Mathematical Analysis
Lecture 1
1.1 Basic properties of $\R$
Definition 1.1.1 (Field Axioms). On the set $\R$ of real numbers there are two operations
$+:\R\times \R \rightarrow \R\;\;$ and $\;\;\boldsymbol{\cdot} :\R\times \R \ra \R$
called addition and multiplication, respectively.
Remark: We usually write $xy$ instead of $x\cdot y$.
1.1 Basic properties of $\R$
👉 $+:\R\times \R \rightarrow \R$ and $\boldsymbol{\cdot} :\R\times \R \ra \R$
These operations satisfy the following properties:
| (A1) | If $x \in \R$ and $y \in \R$, then $x+y \in \R.$ |
| (A2) | (commutativity) $x+y = y+x$ for all $x,y \in \R.$ |
| (A3) $\,$ | (associativity) $(x+y)+z = x+(y+z)$ for all $x,y,z \in \R.$ |
| (A4) $\,$ | There exists an element $0 \in \R$ such that $0+x = x$ for all $x \in \R.$ |
| (A5) $\,$ | For every element $x\in \R$, there exists an element $-x \in \R$ such that $x + (-x) = 0.$ |
1.1 Basic properties of $\R$
👉 $+:\R\times \R \rightarrow \R$ and $\boldsymbol{\cdot} :\R\times \R \ra \R$
| (M1) | If $x \in \R$ and $y \in \R$, then $xy \in \R.$ |
| (M2) | (commutativity) $xy = yx$ for all $x,y \in \R.$ |
| (M3) | (associativity) $(xy)z = x(yz)$ for all $x,y,z \in \R.$ |
| (M4) $\,$ | There exists an element $1 \in \R$ (and $1 \not= 0$) such that $1x = x$ for all $x \in \R.$ |
|
(M5)
|
For every $x\in \R$ such that $x \not= 0$ there exists an element $\dfrac{1}{x} \in \R$ such that $x\left(\dfrac{1}{x}\right) = 1.$ |
| (D) $\,$ |
(distributive law) $x(y+z) = xy+xz\,$
for all $\,x,y,z \in \R.$ |
1.1 Basic properties of $\R$
The previous definition establishes that $\R$ is a field. The properties listed here should be familiar to you. The first five are concerned with addition, the next five with multiplication, and the last one connects the two operations.
In general, if an arbitrary set $F$ with two operations $+$ and $\pd$ satisfies (A1)-(A5), (M1)-(M5) and (D), then $F$ is called a field.
1.1 Basic properties of $\R$
The point of this list of axioms is that all the familiar techniques of algebra can be derived from these eleven properties, in the same sense that the theorems of the Euclidean geometry can be deduced from the five basic axioms stated by Euclid in his Elements.
Example 1.1.1.
Claim: If $x\in \R,$ then $0 x = 0.$
Proof. We can prove this by noting that for every $x\in \R,$
$x x $ $= x(0+x) $ $= 0 x + x x,$
using (A4), (D) and (M2).
Example 1.1.1.
Claim: If $x\in \R$, then $0 x = 0$.
Proof. Then
| $\;x x + (-x x)$ | $ =\;( 0 x + x x) + (-x x)$ |
| $0$ | $ =\;( 0 x + x x)+ (-x x) \;\;\;\;\;\,\text{by (A5)}$ |
| $0$ | $ =\;0 x + \big(x x + (-x x) \big) \;\;\;\;\, \text{by (A3)}$ |
| $0$ | $ =\;0 x + 0\qquad \quad \qquad \quad \text{by (A5)}$ |
| $0$ | $ =0 x \,\;\quad\quad \qquad \qquad \quad \,\,\text{by (A4)}$ |
Therefore, $0 x = 0.$ $\qquad \blacksquare$
Example 1.1.2.
The set of natural numbers $$\N := \{ 1, 2, 3, \ldots\}$$ is not a field. It does not satisfy (A5) and (M5).
Example 1.1.3.
The set of integer numbers $$\Z := \left\{ \ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\right\}$$ is not a field. It does not satisfy (M5).
For instance, there is no $x\in \Z$ such that $2x = 1$.
Example 1.1.4.
The set $$\Q := \left\{ \frac{a}{b} : a,b \in \Z, b\neq 0\right\}$$ of rational numbers satisfies all the Field Axioms. Therefore $\Q $ is a field.
1.1.2 Order Axioms
Other important properties of $\R$ are given by the order relation $\lt$.
1.1.2 Order Axioms
Definition 1.1.2. There is an order relation $\lt$ in $\R$ with the following properties:
| (O1) $\,$ | (trichotomy) For all $x, y \in \R$, exactly one of $x \lt y,$ $x=y,$ or $y \lt x$ holds. |
| (O2) $\,$ | (transitivity) If $x,y,z \in \R$ are such that $x \lt y$ and $y \lt z,$ then $x \lt z.$ |
| (O3) | For $x,y,z \in \R$, $x \lt y$ implies $x+z \lt y+z.$ |
| (O4) | For $x,y \in \R,$ $x > 0$ and $y \gt 0$ implies $xy > 0.$ |
1.1.2 Order Axioms
For $x\in \R$, if $x\gt 0,$ we say that $x$ is positive .
If $x\lt 0,$ we say that $x$ is negative.
We also say $x$ is nonnegative if $x\geq 0,$ and nonpositive if $x\leq 0.$
In general, if an arbitrary field $F$ has also an order relation $\lt$ satisfying the properties (O1)-(O4), then it is called an ordered field.
1.1.2 Order Axioms
More facts that you should know
Theorem 1.1.1. Let $x,y,z,w \in \R$. Then
- If $x > 0,$ then $-x \lt 0$ (and vice versa).
- If $x > 0$ and $y \lt z,$ then $xy \lt xz.$
- If $x \lt 0$ and $y \lt z,$ then $xy > xz.$
- If $x \not= 0,$ then $x^2 > 0.$
- If $0 \lt x \lt y,$ then $0 \lt \dfrac{1}{y} \lt \dfrac{1}{x}.$
- If $0\lt x \lt y,$ then $x^2 \lt y^2.$
- If $x \leq y$ and $z \leq w,$ then $x + z \leq y + w.$
- If $xy > 0,$ then either both $x$ and $y$ are positive, or both are negative.
1.2 The least-upper-bound property
Do the Field and Order Axioms
characterise the real numbers?
What makes $\R$ so special? 🤔
1.2 The least-upper-bound property
Definition 1.2.1. Let $A \subset \R$.
- If there exists a number $b \in \R$ such that $x \leq b$ for all $x \in A,$ then we say $A$ is bounded above and $b$ is an upper bound of $A.$
- If there exists a number $b \in \R$ such that $b \leq x$ for all $x \in A$, then we say $A$ is bounded below and $b$ is a lower bound of $A.$
1.2 The least-upper-bound property
Definition 1.2.1. Let $A \subset \R$.
- If there exists an upper bound $b_0$ of $A$ such that whenever $b$ is an upper bound for $A$ we have $b_0 \leq b,$ then $b_0$ is called the least upper bound or the supremum of $A.$ We write \begin{equation*} \sup\, A := b_0 . \end{equation*}
- Similarly, if there exists a lower bound $b_0$ of $A$ such that whenever $b$ is a lower bound for $A$ we have $ b\leq b_0,$ then $b_0$ is called the greatest lower bound or the infimum of $A.$ We write \begin{equation*} \inf\, A := b_0 . \end{equation*}
Upper & Lower bounds, Supremum & Infimum
When a set $A$ is both bounded above and bounded below, we say simply that $A$ is bounded.
Example 1.2.1.
Let $A:= \left\{x\in \R : 0\leq x \leq 1 \right\} = [0,1].$
Then $\sup\, A = 1 $ and $\inf\, A = 0$ with $1\in A$ and $0 \in A.$
The set $$B:= \left\{x\in \R : 0\lt x \lt 1 \right\} = (0,1).$$ has the same upper and lower bounds $$\sup\, B = 1 \quad \text{and} \quad \inf\, B = 0,$$ but in this case $1\notin B$ and $0 \notin B.$
Example 1.2.2.
The set $A := \{ x \in \Q : x \lt 1 \}$ has a least upper bound of 1, but 1 is not in the set $A$ itself.
The set $B := \{ x \in \Q : x \leq 1 \}$ also has an upper bound of 1, and in this case $1 \in B.$
Example 1.2.3.
The set $A := \{ x \in \Q : x \geq 0 \}$ has no upper bound and therefore it cannot have a least upper bound.
This set does have a greatest lower bound: $0.$
Example 1.2.4.
Let $A:= \left\{ \dfrac{1}{n} : n \in \N \right\}.$
The set $A$ is bounded above and below.
Successful candidates for an upper bound include $$3,\,2,\, \text{ and }\, \frac{3}{2}.$$
For the least upper bound, we claim ${\sup\, A = 1}$.
Example 1.2.4.
Let $A:= \left\{ \dfrac{1}{n} : n \in \N \right\}.$
For the least upper bound, we claim ${\sup\, A = 1}$.
An attempt to prove this claim: Assume we already have other upper bound $b$.
Because $1 \in A$ and $b$ is an upper bound for $A,$ we must have $1\leq b.$
Although we do not quite have the tools we need for a rigorous proof, it should be somewhat apparent that $\inf\, A = 0.$
Example 1.2.5.
Consider now the set $$A:= \left\{ x\in \Q : x^2\lt 2 \right\}$$
and pretend for the moment that our "world" consists only of rational numbers.
The set $A$ is certainly bounded above.
Taking $b = 2$ works, $b = \dfrac{3}{2}$ as well.
But notice what happens as we go in search of the least upper bound.
Example 1.2.5.
Consider now the set $$A:= \left\{ x\in \Q : x^2\lt 2 \right\}$$
and pretend for the moment that our "world" consists only of rational numbers.
We might try $b=\dfrac{142}{100}$, which is indeed an upper bound, but then we discover that $b = \dfrac{1415}{1000}$ is an upper bound that is smaller still.
Is there a smallest one? 🤔
Example 1.2.5.
Consider now the set $$A:= \left\{ x\in \Q : x^2\lt 2 \right\}$$
and pretend for the moment that our "world" consists only of rational numbers.
Is there a smallest one? 🤔
In the rational numbers, there is not. 😥 In the real numbers, there is. 😃
Back in $\R$, we can say that $\sup\, A = \sqrt{2},$ but $\sqrt{2}\notin\Q$.
Example 1.2.5.
Consider now the set $$A:= \left\{ x\in \Q : x^2\lt 2 \right\}$$
So, in $\Q$ we don't always have least upper bounds!
√2 is not rational
In Example 1.2.5 we claimed that $\sqrt{2}\notin \Q$. But how do we know this is true? 🤔
Well, this is a well known fact, it was discovered and proved by the school of Pythagoras around 500 B.C. Aproximately 2500 years ago. 🤯
√2 is not rational
Theorem 1.2.1. There does not exist a rational number $r$ such that $$r^2=2.$$
√2 is not rational
Theorem 1.2.1. There does not exist a rational number $r$ such that $r^2=2.$
Proof. Suppose, on the contrary, that $p$ and $q$ are integers such that $$\left(\frac{p}{q}\right)^2=2.$$ We may also assume that $p$ and $q$ have no common integer factors other than 1, because, if they had one, we could simply cancel it out and rewrite the fraction in lowest terms.
√2 is not rational
Theorem 1.2.1. There does not exist a rational number $r$ such that $r^2=2.$
Proof. Since $\left(\dfrac{p}{q}\right)^2=2$, then $p^2 = 2q^2$.
This implies that $p^2$ is even, which means that $p$ is also even. 🤔 Why?
Because if $p=2n+1$ is odd, then its square
$p^2$ $=(2n+1)^2$ $=4n^2+4n+1$ $=2(2n^2+2n)+1$
is odd.
√2 is not rational
Theorem 1.2.1. There does not exist a rational number $r$ such that $r^2=2.$
Proof. Now, since $p$ is even, then $p = 2m$ for some $m\in \N$, and hence
$p^2 = (2m)^2$ $= 2 q^2,\;$ so that $\;2m^2=q^2$.
This means that $q^2$ is even, and then $q$ is also even.
But this is a contradiction❗ Because we have chosen $p$ and $q$ with no common factors.
√2 is not rational
Theorem 1.2.1. There does not exist a rational number $r$ such that $r^2=2.$
Proof. Thus, since the hypothesis that $\left(\frac{p}{q}\right)^2=2$ leads to the contradictory conclusion that $p$ and $q$ are both even, it must be false.
Therefore, there does not exist a rational number $r$ such that $r^2=2.$ $\;\blacksquare$
1.2 The least-upper-bound property
Definition 1.2.2 An ordered set $S$ has the least-upper-bound property if every subset $E\subset S$ that is bounded above has a least upper bound, that is $\sup E$ exists in $S$.
This is also know as the Axiom of Completeness
Existence of $\R$
The properties from the Field and Order axioms, together with the least-upper-bound property characterise the real numbers $\R$. We state this more precisely as a theorem without proof.
Theorem: There exists a unique ordered field $\R$ with the least-upper-bound property such that $\Q \subset \R$.