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Introduction to Real Analysis pt 1

 Hi everyone, here are my notes for the real analysis course I have been taking. 

The book I have referred to: 

Elementary Analysis: The Theory of Calculus Book by Kenneth A. Ross

Review

Theorem: [ Rational Root Theorem]

Suppose $c_0,c_1,\dots, c_n$ are integers and $r$ is a rational number satisfying the polynomial equation $$ c_nx^n+c_{n-1}x^{n-1}+\dots+c_1x+c_0=0$$ where $n\ge 1,c_n\ne 0, c_0\ne 0.$ If $r=\frac{c}{d}$ where $c,d$ are integers and $(c,d)=1$. Then $c|c_0, d|c_n$. 

Example: Prove that $\sqrt[3]{6}$ is not a rational number.

Proof: Note that $\sqrt[3]{6}$ is solution of $x^3-6=0$. But by rational root theorem, the only possible rational solutions of $x^3-6=0$ are $\pm 1, \pm 2, \pm 3, \pm 6$. But none of these eight numbers satisfies $x^3-6=0$. 

Example: All numbers of the form $5^n-4n-1$ are divisible by $16$.

Proof: Our $n$th proposition is $$P_n: 5^n-4n-1 \text{ is divisible by } 16$$ The basis for induction $P_1$ is clearly true, since $5^1-4\cdot 1-1=0.$ Proposition $P_2$ is also true because $5^2-4\cdot 2-1=16$. For the induction step, suppose $P_n$ is true. Then $$P_{n+1}= 5^{n+1}-4(n+1)-1=5(5^n-4n-1)+16n.$$ Since $5^n-4n-1$ is divisible by $16$ by induction hypothesis, it follows that $P_{n+1}$ is also divisible by $16.$

Absolute  values and inequalities

Define: For numbers $a$ and $b$ we define $dist(a,b)=|a-b|$. It represents distance between $a$ and $b$. 

  •  $|a|\ge 0$ for all $a\in \Bbb{R}$ 
  •  $|ab|=|a|\cdot |b|$ for all $a,b\in \Bbb{R}$ 
  •  Triangle inequality: $|a+b|\le |a|+|b|$ for all $a,b \in \Bbb{R}$ 

Theorem: $$\text{dist}(a,c)\le \text{dist}(a,b)+\text{dist}(b,c)$$ 

Proof:  It is enough to show that $$|a-c|\le |a-b|+|b-c|.$$

Note that $$|(a-b)+(b-c)|\le |a-b|+|b-c|.$$ And we are done.

Exercise:

Problem 1: Prove that $$||a|-|b||\le |a-b|.$$

Proof: It is enough to show that $$(||a|-|b||)^2\le (|a-b|)^2$$ or show $$|a|^2+|b|^2-2|a||b|\le a^2+b^2-2ab$$

or show $$-2|a||b|\le -2ab$$

which is true.

Problem 2: Prove that $$|a+b+c|\le |a|+|b|+|c|.$$

Proof: $$|a+b+c|\le |a+b|+|c|\le |a|+|b|+|c|$$ 

Remark: Similarly $$|a_1+a_2+\dots +a_n|\le |a_1+a_2+\dots+ a_{n-1}|\le \dots |a_1|+|a_2|+\dots+|a_n|.$$

Problem 3: Let $a,b\in \Bbb{R}$. Show if $a\le b_1$ for every $b_1>b$, then $a\le b$.

Proof: Suppose not. Then $a-b=\epsilon_0>0$. Then consider $b_1=b+\epsilon_0/2 $.

The Completeness Axiom

Definition: Let $S$ be a nonempty subset of $\Bbb{R}$.

  •  If $S$ contains a largest element $s_0$ i.e $s_0$ belongs to $S$ and $s\le s_0$ for all $s\in S$, then we call $s_0$ the maximum of $S$ and write $s_0= \text{max } S$.
  •  If $S$ contains a least element $s_0$ i.e $s_0$ belongs to $S$ and $s_0\le s$ for all $s\in S$, then we call $s_0$ the minimum of $S$ and write $s_0= \text{min } S$. \end{definition}

Definition: Let $S$ be a non-empty subset of $\Bbb{R}$

  • If a real number $M$ satisfies $s\le M$ for all $s\in S$, then $M$ is called an $\textit{upper bound}$ of $S$ and the set $S$ is said to be $\textit{ bounded above}$.
  • If a real number $m$ satisfies $s\ge m$ for all $s\in S$, then $m$ is called an $\textit{lower bound}$ of $S$ and the set $S$ is said to be $\textit{ bounded below. }$
  •  The set $S$ is said to be $\textit{bounded}$ if it is bounded above and bounded below. Thus $\exists$ $m,M$ st $S\subseteq [m,M]$. 

Definition:  Let $S$ be a non-empty subset of $\Bbb{R}$.

  • If $S$ is bounded above and has a least upper bound, then we will call it the $\textit{supremum}$ of $S$ or $\text{sup}S$.
  •  If $S$ is bounded below and has a maximum lower bound, then we will call it the $\textit{infimum}$ of $S$ or $\text{inf}S$.


Theorem: [Completeness Axiom for $\Bbb{R}$]Every nonempty subset $S$ of $\Bbb{R}$ that is bounded above has a least upper bound i.e  $\text{Sup}S$ exists.

Remark: [Completeness Axiom fails for $\Bbb{Q}$] Consider the set $A=\{r\in \Bbb{Q}:0\le r\le \sqrt{2}\}$. This doesn't have any supremum over $\Bbb{Q}$.

Corollary: Every nonempty subset $S$ of $\Bbb{R}$ that is bounded below has a greatest lower bound $\text{inf}S$.

Proof: Let $-S:=\{-s:s\in S\}$. Since $S$ is bounded below such that there is a $m\in \Bbb{R}$ such that $m\le s\in S$ , we get that $-S$ is bounded above by $-m$. Hence by $\textbf{Completeness axiom}$, we get that $\text{sup}S$ exists. 

We claim that $\text{inf}S=-\text{sup}S$. 

Let $s_0=\text{sup}(-S)$, we need to prove $-s_0\le s, \forall s\in S$. 

If $\text{inf}S\ne -\text{sup}S$ then $\exists$ $t$ such that $t> s_0, t\not\in S, t\le s,\forall s\in S$. But then again consider the negatives, we done.

Archimedean Property

Theorem: If $a>0$ and $b>0$, then for some positive integer $n$, we have $na>b$.

Proof: [This proof is magical]

Assume that archimedean property fails. Then $\exists a>0, b>0$ such that $na\le b, \forall n\in \Bbb{N}$. Then note that $S=\{na: n\in \Bbb{N}\}$ is bounded by $b$. Using completeness axiom, we get that $\exists s_0$ such that $s_0=\text{sup}S$.

Note that $s_0-a<s_0 \implies \exists n_1\in \Bbb{N}$ such that $s_0-a<n_1a\implies s_0<(n_1+1)a$. Contradiction. 


Useful to know.

Claim: If $a>0,$ then $\frac{1}{n} <a$ for some positive integer $n$

Proof: When $b=1$ in the archimedean property.

Claim: If $b>0$, then $b<n$ for some positive integer $n$

Proof:  When $a=1$ in the archimedean property.

Remark: There are many ordered fields which do not satisfy the above two claims.

Denseness of $\Bbb{Q}$

Theorem: If $a,b\in \Bbb{R}$ and $a<b$, then there is a rational $r\in \Bbb{Q}$ such that $a<r<b$.

We need to show $a<\frac{m}{n}<b$ for some integers $m$ and $n$ where $n>0$, thus we need $an<m<bn$.

Note that $(b-a)>0\implies \exists n$ such that $n(b-a)>1$. So there exists an integer between $nb$ and $na$. And we are done. 

Exercises

Problem 1: Prove that if $a>0$, then there exists $n\in \Bbb{N}$ such that $\frac{1}{n}<a<n$.

Proof: We use the archimedean property which states that 

Claim: If $A>0$ and $B>0$, then for some positive integer $n$, we have $nA>B$

Taking $B=1,A=a$ we get that $\exists n_1$ such that $a>\frac{1}{n_1}$. 

Taking $B=a, A=1$, we get $\exists n_2$ such that $n_2>a$.

Taking $n=\max(n_1,n_2)$ we are done.

Problem 2: Let $A$ and $B$ be nonempty bounded subsets of $\Bbb{R}$, and let $A+B$ be the set of all sums $a+b$ where $a\in A$ and $b\in B$. Prove that $$\text{sup}(A+B)= \text{sup}(A)+\text{sup}(B)$$

Proof: For all $a\in A, b\in B$, $$a+b\le \text{sup}(A+B)\implies a\le \text{sup}(A+B)-b$$

Hence we have $\text{sup}(A+B)-b$ the upper bound of $A$. So $$\text{sup}A\le \text{sup}(A+B)-b\implies b\le \text{sup}(A+B)-\text{sup}A$$ $$ \implies \text{sup}B\le  \text{sup}(A+B)-\text{sup}A $$$$\implies \text{sup}A+\text{sup}B\le \text{sup}(A+B).$$

And for the other direction, we clearly have $a\le \text{sup}A, b\le \text{sup}B\implies \max(a+b)\le \text{sup} A+\text{sup} B$ 

Problem 3: Let $a,b\in \Bbb{R}$. Show that if $a\le b+1/n$ for all $n\in \Bbb{N}$, then $a\le b$.

Proof: Suppose not. Then $a-b>0.$ So by the archimedean property, we get that $a-b>1/n$ for some $n$. Which is a contradiction.

Problem 4: Show $\text{sup}A=\{r\in \Bbb{Q}:r<a\}=a$ for each $a\in \Bbb{R}.$

Proof: If $\text{sup}A=a_0\ne a$, then by denseness of $\Bbb{Q}, \exists a_0<r_0<a.$  

Well, yeah that's it for this blog post! Stay tuned for part two :)!

~Sunaina Pati

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