Skip to main content

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


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.


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. 


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


Popular posts from this blog

Constructions in Number Theory

Hi, I am Emon, a ninth grader, an olympiad aspirant, hailing from Kolkata. I love to do olympiad maths and do some competitive programming in my leisure hours, though I take it seriously. I had written INOI this year. Today, I would be giving you a few ideas on how to Construct in Number Theory . Well, so we shall start from the basics and shall try to dig deeper into it with the help of some important and well-known theorems and perhaps, some fancy ones as well. Okay, so without further delay, let's start off... What do we mean by "Constructions"? If noticed, you can see that you often face with some problems in olympiad saying, "... Prove that there exists infinitely many numbers satisfying the given conditions" or "... Prove that there exists a number satisfying the above conditions." These are usually the construction-problems .  For example, let's consider a trivial example : Problem. Prove that there exist infinitely many integers $a$ such th

EGMO solutions, motivations and reviews ft. Atul, Pranjal and Abhay

The  European Girls' Mathematical Olympiad a.k.a EGMO 2022 just ended. Congrats to Jessica Wan from USA, Taisiia Korotchenko, and Galiia Sharafetdinova for the perfect scores! Moreover, the Indian girls brought home 4 bronze medals! By far, this is the best result the EGMO India Team has ever achieved! To celebrate the brilliant result, here's a compilation of EGMO 2022 solutions and motivations written by my and everyone's favorite IMOTCer Atul ! And along with that, we also have reviews of each problem written by everyone's favorite senior, Pranjal !  These solutions were actually found by Atul, Pranjal,  and Abhay  during the 3-hour live solve. In the live solve, they solved all the 6 problems in 3 hours 😍!!! Okie Dokie, I think we should get started with the problems! Enjoy! Problem 1:  Let $ABC$ be an acute-angled triangle in which $BC<AB$ and $BC<CA$. Let point $P$ lie on segment $AB$ and point $Q$ lie on segment $AC$ such that $P \neq B$, $Q \neq C$ and

Q&A with experts about bashing

Heyy everyone! From the title, you can see that we are going to talk about  BASH.  Yesss, we are going to discuss whether you really need to learn bash or not?  Let me first introduce myself, I am Pranav Choudhary, a 10th grader from Haryana(India). I like to do geo and combo the most :PP. Oki so let's begin! For those of you who don't know what bashing is, lemme give you a brief introduction first. Bashing is basically a technique used to solve Geometry Problems. In general, when you try a geo problem you might think of angles, similarities, and some other techniques (e.g. Inversion, spiral similarity etc). All of which are called synthetic geometry. But sometimes people use various other techniques called "bash" to solve the same problems.  Now there are different kinds of bashing techniques, e.g. - Coordinate bash, Trig Bash, Complex bash, Barycentric Coordinates. Let me give you a brief introduction to each of them.  Coordinate Bash : You set one point as the orig