i]\}$.
{{ math104-s21:s:img_0a0c56a64ed8-1.jpeg?400 |}}
    * __U(f) and L(f)__: Define $U(f) = \inf_{P} U(P,f)$ and $L(f)= \sup_{P} L(P,f)$.
      * Since $f$ is bounded, hence $\exists m,M\isin \Reals$ such that $m\leq f(x)\leq M$ for all $x\isin [a,b]$, then $\forall P$ partition of $[a,b]$, $U(P,f) \leq \sum_{i=1}^{n} \Delta x_i M = M(b-a)$, and $L(P,f) \geq m(b-a)$, and $m(b-a)\leq L(P,f) leq U(P,f) \leq M(b-a)$.
    * __Riemann Integrable__: We say a function $f$ is Riemann integrable if $U(f)=L(f)$,
      *Some sufficient conditions: 
        - If $f$ is continuous, then $f$ is Riemann integrable.
        - If $f$ is monotone, then $f$ is Riemann integrable.

=== Homework 11 ===
  * A function $f$ is convex if for any $x,y\isin\Reals$ and any $t\isin [0,1]$, we have $tf(x)+(1-t)f(y)\geq f(tx + (1-t)y)$.
  * If $f: \Reals\to\Reals$ is a differentiable and convex function, then $f'(x)$ is monotone increasing.
  * Real and bounded function $\neq $ Riemann integrable. 
    * Example: $f(x) =1$ if $x\isin \mathbb{Q}$ and $f(x)=0$ if $x\isin \Reals\setminus\mathbb{Q}$. Then $U(f) = 1$ and $L(f) = 0$, since $U(f) \neq L(f)$, then $f$ is not Riemann integrable even if $f$ is real and bounded.

==== Week 14 ====
=== Lecture 25 (Apr 20) - Covered Rudin Chapter 6 ===
  ***Stieltjes Integral**:
    *__Weight Function__: Let $\alpha: [a.b]\to\Reals$ be a monotone increasing function, then $\alpha$ could be referred to as a weight function for Stieltjes Integral. We refer to $\Delta \alpha _i = \alpha(x_i) - \alpha(x_{i-1})$.
    *__Basic Notions__: Similar to what we defined in Riemann Integral, we define $U(P,f,\alpha) = \sum_{i=1}^{n} M_i \Delta\alpha_i$ and $L(P,f,\alpha) = \sum_{i=1}^{n} m_i \Delta\alpha_i$.
    *__Stilejes Integrable__: If $U(f,\alpha) = L(f,\alpha)$, we say $f$ is integrable with respect to $\alpha$ and write $f\isin \mathscr{R}(\alpha)$ on $[a,b]$.
      *Remark: If $\forall x\isin [a,b]$, $m\leq f(x)\leq M$, then $m (\alpha(b)-\alpha(a)) \leq L(P,f,\alpha) \leq U(P,f,\alpha) \leq M(\alpha(b) - \alpha(a))$.
    *__Refinement__: Let $P$ and $Q$ be 2 partitions of $[a,b]$, then $P$ and $Q$ can be identifies as a finite subset of $[a,b]$. We say $Q$ is a refinement of $P$ if $P\subset Q$ as subsets of $[a,b]$.
      *Example: $[a,b]=[0,10]$. Let $P = \{ 0, 1, 2,3,4,5,6,7,8,9,10\}$, and $Q=\{0, 0.5, 1, 1.5,2,3,4,5,6,7,8,9,9.9,10\}$. We could claim that $Q$ is a refinement of $P$ on $[0,10]$. 
    *__Common Refinement__: Let $P_1$ and $P_2$ be 2 partitions of $[a,b]$, then $P_1 \cup P_2$ is the common refinement of $P_1$ and $P_2$.
    *__Rudin 6.4__: If $P'$ is a refinement of $P$, then $L(P',f,\alpha) \leq L(P,f,\alpha)$ and $U(P',f,\alpha) \leq U(P,f,\alpha)$. 
    *__Rudin 6.5__: $L(f,\alpha) \leq U(f,\alpha)$.
    *__Rudin 6.6(Cauchy Condition)__: $f\isin \mathscr{R}(\alpha) \iff \forall \epsilon >0, \exists P$ partition such that $U(P,f,\alpha)-L(P,f,\alpha) < \epsilon$. 
    *__Rudin 6.7__: 
      *If Rudin 6.6 holds for $P$, then for any refinement $Q$ of $P$, $U(Q,f,\alpha)-L(Q,f,\alpha) < \epsilon$. 
      *If Rudin 6.6 holds for $P$, and let $s_i, t_i\isin [x_{i-1},x_i] \forall i = 1,2,...,n$, then $\sum_{i=1}^{n} \lvert f(s_i) - f(t_i) \rvert \Delta\alpha_i < \epsilon$.
      *If $f\isin\mathscr{R}(\alpha)$ and the above holds, then $\sum_{i=1}^{n} \lvert f(s_i) \Delta\alpha_i - \int fd\alpha \rvert < \epsilon$.
    *__Rudin 6.8__: If $f$ is continuous on $[a,b]$, then $f\isin\mathscr{R}(\alpha)$ on $[a,b]$.
    *__Rudin 6.9__: If $f$ is monotonic on $[a,b]$ and $\alpha$ is continuous, then $f\isin \mathscr{R}(\alpha)$.

=== Lecture 26 (Apr 22) - Covered Rudin Chapter 6 ===
  * **More on Integrations**:
    * __Rudin 6.10__: If $f$ is discontinuous only at finitely many points, and $\alpha$ is continuous where $f$ is discontinuous, then $f\isin \mathscr{R}(\alpha)$.
    * __Rudin 6.11__: Let $f:[a,b]\to [m,M]$ and $\phi:[m,M]\to \Reals$ is continuous. If $f$ is integrable with respect to $\alpha$, then $h=\phi \circ f$ is integrable with respect to $\alpha$.
      * If $f_1,f_2\isin\mathscr{R}(\alpha)$ and $c\isin\Reals$, then $f_1+f_2,cf_1\isin\mathscr{R}(\alpha$, and $\int f_1+f_2 d\alpha = \int f_1 d\alpha + \int f_2 d\alpha$, $\int cf_1 d\alpha = c \int f_1 d\alpha$.
      * If $f,g\isin\mathscr{R}(\alpha)$ and $f(x)\leq g(x)\forall x\isin [a,b]$, then $\int_{a}^{b} fd\alpha \leq \int_{a}^{b} gd\alpha$.
      * If $f\isin\mathscr{R}(\alpha)$ on $[a,c]$, then $f\isin\mathscr{R}(\alpha)$ on $[a,b]$ and on $[b,c]$ if $a < c< b$, and $\int_{a}^{c} fd\alpha = \int_{a}^{b} fd\alpha + \int_{b}^{c} fd\alpha$.
      * If $f\isin\mathscr{R}(\alpha)$ on $[a,b]$, and $\lvert f(x) \rvert \leq M$ on $[a,b]$, then $\lvert \int_{a}^{b} fd\alpha \rvert \leq M(\alpha(b)-\alpha(a))$.
      * If $f\isin\mathscr{R}(\alpha_1)$ and $f\isin\mathscr{R}(\alpha_2)$ and let $c$ be a positive constant, then $f\isin\mathscr{R}(\alpha_1 +\alpha_2)$ and $f\isin\mathscr{R}(c \alpha_1)$ with $\int fd(\alpha_1 + \alpha_2) = \int fd\alpha_1 + \int fd\alpha_2$ and $\int fd(c \alpha_1) = c \int fd\alpha_1$.
    * __Rudin 6.13__:
      * If $f,g\isin\mathscr{R}(\alpha)$, then $fg\isin\mathscr{R}(\alpha)$.
      * If $f\isin\mathscr{R}(\alpha)$, then $\lvert f\rvert \isin\mathscr{R}(\alpha)$ and $\lvert \int_{a}^{b} fd\alpha \rvert \leq \int_{a}^{b} \lvert f\rvert d\alpha$.
    * __Unit Step function__: The unit step function $I$ is defined by $I(x) = 0$ if $x\leq 0$ and $I(x) = 1$ if $x > 0$.
    * __Rudin 6.15__: If $f:[a,b]\to\Reals$ and is continuous at $s\isin [a,b]$ and $\alpha(x) = I(x-s)$, then $\int fd\alpha = f(s)$.
    * __Rudin 6.16__: Suppose $c_n\geq 0$ for $n=1,2,3,...$, $\sum c_n < \infty$, $\{ s_n \}$ is a sequence of distinct points in $(a,b)$, and $\alpha(x) = \sum_{n=1}^{\infty} c_n I(x-s_n)$. Let $f$ be continuous on $[a,b]$, then $\int fd\alpha = \sum_{n=1}^{\infty} c_n f(s_n)$.

=== Homework 12 ===
  * If $f$ is a continuous non-negative function on $[a,b]$ and $\int_{a}^{b} fdx=0$, then $f(x) = 0$ for all $x\isin [a,b]$. 
  * If $f,g\isin\mathscr{R}$ are real and bounded functions and $p,q>0$ such that $\frac{1}{p} + \frac{1}{q} = 1$, then $\int fgdx \leq [\int \lvert f \rvert^p dx]^{\frac{1}{p}} [\int \lvert g \rvert^q dx]^{\frac{1}{q}}$.
    * If $u,v>0$, then $uv \leq \frac{u^p}{p} + \frac{v^q}{q}$.
    * If $f,g$ are non-negative Riemann integrable functions on $[a,b]$, and $\int f^p dx = \int g^q dx = 1$ then $\int fgdx \leq 1$.

==== Week 15 ====
=== Lecture 27 (Apr 27) - Covered Rudin Chapter 6 ===
  ***Integration**:
    *__Rudin 6.17__: Assume $\alpha$ is increasing and $\alpha' \isin \mathscr{R}$ on $[a,b]$. $f$ is a bounded real function on $[a,b]$, then $f\isin\mathscr{R}(\alpha) \iff f\alpha' \isin \mathscr{R}$ and if so, $\int_{a}^{b} fd\alpha = \int_{a}^{b} f\alpha' dx$.
    *__Rudin 6.19 (Change of Variable)__: Suppose $\alpha$ is increasing on $[a,b]$ and $f\isin\mathscr{R}(\alpha)$. Suppose $\phi :[A,B]\to [a,b]$ is a strictly increasing continuous and surjective function. Define $\beta(y) = \alpha(\phi(y))$ and $g(y) = f(\phi(y))$. Then $g\isin\mathscr{R}(\beta)$ and $\int_{A}^{B} gd\beta = \int_{a}^{b} fd\alpha$. 
  ***Relation Between Integration and Differentiation**:
    *__Rudin 6.20__: Let $f\isin\mathscr{R}$ on $[a,b]$. For $a\leq x\leq b$, let $F(x) = \int_{a}^{x} f(t)dt$. Then F is continuous on $[a,b]$; furthermore if $f(x)$ is continuous at $x_o\isin [a,b]$, then $F(x)$ is differentiable at $x_o$, $F'(x_o) = f(x_o)$.
    *__Rudin 6.21 (Fundamental Theorem of Calculus)__: If $f\isin\mathscr{R}$ on $[a,b]$ and if there is a differentiable function $F$ on $[a,b]$ such that $F'=f$, then $\int_{a}^{b} f(x)dx = F(b) - F(a)$.
    *__Rudin 6.22 (Integration by Parts)__: Suppose $F,G$ are differentiable, $F',G'$ are integrable, $f=F'$ and $g=G'$. Then $\int_{a}^{b} F(x)g(x)dx = F(b)G(b) - F(a)G(a) - \int_{a}^{b} f(x)G(x)dx$.

=== Lecture 28 (Apr 29) - Covered Rudin Chapter 7 ===
  ***Uniform Convergence **:
    *__Rudin 7.16__: Let $\alpha$ be monotone increasing on $[a,b]$. Suppose $f_n\isin\mathscr{R}(\alpha)$, and $f_n\to f$ uniformly on $[a,b]$. Then $f$ is integrable and $\int_{a}^{b} fd\alpha = \lim_{n\to\infty} \int_{a}^{b} f_n d\alpha$.
    *__Corollary__: Suppose $f_n\isin\mathscr{R}(\alpha)$ and $F(x) = \sum_{n=1}^{\infty} f_n(x)$, the series converges uniformly, then $F\isin\mathscr{R}(\alpha)$ and $\int_{a}^{b} F(x)d\alpha = \sum_{n=1}^{\infty} \int_{a}^{b} f_n(x)d\alpha$.
    *__Theorem__: Suppose $\{ f_n \}$ is a sequence of differentiable functions on $[a,b]$ such that $f_n'(x)$ converges uniformly to $g(x)$ and $\exists x_o\isin [a,b]$ such that $\{f_n(x_o)\}$ converges. Then $f_n(x)$ converges to some function $f$ uniformly and $f'(x)=g(x)=\lim_{n\to\infty} f_n'(x)$.

==== Questions ====
  - {{math104-s21:s:img_769dec51bb88-1.jpeg?400|}} I understand the visualization of this recursive sequence, but  to $\sqrt{5}$?
  - In general, how to prove a set is infinite(in order to use theorem 11.2 in Ross)?
  - Is there a way/analogy to understand/visualize the closure of a set?
  - Is there a way to actually test if a set is compact or not instead of merely finding finite subcovers for all open covers?
  - Rudin 4.6 states that if $p\isin E$, a limit point, and $f$ is continuous at $p$ if and only if $\lim_{x\to p} f(x) = f(p)$. Does this theorem hold for $p\isin E$ but not a limit point of $E$?
  - {{math104-s21:s:img_0324.jpg?400|}} How is the claim at the bottom proved?
  - Could we regard the global maximum as the maximum of all local minimums?
  - Under what circumstance would Taylor Theorem be essential to our proof, since for the question appeared in HW11 Q2, I really do not think using Taylor Theorem is a better way to prove the claim?
  - In order for a Taylor series to converge ($\sum_{n} c_n z^n$), $\lvert z \rvert < R$ where $R$ is the radium of convergence. But if $\lvert z \rvert = R$, how can we tell?
  - If we are claiming $f$ is continuous on $[a,b]$, , i.e. do we just extend our interval to the left side of $a$ and right side of $b$ to do so?
  - What information can we extract from the line "$f$ has a bounded first derivative (i.e. $\lvert f' \rvert \leq M$ for some $M>0$)"?
  - How sequentially compact without proving that it is compact? (Starting from ms too complicated to take into account all sequences in the set)
  - If $a_{n+1} = \cos (a_n)$ and choose $a_1$ such that $0 < a_1 < 1$, is $a_n$ a ?
  - Does uniform convergence on a sequence of functions $\{f_n\}$ in $F$ to $f$ imply ? 
  - If $\sum f_n$ converges uniformly, does it imply $f_n$ satisfies Weiestrass M-test?
  - For the alternating series test, if instead of sequence of numbers we have sequence of functions and those functions $\{ f_n \}$ satisfies $f_1 \geq f_2 \geq f_3 ...$ and $f_n \geq 0$ for all $x\isin X$, $\lim f_n = 0$, does that mean $\sum_{n} (-1)^n f_n$ converges uniformly?
  - What is measure zero? (Related to Lebesgue measure and volume of open balls)
  - Is there any covering for a compact set that satisfies total length of the finite subcovers for the set is less than $\lvert b-a \rvert$? (Answer is no)
  - Question 16 on Prof Fan's practice exam.
  - This is my solutions towards the practice exam: {{ math104-s21:s:practice_solutions.pdf |}}