[Lecture Summary] 02 Real Anaysis : Field

This contents is based on Lecture of 18.100A of MIT Open Course. Coverage will be Lecture from 4~. link

Overview

All we need to know once we learn about it is

• Useful inequality manipulation!
  • inequality is relevent lubp!

Definition and Symbols

Field

• Two operations matter!
  • Addition and multiplication
  • Some properties should be satisfied

Ordered field

• ordered set $F$ with ordering $< s.t.$
  • $\forall x,y,z\in F, x<y \Rightarrow x+z<y+z$
  • $x>0 \mbox{ and } y>0 , \mbox{ then } xy>0$

$x+A$

• $\lbrace x+a \mid a\in A\rbrace$

$xA$

• $\lbrace xa \mid a\in A\rbrace$

Absolute value

• $\mid x\mid = x$ for $x\ge 0$ o.w. $-x$

Decimal representation(represent by digits)

• $x\in (0,1], d_{j} \in \lbrace 0,1,\cdots 9\rbrace$
• $x = \sup \lbrace 10^{-1}d_{-1}+10^{-2}d_{-2} + \cdots + 10^{-n}d_{-n} \mid n\in \mathbb{N}\rbrace$

Theorem

Let $F$ be an ordered field with the least uppber bound property. If $A \subset F$ is nonempty and bounded below, then $\inf A$ exist in $F$

Approach

• use $-A$ and find $\sup(-A)$ first
  • LUBP is impactful!
• use definition of infimum and supremum.
  • most of relation is based on the fact that it is ordered field.

There exists a unique ordered field, which is $\mathbb{R}$, satisfying that $\mathbb{Q} \subset \mathbb{R}$ and LUBP.

Approach

• Order property
  • prove existence of supremum
  • use the above theorem.
• Algebric property
  • prove uniqueness of supremum
  • uniqueness comes from equality
  • use the fact that it is a field

(Archimedian Property, AP) If $x,y\in \mathbb{R}$ and $x>0$, then $\exists n\in \mathbb{N}$ s.t. nx>y

Assumption

• $x,y\in \mathbb{R}$
• $x>0$
  • $x$ could be any positive value(even $\epsilon$!)

Want to show that

• $\exists n \in \mathbb{N} s.t. n>{y\over x}$

Approah

• Contradiction!
• $\mathbb{N}$ has lupb!
• find otehr value which is bigger than supremum

(Density of Q) If $x,y in \mathbb{R}$ and $x<y$ then $\exists r\in \mathbb{Q} s.t. x<r<y$

so to speak,

• rational numbers are dense in $\mathbb{R}$
• We can find another number which exists between for a pair of given real numbers.

Assumption

• $x,y \in \mathbb{R}$
• $x<y$

Want to show that

• $ny>nx+1$
• find $j\in \mathbb{N} s.t. nx<j, j\le nx+1$

Approach

• Split cases with respect to zero!
  • we only solve the case of positive interval. Negative inteval case will be followed.
• AP does matter!
  • Derive that $n(y-x)>1$
  • Derive $S=\lbrace k\in \mathbb{N}:k>nx \rbrace$
• WOP for the smallest value.

$x = \sup S \iff x$ is upper bounded for $S$ and $\forall \epsilon >0, \exists y \in S s.t. x-e<y\le x$

This is important theorem!

Assumption

• $S \subset \mathbb{R}$
• $S$ is nonempty and bounded above.

$\sup(x+A)=s+\sup A, \sup(xA)=x\sup A$

Assumption

• $x\in\mathbb{R}$
• $A$ is bounded above

Approach

• Derive upper bound for $x+A$
• And then derive equality!
  • Proof for existence of upper bound is not enough
  • We need to prove that exact supremum of set. Use iff relation!

(Trianle Inequality) $\forall x,y \in \mathbb{R}, \left

x+y\right

\le \left

x\right

+\left

y\right

$

Approach

• Derive inequality so that we can use 5th property of absolute value!

Remark

• Reverse Triangle Inequality($\Delta-inequality$)

  • $\forall x,y \in \mathbb{R}, \left

    \left

    x\right

    - \left

    y\right

    \right

    \le \left

    x-y\right

    $

$\forall x \in (0,1], \exists \mbox{ unique digists } \lbrace d_j : i\in \mathbb{N} \rbrace s.t. x=0.d_{-1}d_{-2}\cdots \mbox{ and } 0.d_{-1}d_{-2} < x \le 0.d_{-1}d_{-2}+10^{-n}$

• for the first constraint, there could be several decimal representations
• to defince decimal representation uniquely, the second constraint does matter. Left hand-side strict inequality let the representation uniquely!
• Eg. ${1/over 2} = 0.4999\cdots$
  • $0.50000\cdots$ does not satisfy the second constraint.

(Cantor) $(0,1]$ is uncountable

Approach

• Contradiction!
  • Suppose that there exists bijection.
    • maps from decimal representation to other values in that interval
  • Function that flips a part of decimal representation
  • Find out an element which does not map onto
• Use just the above theorem, which says that there are unique decimal representation

Want to show

• For arbitrary given (unique) real values, there does not exist decimal representations which uses natural number

Remark

• Even though $\mathbb{Q}$ is countable, $\mathbb{R}$ and $\mathbb{R} \setminus \mathbb{Q}$ is uncountable
• Proof of countablity is equivalent with derivation of bijection from $\mathbb{N}$ to a given set.
  • Countability = Same cardinality = Bijection existence