Analysis 1
Real numbers.

We have $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q}$. These sets meet most computational needs, but some geometric entities fall outside them: the length $x$ of the diagonal of a square with side $1$ satisfies $x^2 = 2$, yet it can be shown that $x \notin \mathbb{Q}$.

If $x$ is a solution of $x^2 = 2$, then $x \notin \mathbb{Q}$.

$$x \in \mathbb{Q} \iff x \text{ has a finite or periodic decimal expansion.}$$

$\frac{3}{5}=0{,}6$ (finite), $\frac{1}{3}=0{,}333\ldots$ (periodic): these are rational numbers.

For $x = 12{,}34\overline{2021}$ (periodic), we multiply by powers of $10$ to eliminate the periodic part:

• Closure: $x+y \in \mathbb{R}$ and $x \times y \in \mathbb{R}$.
• Commutativity, associativity, distributivity of $+$ and $\times$.
• Neutral elements: $\exists\, 0, 1 \in \mathbb{R}$, $x+0=x$ and $x \times 1 = x$.
• Opposites / inverses: $\exists\,(-x)$, $(-x)+x=0$; and for $x\neq 0$, $\exists\, x^{-1}$, $x \times x^{-1}=1$.

We say that $(\mathbb{R},+,\times)$ is a commutative field.

• $x=y \iff x-y=0$
• $0 \times x = 0$
• $(-x)y = -(xy) = x(-y)$
• $xy=0 \iff x=0$ or $y=0$
• $(xy)^{-1} = x^{-1}y^{-1}$

• Reflexivity: $x \le y$ or $y \le x$ (in particular $x \le x$).
• Antisymmetry: $x\le y$ and $y \le x \Rightarrow x=y$.
• Transitivity: $x\le y$ and $y\le z \Rightarrow x \le z$.

Compatible with addition: $x \le y \Rightarrow x+z \le y+z$. Partial compatibility with multiplication: $0\le z$ and $x\le y \Rightarrow zx \le zy$.

$x<y$ means $x\le y$ and $x\ne y$. $x \ge y$ means $y \le x$. These can be combined: $x \le y \le z$.

• $x\ge 0 \iff -x \le 0$
• $x\le y \iff -y\le -x$
• $x\le y \iff x-y\le 0$
• $y\ge 0 \Rightarrow x-y \le x \le x+y$
• $x^2 = xx \ge 0$ for all $x$
• $x<0$ and $y>0 \Rightarrow xy<0$

• $x\le y$ and $z>0 \Rightarrow xz \le yz$
• $x>0 \Rightarrow x^{-1}=\frac1x>0$
• $x,y\ge0$: $x\le y \iff x^2 \le y^2$
• $0<x<y \Rightarrow 0<\frac1y<\frac1x$
• $(x\le y$ and $s\le t) \Rightarrow x+s\le y+t$
• $(x\ge y>s$ or $x>y\ge s) \Rightarrow x>s$

• $[a,b]=\{x: a\le x \le b\}$, $\;]a,b[=\{x: a<x<b\}$
• $[a,b[=\{x: a\le x <b\}$, $\;]a,b]=\{x: a<x\le b\}$
• $[a,\infty[$, $\;]a,\infty[$, $\;]-\infty,b]$, $\;]-\infty,b[$
• $\mathbb{R} = \;]-\infty,\infty[\,$, $\;\overline{\mathbb{R}}=[-\infty,\infty]$
• $\mathbb{R}^+=[0,\infty[\,$, $\;\mathbb{R}^{*+}=]0,\infty[\,$, $\;\mathbb{R}^-=]-\infty,0]\,$, $\;\mathbb{R}^{*-}=]-\infty,0[$

$$|x| = \begin{cases} x & \text{if } x\ge 0 \\ -x & \text{otherwise} \end{cases}$$

We always have $|x| = |-x| = \max(x,-x)$.

• $|x|=0 \iff x=0$
• $|xy|=|x||y|$, and $|x|^2=|x^2|=x^2$
• For $y>0$: $|x|\le y \iff -y\le x\le y$
• Triangle inequality: $|x+y|\le|x|+|y|$
• $\big||x|-|y|\big| \le |x-y|$

Show that for all $x,y \in \mathbb{R}$:

$E$ is bounded below by $m$ if $m\le x$ for all $x\in E$. The greatest lower bound is $\inf E$ (infimum).

$E$ is bounded above by $M$ if $x\le M$ for all $x\in E$. The least upper bound is $\sup E$ (supremum).

$E$ is bounded if it is both bounded below and bounded above.

• $E=\{-3,0,4,10\}$: bounded below by $-3$, bounded above by $\sqrt{101}$.
• $E=\,]{-1},\infty[$: bounded below, $\inf E = -1$; not bounded above.
• $E=\{x\in\mathbb{Q}: x^2<2\}$: bounded above by $\sqrt2$, with $\sqrt2 \notin E$ and $\sqrt2 = \sup E$.

$$a=\inf E \iff \big(\forall x\in E,\, a\le x\big) \text{ and } \big(\forall \varepsilon>0,\, \exists x_\varepsilon\in E: a\le x_\varepsilon<a+\varepsilon\big)$$

$$b=\sup E \iff \big(\forall x\in E,\, x\le b\big) \text{ and } \big(\forall \varepsilon>0,\, \exists x_\varepsilon\in E: b-\varepsilon<x_\varepsilon\le b\big)$$

When the lower (or upper) bound of a set exists, it is unique.

If $E\subset\mathbb{R}$ is nonempty and bounded below, show that $F=\{-x : x\in E\}$ is nonempty and bounded above, with $\inf F = -\sup E$.

If $E$ is nonempty and not bounded above (resp. not bounded below), we set $\sup E = \infty$ (resp. $\inf E = -\infty$).

$$\text{If } x,y\in\mathbb{R},\ 0<y<x,\ \text{then} \ \exists\, n\in\mathbb{N}: x<ny$$

We say that $\mathbb{R}$ is Archimedean. This property follows from the least upper bound property.

$\mathbb{Q}$ is Archimedean but does not satisfy the least upper bound property. So it is indeed the least upper bound property that distinguishes $\mathbb{R}$ from $\mathbb{Q}$, not the Archimedean property.

For every $x \in \mathbb{R}\setminus\mathbb{Z}$, there exists a unique $n_x \in \mathbb{Z}$ such that:

A direct consequence of the Archimedean property.

For every $x \in \mathbb{R}$, the integer part of $x$, denoted $E(x)$ or $[x]$, is the unique integer satisfying:

If $x\in\mathbb{Z}$ then $[x]=x$. If $x\notin\mathbb{Z}$, setting $n_x=[x]$, the inequality becomes strict on both sides :

$[2]=2$, $\;[2{,}3]=2$, $\;[-2{,}3]=-3$, $\;[\sqrt2]=1$.

• Show that $n=[x] \iff x-1<n\le x$.
• Show that $[x+n]=[x]+n$ for all $n\in\mathbb{Z}$.
• Show that $[x+y]-[x]-[y] \in \{0,1\}$.

$d\in\mathbb{R}$ is decimal if it has a finite decimal expansion: $d\in\mathbb{D} \iff \exists n\in\mathbb{N}: 10^n d \in \mathbb{Z}$. We have $\mathbb{D}\subset\mathbb{Q}$.

E.g.: $11{,}319$ is decimal ($10^3x\in\mathbb{Z}$); $\frac13=0{,}333\ldots$ is not (infinite period).

1. Set $x_n = [x\times10^n]\times10^{-n}$.
2. This gives the bound $x_n \le x < x_n+10^{-n}$.
3. $x_n$: approximation from below ; $x_n+10^{-n}$: approximation from above.

$$\forall\, x,y\in\mathbb{R},\ x<y,\ \exists\, r\in\mathbb{Q}: x<r<y$$

Every interval of $\mathbb{R}$ containing at least two reals contains infinitely many rationals and infinitely many irrationals.