Math 104 Notes
Spring 2020
Professor: Mariusz Wodjzicki
Scribe: Raymond Feng
Lecture 1, 1/21/2020
$p \in \mathbb{Q}:$ the set of rational numbers. $p$ belongs to $\mathbb{Q}$. $p=\frac{m}{n}$. $\frac{m’}{n’}$, $m,n$ are integers. $\mathbb{Z}, n \neq 0$. $\mathbb{Z} \subseteq \mathbb{Q}$ Equality of fractions.
 $\frac{m}{n}=\frac{m’}{n’}$. relation: $\frac{m}{n} \sim \frac{m’}{n’}$. $mn’=nm’$. Reflextivity of the relation.
 $\frac{m}{n} \sim \frac{m’}{n’}$, then, $\frac{m’}{n’} \sim \frac{m}{n}$ Symmetry
 $\frac{m}{n} \sim \frac{m’}{n’} \sim \frac{m’’}{n’’} \Rightarrow mn’=nm’,m’n’’ =n’m’’$ to show this, do $(mn’)n’’=(nm’)n’’$. This is true by the first equality. Use the associativity of multiplication to get $m(n’n’’)=n(m’n’’)$
To formally define rational numbers, do: $F:=\mathbb{Z}x\mathbb{Z}=\frac{m}{n} m \in \mathbb{Z}, n \in \mathbb{Z}$
$Z_{*}:= \{ n \in \mathbb{Z} n \neq 0 \}$
$\mathbb{Q}:=F/n= \{c \in \mathcal{P}(F) c \text{ is an equivalent class of rel.} \sim \} $ $p=\frac{m}{n}$, this means that $p$ is in the equivalence class of the fraction. $m’ \vert n’$ means that $m’$ divides $n’$ if there exists $d \in \mathbb{Z}$ such that $n’=dm’$ On the set $\mathbb{Z}_{>0}$: divisibility is an order relation.
Assigned reading
Rudin, first section, then to the point of getting to ordered sets.
$A=\{p \in \mathbb{Q}, p^2 \leq 2\}=\{p \in \mathbb{Q}, p^2 < 2\}$ $B=\{p \in \mathbb{Q}, 2 \leq p^2\}=\{p \in \mathbb{Q}, 2 < p^2\}$
 Homework is due weekly:20%
 1st midterm:18%
 2nd midterm:22%
 Final: 40%
Lecture 2, 1/23/2020
Definition: A set $S$ equipped with a binary relation, $\sim$, satisfies the following three properties:
 Reflexivity: for any element $s \in S, s \sim s$.
 Weak antisymmetry: for any $t \in S$, if $s \sim t$ and $t \sim s$, then $s=t$.
 Transitivity: $s,t,u \in s$, if $s \sim t$, and $t \sim u$, then $s \sim u$. Is called an ordered set.
Compact form:
 (R): $\forall x \in S, s \sim s$
 (WA): $\forall s,t \in S, s \sim t \land t \sim s \Rightarrow s=t$
 (T): $\forall s,t,u \in S, s \sim t \land t \sim u \Rightarrow s \sim u$
Aside: The negation of $s \Rightarrow t$ is $s \land \neg t$. Talking about the philosophy of mathematical notation.
If you look at all the relations on a set, then there is something special about the relationship between order relations and the equivalence relation. Mentioned partially ordered set.
What is a partially ordered set?
Partially ordered set means that it is not linearly ordered.
Standard Notation:
 $s \preceq t$, $s$ less or equal than $t$.
 $s \sim t$.
Alternative property: $\forall s,t \in S (s \preceq t \lor t \preceq s)$ Linearly ordered, totally ordered.
Many of the mathematical structures are possible due to partially ordered sets. Linearly ordered structures are very rare. God given ones are even rarer. Real lines are linearly ordered sets. Non sharp orders?
Aside: Rudin’s student solved a big problem. Continuum hypothesis. A problem in set theory and logic. A shock to a lot of mathematicians. Conclusion: Rudin has very good credentials.
Suppose that someone gives you an order relation. Then you can immediately define a new order relation. Given an order relation $\preceq$, let $\prec$ be defined as follows: $s \preceq$ and $s \neq t$. If $\preceq$ is an order relation, then $\prec$ satisfies transitivity, and weak antisymmetry becomes strict symmetry.
$s \preceq t \equiv s \prec t \lor s=t$
Thus, the alternative property becomes: $\forall s,t \in S (s \prec t \lor s=t \lor t \preceq s \lor t=s)$. However the last equality is redundant. This is what is known as the Trichotomy law. You can make this law stronger by modifying it to become: $\forall s,t \in S$, exactly one of the $s \prec t, t \prec s, s=t$ is true.
Aside: Mind exposed to to much rubbish, SNR is too low. But Rudin is elegant and has high SNR.
Need to know how to transfer nonsharp order to sharp order, and vice versa.
Suppose a sharp order relation $\prec$ is given on a set $S$. Define $\preceq$ by $s \preceq t$ if $s \prec t \lor s=t$
Possible midterm problem: If this relation is sharp, then this other relation satisfies the three axioms of nonsharp.
Sharp linear orders will satisfy Trichotomy law, nonsharp orders will satisfy Alternative property. From now on, know the relationship between sharp ($\prec$) and nonsharp ($\preceq$) orders.
Followup: $S$ is equipped with a fixed order relation. Suppose $E \subseteq S$. $s \in S$ is an upper bound/lower bound for $E$ if for $\forall e \in E, e \preceq s$ or $\forall e \in E, s \preceq e$ respectively.
$\mathcal{U}E=\{s \in S \vert \forall e \in E, e \preceq s\}$ $\mathcal{L}E=\{s \in S \vert \forall e \in E, s \preceq e\}$
Figure: light cone pyramid looking thing. $\{t \in S \vert t \preceq s\}=\langle s ]$. $E \subseteq \langle s ]$. $\mathcal{L}\{s\}$. $s \in \mathcal{U}E$.
Suppose that $E$ has the greatest element $m \in E$. $\forall e \in E$, for ever $m \preceq s, e \preceq s$ for all $e \in E$. In this case $\mathcal{U}E=\{ t \in S \vert m \preceq t \} = [ m \rangle$
The situation when $\mathcal{U}E$ is an upper interval $[ \beta \rangle$ is very important. The “lower end” of $\mathcal{U}E$, is referred to as the supremum of $E$, or, the least upper bound for $E$.
Dually, when $\mathcal{L}E$ is a lower interval $\langle \alpha ]$ is equally important. The upper end of $\mathcal{L}E$ is referred to as the infimum of $E$.
Note: he keeps mentioning real lines for some reason.
A couple of very simple observations. $E \subseteq F \Rightarrow \mathcal{U}F \subseteq \mathcal{U}E \Rightarrow \mathcal{LU}E \subseteq \mathcal{LU}F$. $E \subseteq \mathcal{LU}E$. $E \subseteq \mathcal{UL}E$
“Get ready for a surprise”. $\mathcal{ULU}E \subseteq \mathcal{U}E$. $\mathcal{U}E \subseteq \mathcal{ULU}E$
The punchline: $\mathcal{U}E = \mathcal{ULU}E$, and $\mathcal{L}E=\mathcal{LUL}E$. Closure…
$\mathcal{U}(\mathcal{LU}E)=\mathcal{U}(\langle \alpha ])=[ \alpha \rangle$. Finishing analogy. That’s why he’s a mathematician. He likes building technology of the mind. Anybody can use a plane, but some like to build it.
Assigned reading: Read to the end of the complex numbers. Definition of a field. Ordered field means that any number is comparable to zero.
Homework assignment #1: From 1st chapter, exercises 15, problem 6 parts a and b. Due in one week.
HW1
1.1 If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ and $rx$ are irrational.
1.2 Prove that there is no rational number whose square is 12.
1.3 Prove proposition 1.15
 Proposition 1.15: The axioms for multiplication imply the following statements.
 (a) If $x \neq 0$ and $xy=xz$, then $y=z$
 (b) If $x \neq 0$ and $xy=x$, then $y=1$
 (c) If $x \neq 0$ and $xy=1$, then $y = \frac{1}{x}$
 (d) If $x \neq 0$, then $\frac{1}{\frac{1}{x}}=x$
1.4 Let $E$ be a nonempty subset of an ordered set; suppose $\alpha$ is a lower bound of $E$ and $\beta$ is an upper bound of $E$. Prove that $\alpha \leq \beta$
1.5 Let $A$ be a nonempty set of real numbers which is bounded below. Let $A$ be the set of all numbers $x$, where $x \in A$. Prove that $\inf{A} =  \sup{A}$.
1.6 Fix $b > 1$.
Lecture 3: 1/28/20
 Ordered fields. Recall that a field in algebra is an algebraic structure consisting of a set, say, F, equipped with 2 operations. The first one is referred to as addition, the second one is referred to as multiplcation.
 $F,F \Rightarrow (\text{addition and multiplication}) F$
 $x,y \Rightarrow (\text{addition}) x+y$
 $x,y \Rightarrow (\text{multiplication}) xy$
 Both multiplication and addition are associative.
 $x+y+z=x+(y+z)=(x+y)+z$.
 Note that the individual variables are not getting permuted.
 It matters which variable comes first. We are not assuming that $x+y=y+x$.
 $(x,y,z)=(x+y,z)=(x+y)+z=(x,y,z)=(x,y+z)=x+(y+z)$.
 This is defined as an associative operation on the abstract set $F$.
 This is already a very powerful structure.
 Terminology: A set equipped with an associative, binary operation is called a (half)semigroup. Groups were already invented in the 1870s. Associativity is the most important property of an operation.
 Theory of Semigroups
 Present anywhere we encounter evolution (in time).
 $S$: a certain system. Maybe a hybrid system.
 $S_t$ is the state of the system at time $t$. Then $S_0$ will be your initial state.
 $S_0 \rightarrow (H_t) S_t \rightarrow (H_s) S_{s+t}$.
 Then: $S_0 \rightarrow (H_{s+t}=H_s \dot H_t \forall s,t \geq 0) S_{s+t}$
 $H_0$ is the identity operator.
 $H$ is a semigroup operator.
 $(X, \text{structure of a certain kind})$
 The structure must be prescribed and fixed.
 $(X, \text{structure of a certain kind}) \rightarrow (\text{interaction, fromto}) (X’,S’)$.
 very often just a function from $X$ to $X’$
 $X$ and $X’$ are the underlying sets of the structures.
 Compatible with the structures
 $(X, \text{binary operation})$ and $(X’, \text{binary operation}’)$
 $X \rightarrow (f) X’$ such that $x,y \rightarrow f(x),f(y)$
 Def: A function between the underlying sets of semigroups is a homomorphism if the graphic on the board holds.
 Much of this course is the theory of metric spaces
 Metric spaces are a structure
 A set equipped with a distance function
 Isometries strict, less strict is continuous mappings, open mappings.
 Study this on your own or in another class. This was a motivation
 (X,\dot (\text{binary operation})). If there exists a 2sided “identity” (element) sometimes called a “neutral element”
 Denote such an element as $e$.
 $\forall x \in X, xe=ex=x$
 If an identity exists, it is unique.
 Suppose $e’$ is another identity.
 Indeed $e’=ee’=e$. We get this because e is a left identity and $e’$ is a right identity.
 Take any set $X$, consider the function of two variables
 $x,y \rightarrow x$
 Thus $xy=x$ for ever $x,y \in X$
 Drops the second item.
 An identity element will be the identity element.
 Very important type of structure called a monoid
 Dfn: A semigroup with the identity element is called a monoid.
 Dfn: $X$ be a given element in a monoid, there exists an element $y \in X$ s.t. $xy=e=yx$, we say that $x$ is invertible.
 For the unique inverse of $x$, $y=x^{1}$. This does not mean that we have powers! This is simply notation.
 In additive notation: $x$. $x,y \rightarrow x+y$.
 If you have right inverse and left inverse, they are the same, provided that my operation is associative.
 Given $x \in X$, suppose $xy=e=xy’$
 $(yx)y’=(\text{associative})y(xy’)$
 $ey’=ye \Rightarrow y’=y$
 $(X,\dot)$ is a monoid.
 $X^{*} \subset X$ are the invertible elements.
 Closed with respect to multiplication
 $x,y \in X^{*}$
 $x(yy^{1})x^{1}=xex^{1}=xx^{1}=e$
 $(xy)^{1}=x^{1}y^{1}$
 If $X=X^{*}$, we say that $(X,\dot)$ is a group.
 $(X,+,\dot)$ A set, addition, and multiplication
 Suppose that $(X,+)$ is a group, with commutative operation, its identity is defined 0, and is referred as the zero element.
 $(X,\dot)$, a monoid, its identity is denoted 1, and is referred to as the identity element.
 $0+x=x=x+0, 1x=x=x1$
 The whole point of the two operations multiplication and addition is that…
 Suppose that you have two additively written semigroups.
 $(X,+) \rightarrow (f) (X’,+’)$ Is called homomorphism.
 if $f(x+y)=f(x)+’f(y)$. We call such functions “additive”
 If we have a second operation
 $(X,+,\dot) \rightarrow (f) (X’,+’,\dot’)$
 $x+y,z \rightarrow (x+y)z$. This is called the right distributative property.
 $x,y \rightarrow xy$. Multiplication is assumed to be additive in each argument.
 Additivity conditions
 (1) For 1st argument $(x+y)z=xz+yz$
 (2) 2nd argument
 We are essentially freezing $z$.
 Mult by $z$ on the right. Is additive in all of $X$.
 Mult by $z$ on the left is additive for all $z \in X$
 $X \rightarrow X$
 $x \rightarrow xz \leftrightarrow (x+y)z=xz+yz$
 $x \rightarrow zx \leftrightarrow z(x+y)=zx+zy$
 $0x=0=x0$ for every $x \in X$. In particular, 0 cannot be invertible, or else $0=1$.
 If $0=1$, then $x=1x=0x=0$
 If $X \text{set complement} \{0\}$ is closed under multiplication.
 Example: $x=\mu_2{\mathbb{Z}}$
 A ring is very important, mobile phones and stuff.
 A ring is a group…?
 This structure $(X,+,\dot)$ has a division ring, then it is a field.
HW1
1.1 Prove that if $r$ is rational and $r \neq 0$ and $x$ is irrational, then $rx$ and $r+x$ is irrational.
 Suppose the oppposite, suppose $rx$ was rational, then because $\mathbb{Q}$ is closed under multiplication, we would have taht $r^{1}rx=x$ is also rational, which is a contradiction. Furthermore, assume that $r+x$ is rational. Because $\mathbb{Q}$ is closed under addition, then that means that $r+x+(r)=x$ must also be rational, which is a contradiction.
1.2 Prove that there is no rational number whose square is 12.
 $\sqrt{12}=2\sqrt{3}$. So then you just need to show that $\sqrt{3}$ is irrational. Suppose the contrary. Suppose that there is a $r^2=\frac{a^2}{b^2}=3$, where $a$ and $b$ are relatively prime. Then $a^2=3b^2$. Because $a^2$ is divisible by 3, $a$ is also divisible by 3, and can be written as $a=3k$. Substituting this back in, we get $9k^2=3b^2 \Rightarrow 3k^2=b^2$. This means that $b$ is also divisible by 3. This contradicts the assumption that $a$ and $b$ have no common factor.
1.3 Prove the following statements from Proposition 1.15: (a) If $x \neq 0$
 (a) We know that the multiplicative inverse $\frac{1}{x}$ exists, and so multiplying on both sides by this inverse we have $\frac{1}{x}(xy)=\frac{1}{x}(xz)=(\frac{1}{x}x)y=(\frac{1}{x}x)z=1y=1z=y=z$.
 (b) You can apply (a) with $z=1$ to get this result
 (c) You can apply (a) with $x=\frac{1}{x}$ to get this result
 (d) Set the variables in (a) to be $x \leftarrow \frac{1}{x}$, $y \leftarrow \frac{1}{\frac{1}{x}}$, $z \leftarrow x$.
1.4 Let $E$ be a nonempty subset of an ordered set; suppose $\alpha$ is a lower bound of $E$, and $\beta$ is an upper bound of $E$. Prove that $\alpha \leq \beta$.
 By the definition of GLB and LLB, we know that for some element $\forall x \in E, \alpha \leq x \leq \beta$. Then, by the definition of an ordered set, either $\alpha < \beta$, or $\alpha = x = \beta$.
1.5 Let $A$ be a nonempty set of real numbers which is bounded below. Let $A$ be the set of all numbers $x$, where $x \in A$. Prove that $\inf A =  \sup{A}$.
 Suppose $\alpha = \sup{A}$. We want to show that $\forall x \in A, \alpha \leq x$ and $\beta \text{ is lower bound of } A \Rightarrow \alpha \geq \beta$. To prove that $\alpha$ is a lower bound, $x \in A \Rightarrow x \in A \Rightarrow x \leq \sup{A} \Rightarrow x \geq \sup{A} \Rightarrow x \geq \alpha$. To show that $\alpha$ is the greatest lower bound, i.e. $\forall x \in A, \beta \leq x \Rightarrow \forall x \in A, \beta \geq x$. This means that $\beta$ is an upper bound of $A$. Thus $\beta \geq \sup{A} \Rightarrow \beta \leq \sup{A}$. So $\sup{A}$ is the GLB of $A$.
1.6 Fix $b > 1$. (a) If $m,n,p,q$ are integers, $n>0,q>0,$ and $r=\frac{m}{n}=\frac{p}{q}$, prove that $(b^m)^{\frac{1}{n}}=(b^p)^{\frac{1}{q}}$. (b) Prove that $b^{r+s}=b^rb^s$ if $r,s$ are rational.
 To do (a), define $a=mq=pn$. Then, we show that taking both of these values to the $nq$th power, we will get $((b^m)^{\frac{1}{n}})^{nq}=((b^p)^{\frac{1}{q}})^{nq} \Leftrightarrow (b^m)^q=(b^p)^n \Leftrightarrow b^a=b^a$. We have reached a true statement, meaning that the first statement that we started from must’ve been true. Because $(b^m)^{\frac{1}{n}}$ and $(b^p)^{\frac{1}{q}}$ are both positive, we know that if they become equivalent when taken to the $a$th power, then they themselves are equivalent.
 For (b), assume that $r=\frac{a}{b}$ and $s=\frac{c}{d}$. Then adding them we get $r+s=\frac{ad+cb}{bd}$. Then, replacing this into our expression we get: $b^{\frac{ab+cd}{bd}}=(b^{ab}b^{cd})^{\frac{1}{bd}}=b^{\frac{ab}{bd}}b^{\frac{cd}{bd}}=b^rb^s$.
Lecture 4: 1/30/2020
 Ordered fields: A field, say $F$, equipped with a linear order compatible with the operations is called an ordered field.
 Compatibility conditions:
 Addition: For every element $a \in F$ adding $a$ preserves the order. For any two elements $x,y \in F$, if $x < y \Rightarrow x + a < y + a$.
 Warning: In an ordered set, say $S$, we can think of the ordering relation as a sharp order, or nonsharp order. $\prec \leftrightarrow \preceq$. We have a faithful translation. Sharp is never reflexive, nonsharp is always reflexive.
 $S(\prec) \rightarrow(f) S’(\preceq)$. If $f$ respects the sharp ordering, i.e. $\forall s,t \in S$, (if $s < t$, then $f(s) <’ f(t))$.
 $\forall s,t \in S, s < t \Rightarrow f(s) <’ f(t)$.
 The truth value of this expression is…unknown?
 (There exists) $\exists s,t \in S (s < t \Rightarrow f(s) <’ f(t))$
 Compatibility conditions:
 $(X, d) \leftrightarrow (X’,d’)$ with $X,X’$ are sets.
 $d:X,X’ \rightarrow [0,\infty)$, the distance function
 The distance function is an abstract function that must satisfy the following properties
 $\forall x,y,z \in X, d(x,z) \leq d(x,y) + d(y,z)$ The most fundamental principle, the triangle inequality. Wodzicki thinks it’s the most fundamental, but Rudin and others will usually list this last.
 Symmetry: $\forall x,y \in X, d(x,y)=d(y,x)$, apparently…not that important?
 Nondegeneracy: $\forall x,y \in X (d(x,y)=0 \Leftrightarrow x=y)$
 This is the definition of a metric space.
 $X=\mathbb{R}^2$. $x=(x_1,x_2)^T, y=(y_1,y_2)^T$. $d_m(x,y)=x_1y_1+x_2y_2$. This is the Manhattan distance.
 $d_e(x,y)=\sqrt{x_1y_1^2+x_2y_2^2}$
 In general $d_{L_p}(x,y)=(x_1y_1^p+x_2y_2^p)^{\frac{1}{p}}$
 Manhattan is when $p=1$, Euclidean is when $p=2$.
 $F$ is either $\mathbb{R}$ or $\mathbb{C}$. $V=\mathbb{F}^n$. $d_{L_p}(x,y)=(\sumx_iy_i^p)^{\frac{1}{p}}$.
 In this case, the Triangle Inequality is the famous classical inequality. It is not easy to prove. This is called Holder’s Inequality.
 The vector space of continuous $F$ ordered functions on the interval $V=C([a,b],F)$.
 Weird diagram on board. Distance between two functions. The functions don’t even have to be continuous.
 $d_{L_p}(f,g)=(\int_a^b f(x)g(x)^p dx)^{\frac{1}{p}}$. All you need to know is that $f(x)g(x)^p$ is integrable. Ensuring that $f$ and $g$ are continuous gives you this.
 What does the Riemann Integral have to do with all of this? Cauchy could not prove that the every continuous function is Riemann Integrable.
 Lebesgue integrable $\Rightarrow$ Riemann Integrable, Math 105, more demanding, Honors class flavor.
 $L_p([a,b],F)$. Linear algebra in infinite funcion spaces is called functional analysis.
 This course was supposed to lead you to…something something metric
 Next Homework Assignment: Chapter 1, Problem 6cd, 7, 8, 9, 10 is optional
 First midterm every four five weeks?
 Metric spaces, possibly the number one definition in the whole course
 Suppose you have insert the drawing of the two sets…
 Dfn: We say that the function $f: X \rightarrow X’$ between metric spaces, is continuous at $x \in X$ if $\forall \epsilon > 0 \exists \delta > 0 \forall y \in X (d(x,y) < \delta \Rightarrow d’(f(x),f(y)) < \epsilon)$.
 This is the $\epsilon$$\delta$ “epsilondelta” definition of continuity.
 This will be a great quiz problem.
 Note that this statement does not commute.
 The negation of this statement is:
 $\exists \epsilon > 0 \forall \delta > 0 \exists y \in X (d(x,y) < \delta \land d’(f(x),f(y)) \geq \epsilon)$
 Why $\epsilon$ and why $\delta$? $\epsilon$ means error, and $\delta$ is the next letter in the alphabet.
 Midterm: Chapter 1, portions of Chapter 2.
 Reading: Until the end of compact sets.
Lecture 5: 2/4/2020
Ordered fields
 Dfn: A field $F$ that is equipped with a linear order relation, $\preceq$, which is compatible with the two operations is called an ordered field.
 Compatibility Conditions:
 Compatibility with addition: $\forall x,y,z \in F, (x < y \Rightarrow x+z < y+z)$
 Compatibility with multiplication: It is not possible that ever element is invertible, because of course, zero is not invertible.
 Dfn: $F_{+} := \{x \in F  0 < x\}$ is called the “positive cone” (the set of all positive elements in the ordered field $F$).
 Dfn: $F_{}:=\{x \in F  x < 0\}$
 Thus $F$ is the union of 3 distinct disjoint sets $F_{} \bigcup \{0\} \bigcup F_{+}$. In other words, every $x \in F$ is negative, zero, or positive, and only one of the three hold.
 Observation: $x \in F_{+} \Leftrightarrow x \in F_{}$. It’s enough to show the $\Rightarrow$ implication because $(x)=x$.
 Proof: $x+(x)=0 < x \Rightarrow 0+(x) < x+(x) \Rightarrow x < 0$. Scratch this…
 Proof:
 Hypothesis: $0 < x$
 Then: $0+(x) < x+(x)$
 Then: $x < 0$
 What is the importance of linearly ordered fields? They are rare.
 The field of rational numbers.
 How do we order rational numbers?
 $p < q \Leftrightarrow 0 < qp$
 Suppose you lost the order relation. Suppose you only know $F_{+}$. Then you know the order relation by simply checking if subtraction will bring you into $F_{+}$! This is cool.
 $x,y \in F_{+} \Rightarrow x+y \in F_{+}$
 Proof: $0 < x \Rightarrow y < x + y$. Since $0 < y$, but the transitivity property of the order relation $\preceq$, $0 < x + y$.
 Cones are very important in applied math.
 “A very easy problem.”  Mariusz Wodjzicki
 “If you fail this test, may god help you.”  Big M
 Problem: Let $C \subset F$ be a subset which is closed under addition, multiplication and $0 \notin C$. $C = F_{+}$ for some linear order on $F$, making $F$ an ordered field.
 $\Leftrightarrow$
 $\forall x \in C, x \notin C$, Let $C_{}:=\{x \in F  x \in C\}$. The above condition means that $C \bigcap C = \emptyset$.
 $F=C \bigcup C_{} \bigcup \{0\}$
 This is the standard way to order any field.
 $\mathbb{Q_{+}}=\{q \in \mathbb{Q}  q \text{is represented by a fraction } \frac{m}{n} \text{ with } m,n > 0\}$.
 In this case, we have $\mathbb{Q_{}}=\{ q \in \mathbb{Q}  q \text{ is }\}$
 $1 \in F$ and $(1)(1)=1$. For any $x \in \mathbb{F_{}}, (x)(x) \in F_{+}$
 $\forall x \neq 0, x^2 \in F_+$
 On a field $1 \neq 0$, but $(n)(1)$ can be zero for a positive integer. $1+…+1=n * 1$, $n$ is a positive integer.
 Listen: $1 > 0$, need to prove this.
 By property A, $F_+$ is closed with respect to addition. $n * 1 \in F_+$ for any positive integer $n$. In other words, an ordered field has characteristic zero.
 Def: If $F$ is any field, its characteristic is the smallest positive integer, say $p$, such that $p * 1 = 0$. If no such $p$ exists, we say that $F$ has zero characteristic.
 If $\exists m,n > 0 \in \mathbb{Z}, n * 1 = m * 1$. If $n \neq m, n > m \lor n < m$. Hence $(nm) * 1 = 0$.
 Can you show that $n * 1  m * 1 = (nm) * 1$
 Gotcha! Distributivity holds when all elements are of set $F$, but not of the integer set.
 Computers understand in terms of induction: $nx \Rightarrow (n+1)x=x+nx$
 Mclain, Introduction to Abstract Algebra: Just Algebra
 There is no order relation on it that makes a field an ordered field, a field of 80 elements
 If there exists any ordering on a field, $n * 1$…
 $e_n := n * 1$, a sequence, defined in every field $F$.
 $1,1+1,…$. Axiom of choice, well ordering. If there exists order relation on $F$, making $F$ an ordered field, then all terms of this sequence are distinct. Why?
 If for some $m \neq n \land m * 1 = n * 1 \Rightarrow (m  n) * 1 = 0 \lor (n  m) * 1 = 0$
 Digression: If a negative integer, one can define…something about Abelian groups and stuff.
 Some serious things
 Every ordered field $F$ contains “a copy” of the ring of integers.
 Canonical Embedding
 Homomorphisms of rings.
 Fractions represent an attempt to divide m by n.
 Every ordered field contains a copy of rational numbers!
 There are other ordered fields that are not real numbers and all inclusive.
 Rudin is optimal.
 $\mathbb{Z} \subset F$ bounded above and below
 $A \subset F \text{ bounded above} \Leftrightarrow A \text{ bounded below}$
 Def: We say that an ordered field is Archimedean if $\mathbb{Z}$ is not bounded.
 Observation: If $F$ is not Archimedean, then $\mathbb{Z}$ is bounded above…
 Incompleteness: Just the most basic set of integers is without supremum.
 Suppose that $\mathbb{Z}$ is bounded above, and $\alpha = \sup{\mathbb{Z}} \in F$. Then $\alpha  1$ being
HW 2: CH 1, 6cd, 7,8,9, (10 is optional)
Problem 6cd
 (c) We know that $b^r$ is an upper bound of $B(r)$, so just need to show that it is the least such upper bound. To do this,
 (d)
Problem 7
 (a) Use induction. The base case is $n=1$, and we have $b^11 \geq 1(b1)$. Then, the inductive step gives us $(b^{n+1}1) \geq (n+1)(b1)$. $(b^nb1) \geq n(b1)+(b1) \Leftrightarrow b^nb \geq n(b1)+b$. Which is true from the inductive hypothesis.
 (b) If you substitute $b$ with $b^{\frac{1}{n}}$ in part (a), you will get this as the result.
 (c) The inequality is equivalent to $n(t1) > (b1)$. We can then use the previous part and the transitive property to say that $n(t1) > n(b^{\frac{1}{n}}1) \Rightarrow t > b^{\frac{1}{n}}$.
 (d) Plugging in $t=yb^{w}$ to the previous part.
 (e) Plugging in $t=b^w(\frac{1}{y})$ to part c.
 (f) Either $b^x < y \lor b^x > y \lor b^x = y$, there is no other possibility. If $b^x < y$, then that means that $x+(\frac{1}{n}) \in A$ for some $n$, by part d, meaning that $x$ is not a UB of $A$. If $b^x > y$, then (e) tells us that $x\frac{1}{n}$ is an upper bound, meaning that $x$ cannot be the LUB of $A$. Thus, it must be that $b^x=y$.
 (g) $y \neq x$. WLOG $y > x$. Then, $b^y=b^{x+(yx)}=b^xb^{yx} > b^x$.
Problem 8
Because $i \neq 0$ by definition, we know that $i \in C_+ \lor i \in C_+$. Then, $i^2=1$ must be positive as well. But this is a contradition because then both 1 and 1 would be positive. Because $i$ cannot be either zero, or in either the positive or negative cones, the complex field cannot be ordered.
Problem 9
To show that the set is ordered, we need to show that $\forall a,b \in C, a=b \lor a < b \lor b < a$. Suppose $a = c + di, b = e + fi$, where $c,d,e,f$ are real numbers. Because the real numbers are an ordered set, we can use that to split into three cases:  (1) $c < e \Rightarrow a < b$  (2) $c > e \Rightarrow a > b$  (3) $c = e$, we can further split this into three cases * $d < f \Rightarrow a < b$ * $d > f \Rightarrow a > b$ * $d = f \Rightarrow a = b$ However, this ordering does not give the ordered set the least upper bound property. Suppose we have the subset $E \subset C$, where $E = \{ik  k \in \mathbb{R}\}$. This subset is upper bounded by $1$, but does not have a least upper bound, because for any $(a,b) \in C$ that is an upper bound of $E$, then $(\frac{a}{2},b)$ would be a smaller upper bound.
Lecture 6
The next homework: Chapter 2, Basic Topology, Problems 17
 A NonArchimedean ordered field
 Consider the set of rational functions $f$. These functions are represented by fractions $\frac{P}{Q}$, where $P,Q$ polynomials, $Q \neq 0$.
 Given two fractions $\frac{P}{Q} \sim \frac{P’}{Q’}$ if $PQ’=QP’$. Multiplication is defined by $\frac{P}{Q} * \frac{P’}{Q’} := \frac{PP’}{QQ’}$.
 Addition is defined by $\frac{P}{Q}+\frac{P’}{Q’}:=\frac{PQ’+QP’}{QQ’}$.
 Fractional equivalence relation…I think I am missing the insight here. What is special about this?
 $[\frac{0}{Q}] \sim 0$
 Claim: there is an ordering on the field of rational functions.
 If you can order a field, then you can introduce a positive element on that field.
 Visualization:
 $f$ and $g$ is on the board.
 What is the meaning of being positive? $\forall x, f(x) > 0$. But this ordering is not linear.
 $f  g$ being positive is really looking at …
 The zeros of $f$ is where $f$ changes from positive to negative.
 The great idea: Maybe David Hilbert came up with this?
 When $f$ is rational, it vanishes at finitely many points. Zeros of $f$, if $f=\frac{P}{Q}$, is contained in the roots of $P$.
 $f(x) = \frac{x^21}{x1} \sim \frac{x+1}{1}$.
 There are points where $Q$ vanishes, and points where $P$ vanishes. Both these points are finite though.
 Hilbert’s Big Idea
 Check the right half line, after the last occurence of one of these points.
 Define, for rational functions $f$, $g$, $f < g$ if there exists $a \in \mathbb{R}$, such that $f(x) < g(x)$ for all $x \in a$.
 Prove that this is a sharp order, prove you have an ordered field
 But it is not Archimedean.
 Prove that the field of rational functions is not complete.
 If it were complete, it would have to be Archimedean
 How would you prove a field is Archimedean
 If $F$ is a complete ordered field, then $F$ is Archimedean.
 Proof: Let $A = \{1,2,…\} \subset F$. If $A$ is bounded above, then, by completeness of $F$, it has a supremum. Denote that supremum by $\alpha = \sup{F}$. Then, $\alpha  1$ is not an upper bound for $A$.
 Hence, $\exists n \in A, n \nleq \alpha  1$. But in an ordered field, any two elements are compatible.
 $\exists n, n > \alpha  1 \Leftrightarrow \alpha < n + 1$ But $n+1 \in A$, contradiction.
 This only works with functions with finite number of zeroes.
 If we want to show that the ordered field is not complete, we want to show that it has no supremum.
 In Archimedean field, $\forall x,y \in F, \exists r \in Q, x < r < y$
 Let $F$ and $G$ be two complete ordered fields. $F \subset Q, G \subset Q$. Let $F$ be a function from $F$ to $G$ which respects the two operations and the order. If you have a function that goes from a group to a group, and perserves the two operations, it must take zero to zero.
 $\forall x \in F, f(nx) = nf(x), f(x+…+x)=f(x)+…+f(x)$, $f(1)=1$. $f(\frac{m}{n})=\frac{f(m)}{f(n)}$. $f(\frac{m}{n})=f(m*n^{1})=f(m)f(n^{1})$.
 Thus, we have shown that $\forall r \in Q, f(r)=r$.
 $r < x < s$, $f(r) = r < f(x) < f(s) = s$
 Both these fields contains copies of rational numbers.
 $r < f(x) < s, \forall r < x < s$.
 Every complete ordered field, every element $x$ is $\sup{r \in Q  r < x}$, and is $\inf{r \in Q  r > x}$.
 In a complete set, this number exists, but in an incomplete set, this number may not exist.
 For an element $x \in F$, let $A_x = \{r \in Q  r < x\}$, $B_x = \{s \in Q  x < s\}$, Note: $x = \sup{A_x} = \inf{B_x}$. If $f$ is a homomorphism of fields and preserves order, then $\sup{A_x} \leq f(x) \leq \inf{B_x}$.
 Suppose $\exists y,z, y \neq z, r < y < z < s…$
 Prove that in every Archimedean field, there exists one such element where
 You are building the set of reals as the subset of the power set of rationals.
Lecture 7 2/11/2020, Basic concepts of the Theory of Metric Spaces
 $(X,d)$. $X$ is a set, $d$ is a distance function.
 $d: X,X \rightarrow [,\infty)$
 Triangle Inequality: $\forall p,q,s \in X d(p,s) \leq d(p,q) + d(q,s)$
 Symmetry: $\forall p,q \in X d(q,p) = d(p,q)$
 Nondegeneracy: $\forall p,q \in X d(p,q)=0 \Leftrightarrow p=q$
 $p \in X, E \subseteq X, E^C=X \setminus E$. $p \in E \oplus p \notin E$, $p \notin E \Leftrightarrow p \in E^C$.
 Properties of (placement of) a point within a subset (in a given metric space).
 Def 1. We say that $p \in X$ is an interior point of a subset $E$ if $\exists \tau > 0 p \in \{q \in X  d(p,q) < \tau\} \subseteq E$. Rudin: $N_r(p)$: All points in the ball $r$. There are open balls and closed balls.
 The neighborhood of any point $p$ of a metric space $X$ form a family of subsets of $X$ indexed by the set of positive reals. $N_r(p)$, $r \in (0, \infty)$.
 Terminology: $S \rightarrow^{f} T$. functions (domain) with the source $S$ and the target $T$ (codomain).
 $S \mapsto$ the value of $f$ at $s$. $f(s)$
 Special scenarios:
 $S$ a subset of the set of natural numbers. $S \rightarrow X$, especially if $S = \{0,1,2,3,…\}$ or $\{1,2,…\}$.
 $x \mapsto 3x^22x+5  y=3x^22x+5$.
 Sets with operations: Algebra, Sets with families of subsets: Topology
 $n \mapsto x_n \in X$
 $i,j,k,l$, you can use the notation $x(n)$.
 $s \rightarrow \mathcal{P}(X)$ sequences of elements of the power set of a set $X$. We are talking: sequences of subsets of $X$.
 Given any set $S$, functions $S \rightarrow \mathcal{P}(X)$ are referred to as families of sebsets of $X$. $S \mapsto E_S$. $(E_s)_{s \in S}$.
 $S=\{1,…,n\} \rightarrow X$. $n=2$ is a pair, $n=3$ is a tuple. $\{x_1,…,x_n\}$ ntuple of elements of ste $X$. $(x_n)_{1 \leq n \leq n}$.
 Rudin uses often, for general families, Greek letters, $x_{\alpha}, x_{\lambda}$. $\alpha \in \Lambda, \lambda \in \Lambda$.
 Standard notation for today for $N_r(p)$ “ball”. $B_n(p)$ or $B_p(r)$ or $B(p,r)$.
Terminology
 Bounded: $E \subseteq X$, $E$ is bounded if there is a real number $M$ and a point $q \in X$ such that $d(p,q) < M$ for all $p \in E$.
 $\forall p \in E \exists M \in \mathbb{R} \exists q \in X, d(p,q) < M$.
 Closed: $E \subseteq X$, $E$ is closed if every limit point of $E$ is a point of $E$.
 Limit Points: $E \subseteq X$. A point $p$ is a limit point of the set $E$ if every neighborhood of $p$ contains a point $q \neq p$ such that $q \in E$.
Homework 3: Ch2 17
2.1 If $E$ is the empty set and $A$ is an arbitrary set, we want to show that $E \subseteq A$. To do this, we just need to show that $\forall e \in E, e \in A$. However, the hypothesis of this implication is always false, and so the implication is vacuously true.
2.2 Not sure how to use the hint. The set of all integer coefficient polynomials with degree at most $n$ is countable, because they can be represented by set of all $n+1$ tuples of integers. Thus, all possible integer polynomials can be represented by the union of all degree at most $n$ polynomials. The union of countably many countable sets is countable. Furthermore, each polynomial in our set must have at most $n$ roots, and thus, the set of all algebraic numbers are countable.
2.3 The reals are uncountable. If there were no reals that are not algebraic, then you have a 1 to 1 mapping from reals to algeraics, meaning that the reals are countable. This is a contradiction, so there must be reals that are not algebraic.
2.4 No. Suppose that the set of irrational numbers were countable. We know that the set of rational numbers are countable as well. This means taht we can easily find a one to one relationship between the natural numbers and the reals. This is because a real number is either rational or irrational. However, the reals are uncountable. Thus, irrational numbers must be uncountable.
2.5 Wouldn’t simply (0,1] $\bigcup$ (2,3) suffice? The limit points are 0,1,3, and this set is clearly bounded.
Lecture 8
 $X$ is a set
 $\mathcal{P}(X) \rightarrow ()^C \mathcal{P}(X)$, $E \mapsto E^C$.
 This is the god given operation on the powerset.
 The complement
 $S \rightarrow (\mu) S$. A unary operation.
 $0^0$ makes sense here because we have the identity function.
 If $X$ has metric structures. If $(X,d)$ indeed is a metric space structure.
 $E \mapsto \bigcup_{G \in E \text{ open}} G$.
 $E \subset X$, $E’ = \{p \in X  p \text{ is a limit point of } E \text{ in } X\}$
 We say that $E$ is closed if $E’ \subseteq E, (E’=E$) \text{ (perfect)}$
 Associate with $E$ the subset of $X: \mathcal{Cl}(E) := E \bigcup E’$ the closure of $E$.
 $E \mapsto \mathcal{Cl}(E)=\bigcap_{E \subset F \text{ closed}}F$
 $\bigcap_{\alpha \in A}F_{\alpha}$

$A \mapsto \mathcal{P}(X), \alpha \mapsto F_{\alpha}$
 Next week’s hw: 814 Chapter 2
Lecture 9
 $(E_{\alpha}){\alpha \in A}, E{\alpha} \subset X, K \subset X$ covers $K$ if (def) $K \subset \bigcup_{\alpha \in A} E_{\alpha}$. $\mathbb{A} \mapsto A$.
 Berkeley is prestigious math university
 Tangent on Measure Theory. Dynamic and evolution of probabilistic spaces.
 Probability theory, makes the space compact.
 Def: $K4 is compact if every open cover $(G_{\alpha})$ of $K$ has a finite subcover. $(E_{\alpha})_{\alpha \in B}$. $B$ is finite.
Lecture 10: Compactness Part 2
 $X$ is a set, and $(E_{\alpha}){\alpha \in A}$ is a family of subsets. Then we have a family of the complements of $E{\alpha}$. $(E_{\alpha}^C)_{\alpha \in A}$.
 You can go in the forward direction, and you can also go in the backward direction.
 Suppose that the first family is a cover of $X$.