Lecture 1 - Introduction & Real numbers

Syllabus

• Real numbers
• Sequences
• Series of real numbers
• Power series
• Functions of one variable: limits, continuity, differentiability
• Higher Order derivatives
• Taylor series
• Riemann integrals
• Improper integrals

Real numbers

Definition

Let $A\subseteq \mathbb{R}$. We define $x\in \mathbb{R}$ to be - a lower bound for $A$ if $x\le a,\forall a\in A$ - an upper bound for $A$ if $x\ge a,\forall a\in A$

• $lb(A):=\{x\in \mathbb{R}|x\le a,\forall a\in A\}$, the set of lower bounds of A.
• $ub(A):=\{x\in \mathbb{R}|x\ge a,\forall a\in A\}$, the set of upper bounds of A.

Definition

A set $A\subseteq \mathbb{R}$ is defined to be

• bounded (from) below if $lb(A)\ne \varnothing$
• bounded (from) above if $ub(A)\ne \varnothing$

Definition

We say that $x\in \mathbb{R}$ is the supremum of $A\subseteq \mathbb{R},x:=sup(A)$, if and only if

1. $x\ge a,\forall a\in A$, that is $x\in ub(A)$
2. if $u$ is an upper bound for $A$, then $x\le u$ The supremum is the least upper bound, i.e. $sup(A):=min(ub(A))$

Definition

We say that $x\in \mathbb{R}$ is the infimum of $A\subseteq \mathbb{R},x:=inf(A)$, if and only if

1. $x\le a,\forall a\in A$, that is $x\in lb(A)$
2. if $u$ is an lower bound for $A$, then $x\ge u$ The supremum is the greatest lower bound, i.e. $inf(A):=max(lb(A))$

Example

Find $sup(A),max(A),inf(A),min(A)$ for the following:

1. $A=\{\frac{1}{n}|n\in \mathbb{N}\}$
2. $A=\{x\in \mathbb{Q}|x^2\le 2\}$

Solution

1. $A=\{\frac{1}{n}|n\in \mathbb{N}\}=\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4}\dots \}$
2. $sup(A)=1=max(A)$
3. $inf(A)=0$
4. $lb(A)=(-\infty ,0]$
5. $\nexists min(A)$
6. $A=\{x\in \mathbb{Q}|x^2\le 2\}=[-\sqrt{2},\sqrt{2}]\cap \mathbb{Q}$ >> 1. $sup(A)=\sqrt{2},\nexists max(A)$ >> 2. $inf(A)=-\sqrt{2},\nexists min(A)$

Definition

Completeness Axiom Every set $A\subseteq \mathbb{R}$ that is bounded above has a supremum. Similarly, every set $A\subseteq \mathbb{R}$ that is bounded below has an infimum.

Theorem

Let $A\subseteq \mathbb{R}$ be a bounded set. For $sup(A)$ and $inf(A)$ the following are true: $\forall \epsilon >0,\exists x\in A\text{ such that }sup(A)-ϵ<x$ $\forall \epsilon >0,\exists x\in A\text{ such that }x<inf(A)+ϵ$

$y<sup(A)$, say $y=sup(A)-\epsilon$ with $\epsilon >0$, we have that $y\notin ub(A)$. Hence there exists $x\in A$ such that $y<x$. Similar proof for $inf(A)$.

By definition, for any

Proposition

Let $A\subseteq B\subseteq \mathbb{R}$ be (nonempty) bounded sets. Then $inf(B)\le inf(A)\le sup(A)\le sup(B)$
and $\begin{array}{r}sup(A\cup B)=max\{sup(A),sup(B)\}\\ inf(A\cup B)=min\{inf(A),inf(B)\}\end{array}$

Definition

Define the extended real line $\overline{\mathbb{R}}:=\mathbb{R}\cup \{-\infty ,\infty \}$ where $\infty$ and $-\infty$ are such that $\forall x\in \mathbb{R},-\infty <x<\infty$ If a set $A$ is not bounded above, we define $sup(A):=\infty$ If a set $A$ is not bounded below, we define $inf(A):=-\infty$

Definition

A set $V\subseteq \mathbb{R}$ is a neighborhood (vecinity) of $x\in \mathbb{R}$ if $\exists \epsilon >0\text{ such that }(x-\epsilon ,x+\epsilon )\subseteq V$ A set $V\subseteq \mathbb{R}$ is a neighborhood of $\infty$ if $\exists a\in \mathbb{R}\text{ such that }(a,\infty )\subseteq V$

A set $V\subseteq \mathbb{R}$ is a neighborhood of $-\infty$ if $\exists a\in \mathbb{R}\text{ such that }(-\infty ,a)\subseteq V$ $\mathcal{V}(x):=\{V\subseteq \mathbb{R}|\text{V is a neighborhood of x}\}$

Definition

Let $A\subseteq \mathbb{R}$. the following set is called the interior of $A$ $int(A):=\{x\in \mathbb{R}|\exists V\in \mathcal{V}(x)\text{ such that }V\subseteq A\}$ and the following set is called the closure $cl(A):=\{x\in \mathbb{R}|\forall V\in \mathcal{V}(x)\text{ such that }V\cap A\ne 0\}$

Proposition

For any $A\subseteq \mathbb{R}$, it holds that $int(A)\subseteq A\subseteq cl(A)$.

Proof $int(A)\subseteq A$ we prove that if $x\in int(A)$, then $x\in A$. Let $x\in int(A)$, then $\exists \epsilon >0$ such that $(x-\epsilon ,x+\epsilon )\subseteq A$. Since $x\in (x-\epsilon ,x+\epsilon )$, we have that $x\in A$. To prove that $A\subseteq cl(A)$ we show that if $x\in A$, then $x\in cl(A)$. Let $x\in A$. Then for any $V\in \mathcal{V}(x)$ it holds that $x\in V$, giving that $x\in V\cup A$. Hence $x\in cl(A)$ since $V\in A\ne \varnothing$.

To prove that

Definition

If $A=int(A)$ then $A$ is called open. If $A=cl(A)$, then $A$ is called closed.

Remark

To prove that a set $A$ is open, it is sufficient to prove that $A\subseteq int(A)$. To prove that a set $A$ is closed, it is sufficient to prove that $cl(A)\subseteq A$.

Proposition

The following statements are true

• The complement of an open set is closed.
• The complement of a closed set is open.

Proof $A$ an open set, i.e. $A=int(A)$, and denote by $A^c=\{x\in \mathbb{R}|x\notin A\}$ its complement. To prove that $A^c$ is closed, we prove that $cl(A^c)\subseteq A^c$. Consider $x\in cl(A^c)$ and let’s assume that $x\notin A^c$, i.e. $x\in A$, aiming to obtain a contradiction. Since $A$ is open, there exists $V\in \mathcal{V}(x)$ such that $V\subseteq A$, giving that $V\cup A^c=\varnothing$: contradiction with $x\in cl(A^c)$. Hence the assumption $x\in A^c$ is false, and we have that if $x\in cl(A^c)$, then $x\in A^c$. In other words, $cl(A^c)\subseteq A^c$

Let us prove the first statement, the other one being similar. Consider

Proposition

Any union of open sets is open. Any intersection of closed sets is closed. Any finite intersection of open sets is open. Any finite union of closed sets is closed.