Highlights of Chapter 1

  • 1.1 Rational numbers. $Q$ is incomplete. $p^2=2$ for $p\in Q$ has no solution
  • 1.5 Ordered Set. Rudin calls a strict total order an order on S
  • 1.7 Bounds
  • 1.8 Least Upper Bound (infimum). Greatest Lower Bound (supremum). Definitions.
  • 1.10 The Least Upper Bound Property
  • 1.10 Equivalence of LUB/GLB Properties
  • 1.12 Field definition and axioms
  • 1.13 Notation and $Q$ is a field
  • 1.14-16 Familiar results for fields
  • 1.17 Ordered Fields
  • 1.18 Familiar results for ordered fields
  • 1.19 The Real Field, Existence Theorem. Proof in Chapter 1's Appendix
  • 1.20 Archimedean Property of Real Numbers
  • 1.20 Density of $Q$ in $R$
  • 1.21 Reals are complete, $y^n=x$ for $x\in R^+$ and $n\in J$ has a unique solution.
  • 1.22 Decimals
  • 1.23 Extended Reals
  • 1.24-34 Complex Number Field
  • 1.35 Schwarz Inequality
  • 1.36 Euclidean Vector Space. The Real Line. The Complex Plane. Euclidean space is also a Metric Space.
  • APPENDIX. Proof of 1.19 Existence Theorem, due to Dedekind and Cantor.

Introduction

1.1 Rudin Example. No rational number $p$ exists such that \begin{equation} \label{eq_sqrt2} p^2 = 2 \end{equation}

Proof. Suppose such $p$ exists. Then there are integers $m$ and $n$ such that $p = m/n,$ $n\neq 0$ and either $2 \nmid m$ or $2 \nmid n,$ and by \eqref{eq_sqrt2} \begin{equation} {m^2} = 2 {n^2} \end{equation} Thus, $2 \mid m^2$ so $2 \mid m.$ (If $m$ were odd $m^2$ would be.) So $4 \mid m^2$ so $4 \mid 2 {n^2}$ so $2 \mid n^2$ so $2 \mid n,$ a contradiction.

1.1.1 Rudin Lemma. Consider the sets \[ A = \{ p \in Q : p^2 < 2 \} \] and \[ B = \{ p \in Q : p^2 > 2 \} \] Then $A$ contains no largest member and $B$ contains no smallest. That is, for each rational $p$ in $A,$ we can find a $q$ in $A$ such that $p < q$ and for each $p$ in $B$ we can find a $q$ in $B$ such that $q < p.$

Proof.

Let $q$ be defined in terms of $p>0$ as follows: \begin{equation} \label{rud_q_from_p} q = p - \frac{p^2 - 2}{p + 2} = \frac{2p + 2}{p + 2} \end{equation} Then \begin{equation} \label{rud_sq_q_from_p} q^2 - 2 = \frac{2(p^2 - 2)}{(p+2)^2} \end{equation}

If $p\in A$ then $p^2-2\lt0,$ \eqref{rud_q_from_p} shows that $q\gt p,$ and \eqref{rud_sq_q_from_p} shows that $q^2\lt2.$ Thus, $q\in A$ .

If $p\in B$ then $p^2-2\gt0,$ \eqref{rud_q_from_p} shows that $0\lt q\lt p$ and \eqref{rud_sq_q_from_p} shows that $q^2\gt2.$ Thus, $q\in B.$

1.2 Rudin Remark. Rudin 1.1 shows that the rational numbers have certain "gaps", despite the fact that between any two rational numbers, there is another. The real numbers fill these gaps.

1.3 Rudin Definition. If $A$ is any set (whose elements may be numbers or any other objects), we write $x \in A$ to indicate that $x$ is a member (or an element ) of $A.$ If $x$ is not an element of $A,$ we write $x \notin A.$ The set which contains no members will be called the empty set or $\varnothing.$ If a set has at least one element, it is called nonempty . If $A$ and $B$ are sets and every element of $A$ is an element of $B,$ then we say $A$ is a subset of $B,$ or $B$ is a superset of $A,$ and write $A \subset B$ or $B \supset A,$ respectively. If, in addition, there is an element of $B$ that is not in $A,$ then $A$ is called a proper subset of $B,$ and we may write $A \subsetneq B.$ Note that $A \subset A$ for all $A.$ If $A \subset B$ and $B \subset A$ then we write $A = B.$ Otherwise, $A \neq B.$ That is, set equality is now defined.

1.4 Rudin Definition. The set of rational numbers will be denoted by $Q.$

Ordered Sets

1.5 Rudin Definition. Let $S$ be a set. An order on $S$ is a relation, denoted by $\lt,$ with the following two properties:

  1. Trichotomy. If $x\in S$ and $y\in S,$ then exactly one of the following is true: $x\lt y,$ $x=y,$ $x\gt y.$
  2. Transitivity. If $x,y,z\in S$ and $x\lt y$ and $y\lt z$ then $x < z.$

1.6 Rudin Definition. An ordered set is a set in which an order is defined.

1.7 Rudin Definition. Suppose $S$ is an ordered set, and $E\subset S.$ If there exists $\beta\in S$ such that $x\leq\beta$ for every $x\in E,$ we say that $E$ is bounded above and call $\beta$ an upper bound of $E.$ bounded below and lower bound are defined similarly with $\leq$ replaced with $\geq.$

1.8 Rudin Definition. Suppose $S$ is an ordered set, $E\subset S$ and $E$ is bounded above. Suppose there exist $\alpha,\gamma\in S$ with the following properties:

  1. $\alpha$ is an upper bound of $E.$
  2. if $\gamma\lt\alpha$ then $\gamma$ is not an upper bound of $E.$
Then $\alpha$ is called the least upper bound of $E,$ or the supremum of $E$ , or $\alpha=\sup{E},$ and $\beta$ the greatest lower bound , infimum of $E,$ $\beta=\inf{E}.$

1.9 Rudin Example.
  1. Consider the sets from Rudin 1.1 $A = \{ p \in Q : p^2 < 2 \}$ and $B = \{ p \in Q : p^2 > 2 \}.$ $A$ has no least upper bound in $Q$ and $B$ has no greatest lower bound in $Q.$
  2. If $\alpha \in \sup{E}$ exists, then $\alpha$ may or may not be a member of $E.$

1.10 Rudin Definition. An ordered set $S$ is said to have the least-upper-bound property if $\sup{E}$ exists in $S$ for every $E$ such that $E \subset S,$ $E$ is not empty, and $E$ is bounded above. The greatest-lower-bound property is defined similarly with $\sup{E}$ replaced with $\inf{E}.$

1.11 Rudin Theorem. If $S$ is an ordered set with the least upper bound property and $E$ is a nonempty, bounded subset of $S,$ then $\inf{E}$ exists in $S.$

Proof. Since $E$ is bounded below, the set of all lower bounds (in $S$) of $E$ is nonempty. Denote this nonempty subset of $S$ by $L.$ We first show that $L$ has the least upper bound. Since $L$ is the set of all lower bounds of $E,$ that means $\forall x\in L:x\leq y$; but this equivalently means that $E$ is the set of all upper bounds of $L.$ Since $E$ is nonempty it follows that $L$ is bounded above, i.e. $\exists y\in E:\forall x\in L: y\geq x.$ We have shown that $L$ is a nonempty subset of $S$ and $L$ is bounded above. Since $S$ has the least upper bound property, it follows that $L$ has the least upper bound in $S.$ Denote it by $\alpha=\sup{L}.$ If $\gamma\lt\alpha,$ then $\gamma$ is not an upper bound of $L,$ since $\alpha=\sup{L}.$ So $\gamma\notin E.$ Thus, $\forall x\in E:\alpha\leq x,$ so $\alpha$ is a lower bound of $E,$ ($\alpha \in L$). If $\gamma\gt\alpha$ then $\gamma\notin L,$ since $\alpha$ is an upper bound of $L,$ ($\forall x\in L: x\leq\alpha$); but $\gamma\notin L$ means that $\gamma$ is not a lower bound of $E.$ Thus, if $\gamma\gt\alpha$ then $\gamma$ is not a lower bound of $E.$ Since $\alpha$ is a lower bound of $E$ and $\gamma\gt\alpha$ implies $\gamma$ is not a lower bound of $E,$ it follows that $\alpha=\inf{E}.$

Fields

1.12 Rudin Definition. A field is a set $F$ with two operations, called addition and multiplication , with the following properties, called the field axioms:

  1. Axioms for Addition
    1. Closure. $F$ is closed under Addition.
      \( (\forall x,y\in F) (x+y\in F). \)
    2. Commutative. Addition is commutative in $F.$
      \( (\forall x,y\in F) (x+y=y+x). \)
    3. Associative. Addition is associative in $F.$
      \( (\forall x,y,z\in F) [(x+y)+z=x+(y+z)]. \)
    4. Additive Identity
      \( (\exists 0\in F) (\forall x \in F) (0+x=x). \)
    5. Additive Inverse
      \( (\forall x\in F) (\exists (-x)\in F) (x+(-x)=0). \)
  2. Axioms for Multiplication
    1. Closure. $F$ is closed under Multiplication.
      \( (\forall x,y\in F) (xy \in F) \)
    2. Commutative. ) Multiplication is commutative in $F.$
      \( (\forall x,y\in F) (xy=yx). \)
    3. Associative. Multiplication is associative in $F.$
      \( (\forall x,y,z\in F) [(xy)z=x(yz)]. \)
    4. Multiplicative Identity.
      \( (\exists 1 \in F) (\forall x \in F) (1x=x). \)
    5. Multiplicative Inverse
      \( (\forall x\in F) (\exists (1/x)\in F) (x(1/x)=1). \)
  3. The Distributive Law
    1. Distributive. Multiplication is distributive over addition in $F.$
      \( (\forall x,y,z\in F) [x(y+z)=xy+xz)] \)

The Real Field

1.19 Rudin Theorem. The Existence Theorem. There exists an ordered field $R$ which has the least-upper-bound property. Moreover, $R$ contains $Q$ as a subfield.

1.20(a) Rudin Theorem. The Archimedean Property of Real Numbers. If $x,y\in R$ and $x\gt0$ then there is a positive integer $n$ such that \[ nx\gt y \]

1.20(b) Rudin Theorem. Q is dense in R. If $x,y\in R$ and $x\lt y,$ then there exists a $p\in Q$ such that $x\lt p\lt y.$ That is, between any two real numbers there is a rational one.

Proof. Since $x\lt y$ we have $y-x\gt 0$ and the Archimedean property furnishes a positive integer $n$ such that \[ n(y-x)\gt 1. \] Apply Archimedes again to obtain positive integers $m_1$ and $m_2$ such that $m_1\gt nx$ and $m_2\gt -nx.$ Then \[ -m_2\lt nx \lt m_1. \] Hence there is an integer $m$ (with $-m_2\le m\le m_1$) such that \[ m-1\le nx \lt m. \] If we combine these inequalities we obtain \[ nx\lt m\le 1+nx\lt ny. \] Since $n\gt 0,$ it follows that \[ x\lt m/n\lt y. \] This proves the theorem with $p=m/n.$

1.21 Rudin Theorem. Positive real nth roots are unique. For every real $x\gt0$ and every integer $n\gt 0$ there is one and only one positive real $y$ such that $y^n=x.$

The Extended Real Number System

1.23 Rudin Definition. The extended real number system consists of the real field $R$ and two symbols, $+\infty$ and $-\infty,$ called positive infinity and negative infinity. We preserve the original order in $R,$ and define \[ -\infty\lt x\lt+\infty \] for every $x\in R.$

The Complex Field

1.27 Rudin Definition. \( i=(0,1) \)

Euclidean Spaces

Guevara Supplement for Chapter 1

1.1 Guevara Note. (Rudin 1.1) Why does Rudin choose integers $m,n$ not both even? Consider the possibilities. If $p^2 = 2$ and $p=m_0/n_0,$ then $m_0$ is even when $n_0$ is odd, since $m_0^2 = 2n_0^2.$ (See Rudin's proof for why this is so.) Therefore, it is impossible for both $m_0, n_0$ to be odd if $p^2 = 2.$ On the other hand, it is possible for both $m_0,n_0$ to be even, but then $m_0/n_0$ can be reduced to $m/n$ where $m,n$ are not both even. Therefore, Rudin safely chooses $m,n$ "not both" even, with no loss of generality. If $m,n$ are not both even, then it follows that $m$ is necessarily even, and $n$ necessarily odd. Rudin could have made this stronger assumption about $m,n$ with no loss of generality, but doing so does not change his proof.

1.2 Guevara Note. (Rudin 1.1.1) Rudin does not offer motivation for his choice of $q.$ For more information on this see Choice of q in Baby Rudin's Example 1.1

1.3 Guevara Theorem. If $S$ is a set, then $\varnothing$ is bounded above and below by every element of $S.$

Proof. Suppose not and let $\alpha\in S.$ Then \( \neg\forall x (x\in\varnothing\Rightarrow x\leq\alpha), \) (Rudin 1.7), so \( \exists x (x\in\varnothing\wedge x\gt\alpha), \) a contradiction. Bounded below proceeds similarly.

1.4 Guevara Theorem. If $S$ has a least element then $\varnothing$ has a least upper bound.

Proof. Let $\alpha$ denote the least element of $S.$ Then $\alpha=\inf{S}.$ So $\forall x\in S:\alpha\lt x,$ and $\alpha$ is an upper bound of $\varnothing$ (Guevara 1.3). Suppose $\gamma\lt\alpha$ and that $\gamma$ is an upper bound of $\varnothing.$ Then $\gamma\in S$ so $\alpha\le\gamma,$ contradiction. Therefore, if $\gamma\lt\alpha$ then $\gamma$ is not an upper bound of $\varnothing.$

1.5 Guevara Corollary. If $S$ is an infinite set, then $\sup{\varnothing}$ and $\inf{\varnothing}$ do not exist in $S,$ or equivalently, if $\sup{E}$ or $\inf{E}$ exist in $S$ for $E \subset S,$ then $E\neq\varnothing.$

Proof. TBD

1.6 Guevara Note. A set $S$ has the least-upper-bound property if every nonempty subset of $S$ that is bounded above has a least upper bound (Rudin 1.10). Note that $\varnothing$ is bounded above and yet demonstrably has no least upper bound if $S$ has no least element. Thus, if every nonempty, bounded-above subset of a set $S$ had a least upper bound and $S$ had a least element, then $S$ would not have the least-upper-bound property if the property required every subset (including $\varnothing$) to satisfy its specified conditions. Since the definition does stipulate that the subsets be nonempty, such sets $S$ still have the least-upper-bound property.

1.7 Guevara Note. Rudin 1.8 fails to state $\gamma\in S,$ so it is given correctly in this document's version. The proof of Rudin 1.11 is only valid under that assumption.

1.8 Guevara Note. Rudin's proof of Rudin 1.20(b) doesn't offer an explanation why there is an integer $m$ (with $-m_2\le m\le m_1)$ such that $m-1\le nx\lt m.$ Guevara 1.9 fills in those details.

1.9 Lemma. If $x\in(m,n)$ then there is a $k\in[m,n]$ such that $x\in[k-1,k).$

Proof. Suppose the contrary. Then for some $x,$ $x\in(m,n)$ and $x\not\in[k-1,k)$ for any $k\in[m,n].$ Thus, $x\not\in\bigcup_{k\in[m,n]}[k-1,k)$ $=[m-1,n).$ Thus, $x\not\in(m,n)\subseteq[m-1,n),$ a contradiction.