CommeUnJeu · L1 PCSI

Properties of $\mathbb{R}$

⌚ ~107 min ▢ 13 blocks ✓ 52 exercises ➣ Prerequisites : Sets, Number Sets

Real numbers come with an order. Four tools formalize the use of that order: inequalities between real numbers (and their compatibility with arithmetic operations), the absolute value (a measure of distance to zero), the floor function (a discretization of $\mathbb{R}$ into $\mathbb{Z}$), and the supremum (least upper bound), which expresses the completeness of $\mathbb{R}$ and prepares the limit chapter. The chapter is short and ground-laying: every analysis chapter that follows --- sequences, limits, continuity, derivatives --- rests on these four tools.

I Order and inequalities on $\mathbb{R}$

The order on $\mathbb{R}$ is total: any two real numbers are comparable. The compatibility rules state how the order interacts with addition and multiplication. Two warnings: multiplying by a negative number reverses the inequality, and inversion (passing from $x \le y$ to $1/x \ge 1/y$) requires the reals to have the same sign.

Proposition — Compatibility of $\le$ with operations

For $x, y, z, t \in \mathbb{R}$:

Addition: $\textcolor{colorprop}{x \le y \implies x + z \le y + z}$, and $\textcolor{colorprop}{x \le y \text{ and } z \le t \implies x + z \le y + t}$.
Positive scaling: $\textcolor{colorprop}{x \le y \text{ and } \lambda \ge 0 \implies \lambda x \le \lambda y}$.
Negative scaling: $\textcolor{colorprop}{x \le y \text{ and } \lambda \le 0 \implies \lambda x \ge \lambda y}$ (reversal).
Inversion (same sign): for $x, y > 0$, $\textcolor{colorprop}{x \le y \implies 1/x \ge 1/y}$.
Squaring (positive): for $x, y \ge 0$, $\textcolor{colorprop}{x \le y \iff x^2 \le y^2}$.

Proof

Each property is a direct consequence of the order axioms ($a \le b \iff b - a \ge 0$ and $a, b \ge 0 \implies ab \ge 0$).

Addition. If $x \le y$, then $(y + z) - (x + z) = y - x \ge 0$, so $x + z \le y + z$. The two-sided version follows by chaining: $x + z \le y + z \le y + t$.
Positive scaling. If $x \le y$ and $\lambda \ge 0$, then $\lambda y - \lambda x = \lambda (y - x)$ is a product of two non-negative reals, hence non-negative; so $\lambda x \le \lambda y$.
Negative scaling. If $\lambda \le 0$, then $-\lambda \ge 0$ and the previous item gives $-\lambda x \le -\lambda y$, that is, $\lambda y \le \lambda x$.
Inversion. For $x, y > 0$ with $x \le y$, $1/y - 1/x = (x - y)/(xy)$ has numerator $\le 0$ and denominator $> 0$, hence $1/y \le 1/x$.
Squaring. For $x, y \ge 0$: $y^2 - x^2 = (y - x)(y + x)$. Both factors have the same sign as $y - x$ (since $y + x \ge 0$), so $x \le y \iff x^2 \le y^2$.

Example

Show that $\dfrac{2}{3} < \dfrac{5}{7}$ by cross-multiplication.

Answer

Both denominators are positive, so multiplying both sides by $21 = 3 \cdot 7 > 0$ preserves the inequality: $2/3 < 5/7 \iff 2 \cdot 7 < 5 \cdot 3 \iff 14 < 15$, which is true.

Example

Solve the inequality $-2x + 3 \le 7$ in $\mathbb{R}$.

Answer

$-2x + 3 \le 7 \iff -2x \le 4 \iff x \ge -2$ (division by $-2$ reverses the inequality). Solution: $x \in [-2, +\infty[$.

Skills to practice

Solving inequations by sign tables
Proving inequalities by algebraic manipulation

II Intervals of $\mathbb{R}$

Intervals are the connected subsets of $\mathbb{R}$. They appear in every analysis result --- as the natural domain of continuity, as the image of a continuous function on a segment, as the solution set of inequalities. We fix the notation here.

Definition — Bounded intervals

For $a, b \in \mathbb{R}$ with $a \le b$:

$\textcolor{colordef}{[a, b]} = \{x \in \mathbb{R} \mid a \le x \le b\}$ (closed interval, both endpoints included);
$\textcolor{colordef}{[a, b[} = \{x \in \mathbb{R} \mid a \le x < b\}$;
$\textcolor{colordef}{]a, b]} = \{x \in \mathbb{R} \mid a < x \le b\}$;
$\textcolor{colordef}{]a, b[} = \{x \in \mathbb{R} \mid a < x < b\}$ (open interval).

A segment of $\mathbb{R}$ is a closed bounded interval $[a, b]$ with $a \le b$.

Example

The four bounded intervals from $0$ to $1$ are: $[0, 1]$ (closed segment, includes both endpoints), $]0, 1[$ (open interval, excludes both), $]0, 1]$ (excludes $0$, includes $1$), $[0, 1[$ (includes $0$, excludes $1$). They share the same length $1$ but differ in which endpoints belong.

Definition — Unbounded intervals

For $a \in \mathbb{R}$:

$\textcolor{colordef}{[a, +\infty[} = \{x \in \mathbb{R} \mid x \ge a\}$, $\textcolor{colordef}{]a, +\infty[} = \{x \in \mathbb{R} \mid x > a\}$;
$\textcolor{colordef}{]-\infty, a]} = \{x \in \mathbb{R} \mid x \le a\}$, $\textcolor{colordef}{]-\infty, a[} = \{x \in \mathbb{R} \mid x < a\}$;
$\textcolor{colordef}{]-\infty, +\infty[} = \mathbb{R}$.

The symbols $-\infty$ and $+\infty$ are not real numbers; the bracket facing them is always open.

Example

Three unbounded intervals: $[0, +\infty[$ is the set of non-negative reals; $]-\infty, 1[$ is the set of reals strictly less than $1$; $]-\infty, +\infty[ = \mathbb{R}$ is all of $\mathbb{R}$. The brackets facing $\pm\infty$ are always open since $\pm\infty \notin \mathbb{R}$.

Skills to practice

Deciding whether a subset is an interval
Determining intersections and unions of intervals

III Absolute value and distance

The absolute value $|x|$ is the distance from $x$ to $0$ on the real line. It strips the sign and yields a non-negative real. Its multiplicative property and the triangle inequality are the two facts you will use most.

Definition — Absolute value

For $x \in \mathbb{R}$, the absolute value of $x$ is $$ \textcolor{colordef}{|x|} = \begin{cases} x & \text{if } x \ge 0, \\ -x & \text{if } x < 0. \end{cases} $$ Equivalently, $|x| = \max(x, -x) = \sqrt{x^2}$.

Example

$|{-3}| = 3$ and $|2| = 2$. The absolute value is the distance to $0$ on the real line.

Example

Quick computations: $|3| = 3$, $|{-5}| = 5$, $|0| = 0$, and $|3 - 7| = |{-4}| = 4$. In particular, $|3 - 7| = |7 - 3|$: the absolute value of a difference does not depend on the order of the terms.

Proposition — Properties of $|\cdot|$

For all $x, y \in \mathbb{R}$:

$\textcolor{colorprop}{|x| \ge 0}$ and $\textcolor{colorprop}{|x| = 0 \iff x = 0}$.
$\textcolor{colorprop}{|{-x}| = |x|}$.
$\textcolor{colorprop}{|xy| = |x|\cdot|y|}$.
$\textcolor{colorprop}{|x+y| \le |x| + |y|}$ (triangle inequality).
$\textcolor{colorprop}{\big||x| - |y|\big| \le |x - y|}$ (reverse triangle inequality).
For $r \ge 0$: $\textcolor{colorprop}{|x| \le r \iff -r \le x \le r}$.

Proof

Non-negativity. By the case definition, $|x| \in \{x, -x\}$, and in each case the chosen branch is $\ge 0$. If $|x| = 0$, then $x = 0$ (if $x > 0$, $|x| = x > 0$; if $x < 0$, $|x| = -x > 0$).
Symmetry $|{-x}| = |x|$. If $x \ge 0$, $-x \le 0$ so $|{-x}| = -(-x) = x = |x|$. If $x < 0$, $-x > 0$ so $|{-x}| = -x = |x|$.
Multiplicativity. Since $|x| = \sqrt{x^2}$, $|xy| = \sqrt{(xy)^2} = \sqrt{x^2} \sqrt{y^2} = |x| \, |y|$.
Triangle inequality. Both sides are non-negative, so it suffices to compare their squares: $(|x| + |y|)^2 - (x + y)^2 = x^2 + 2 |x| |y| + y^2 - x^2 - 2xy - y^2 = 2(|xy| - xy) \ge 0$, since $|xy| \ge xy$.
Reverse triangle. Writing $x = (x - y) + y$ and applying the triangle inequality, $|x| \le |x - y| + |y|$, so $|x| - |y| \le |x - y|$. By symmetry, $|y| - |x| \le |x - y|$, hence $\big||x| - |y|\big| \le |x - y|$.
Bracket form. If $|x| \le r$, then $x \le |x| \le r$ and $-x \le |x| \le r$, so $-r \le x \le r$. Conversely, if $-r \le x \le r$, then both $x \le r$ and $-x \le r$, hence $|x| = \max(x, -x) \le r$.

Proposition — Distance on the real line

For $x, y \in \mathbb{R}$, the distance between $x$ and $y$ is $\textcolor{colordef}{|x - y|}$. In particular, for $a \in \mathbb{R}$ and $r \ge 0$: $$ \textcolor{colorprop}{|x - a| \le r \iff a - r \le x \le a + r \iff x \in [a - r \,;\, a + r]}. $$

Proof

The three defining properties of a distance follow from the properties of $|\cdot|$:

Positivity. $|x - y| \ge 0$, with equality iff $x - y = 0$, that is, $x = y$.
Symmetry. $|y - x| = |{-(x - y)}| = |x - y|$ (P symmetry of $|\cdot|$).
Triangle inequality. $|x - z| = |(x - y) + (y - z)| \le |x - y| + |y - z|$.

For the bracket form: $|x - a| \le r \iff -r \le x - a \le r \iff a - r \le x \le a + r$, applying the bracket form of $|\cdot|$ to $x - a$.

Example

The distance between $2$ and $5$ is $d(2, 5) = |2 - 5| = 3$. The distance between $-1$ and $4$ is $d(-1, 4) = |{-1} - 4| = |{-5}| = 5$. As expected, swapping the arguments leaves the result unchanged: $d(5, 2) = |5 - 2| = 3$.

Example

Solve $|2x - 3| \le 5$ in $\mathbb{R}$.

Answer

$|2x - 3| \le 5 \iff -5 \le 2x - 3 \le 5 \iff -2 \le 2x \le 8 \iff -1 \le x \le 4$. Solution: $x \in [-1, 4]$.

Skills to practice

Solving equations and inequations with $|\cdot|$
Using the triangle inequality

IV Floor function and Archimedean property

The floor function $\lfloor x \rfloor$ maps a real number to the greatest integer less than or equal to it. (It is not the nearest integer: $\lfloor -2.3 \rfloor = -3$, not $-2$.) It is the bridge between $\mathbb{R}$ and $\mathbb{Z}$ used everywhere in number theory, in numerical analysis, and in the definition of the fractional part.

Proposition — Archimedean property of $\mathbb{R}$

\textcolor{colorprop}{For every $x \in \mathbb{R}$, there exists $N \in \mathbb{N}$ such that $N > x$.}
Equivalently: $\mathbb{N}$ is not bounded above in $\mathbb{R}$. (Admitted; this property is what guarantees the existence of the floor of any real number.)

Definition — Floor function

For $x \in \mathbb{R}$, the floor of $x$, denoted $\textcolor{colordef}{\lfloor x \rfloor}$, is the unique integer $n \in \mathbb{Z}$ such that $$ n \le x < n+1. $$ The existence of such an $n$ rests on the Archimedean property of $\mathbb{R}$ (admitted: for every $x \in \mathbb{R}$, there exists $N \in \mathbb{N}$ with $N > x$). The fractional part of $x$ is $\{x\} = x - \lfloor x \rfloor \in [0, 1[$.

Example

$\lfloor 3.7 \rfloor = 3$, $\lfloor 5 \rfloor = 5$.
$\lfloor -2.3 \rfloor = -3$ (rounding toward $-\infty$).
$\lfloor \pi \rfloor = 3$, $\lfloor -\pi \rfloor = -4$.

Proposition — Properties of $\lfloor \cdot \rfloor$

For $x \in \mathbb{R}$ and $n \in \mathbb{Z}$:

$\textcolor{colorprop}{\lfloor x \rfloor \le x < \lfloor x \rfloor + 1}$ and $\textcolor{colorprop}{x - 1 < \lfloor x \rfloor \le x}$.
$\textcolor{colorprop}{\lfloor x + n \rfloor = \lfloor x \rfloor + n}$ (translation by an integer).
$\textcolor{colorprop}{\lfloor x \rfloor = n \iff n \le x < n+1}$.

Proof

Bracketing inequalities. By Definition, $\lfloor x \rfloor$ is the unique integer with $\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$. Subtracting $1$ from both sides of the left inequality of the next-integer version gives $x - 1 < \lfloor x \rfloor \le x$.
Integer translation. Let $n \in \mathbb{Z}$. Adding $n$ to the bracketing inequality $\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$ yields $\lfloor x \rfloor + n \le x + n < \lfloor x \rfloor + n + 1$. Since $\lfloor x \rfloor + n \in \mathbb{Z}$, by uniqueness in the Definition of the floor, $\lfloor x + n \rfloor = \lfloor x \rfloor + n$.
Characterization $\lfloor x \rfloor = n$. ($\Rightarrow$) By the bracketing inequality with $\lfloor x \rfloor = n$. ($\Leftarrow$) If $n \le x < n + 1$ with $n \in \mathbb{Z}$, then $n$ satisfies the defining property of $\lfloor x \rfloor$; by uniqueness, $\lfloor x \rfloor = n$.

Skills to practice

Solving floor equations
Proving identities involving $\lfloor \cdot \rfloor$

V Bounded subsets of $\mathbb{R}$

A subset of $\mathbb{R}$ is bounded above if there is a real number lying above every element, bounded below if there is a real number lying below every element, and bounded if both conditions hold. The smallest upper bound is the supremum; the largest lower bound is the infimum. The fundamental property of $\mathbb{R}$ --- which we admit --- is that every non-empty bounded-above subset has a supremum.

Definition — Bounded above$\virgule$ bounded below$\virgule$ bounded

Let $A \subset \mathbb{R}$, $A \ne \emptyset$.

$A$ is bounded above if $\exists M \in \mathbb{R}, \forall x \in A, x \le M$. Such an $M$ is an upper bound (majorant).
$A$ is bounded below if $\exists m \in \mathbb{R}, \forall x \in A, x \ge m$. Such an $m$ is a lower bound (minorant).
$A$ is bounded if it is both bounded above and bounded below.

Example

The set $A = ]0, 1]$ is bounded: $0$ is a lower bound ($\forall x \in A, x > 0$) and $1$ is an upper bound ($\forall x \in A, x \le 1$). Note that $1 \in A$ but $0 \notin A$. By contrast, $\mathbb{N} \subset \mathbb{R}$ is bounded below (by $0$) but not bounded above (Archimedean property), hence not bounded.

Skills to practice

Computing $\sup$$\virgule$ $\inf$$\virgule$ $\max$$\virgule$ $\min$
Proving sup/inf properties

VI Supremum and infimum

Among the upper bounds of a non-empty bounded-above subset, there is always a smallest one --- the supremum. This is the completeness axiom of $\mathbb{R}$, the property that distinguishes $\mathbb{R}$ from $\mathbb{Q}$ (in $\mathbb{Q}$, the set $\{x \in \mathbb{Q} \mid x^2 < 2\}$ has no $\sup$; in $\mathbb{R}$, it has $\sqrt{2}$).

Definition — Supremum$\virgule$ infimum$\virgule$ maximum$\virgule$ minimum

For $A \subset \mathbb{R}$ non-empty and bounded above:

The supremum (or least upper bound) of $A$, denoted $\textcolor{colordef}{\sup A}$, is the smallest upper bound: it satisfies $\sup A$ is an upper bound, and for every upper bound $M$ of $A$, $\sup A \le M$.
If $\sup A \in A$, it is the maximum, denoted $\textcolor{colordef}{\max A}$.

Similarly, for $A$ bounded below: $\textcolor{colordef}{\inf A}$ (greatest lower bound), $\textcolor{colordef}{\min A}$ if $\inf A \in A$.

Example

Let $A = \{1 - 1/n : n \in \mathbb{N}^*\} = \{0, 1/2, 2/3, 3/4, \ldots\}$. Then $\inf A = 0 = \min A$ (attained at $n = 1$) and $\sup A = 1$, but $1 \notin A$, so $A$ has no maximum. Symmetrically, $B = \{1/n : n \in \mathbb{N}^*\}$ satisfies $\sup B = 1 = \max B$ (attained at $n = 1$) and $\inf B = 0$, but $0 \notin B$, so $B$ has no minimum.

Theorem — Existence of $\sup$ and $\inf$ in $\mathbb{R}$

\textcolor{colorprop}{Every non-empty bounded-above subset of $\mathbb{R}$ has a supremum in $\mathbb{R}$. Every non-empty bounded-below subset has an infimum in $\mathbb{R}$.}
(Admitted; this is a defining property of $\mathbb{R}$, the completeness axiom.)

Skills to practice

Determining sup and inf of a finite or interval set
Proving sup and inf properties

VII $\varepsilon$-characterizations of $\sup$ and $\inf$

The Existence theorem says some real number plays the role of $\sup A$, but does not say how to recognize it. The next Proposition gives the practical $\varepsilon$-characterization, the workhorse of every limit / continuity / sequence proof in this course: $s = \sup A$ iff $s$ majores $A$ and every value strictly below $s$ fails to majorize.

Proposition — Characterization of $\sup$ ($\varepsilon$-form)

Let $A \subset \mathbb{R}$ be non-empty and bounded above, and let $s \in \mathbb{R}$. Then $s = \sup A$ if and only if:

$s$ is an upper bound: $\forall x \in A, x \le s$;
$s$ is approached arbitrarily closely from inside $A$: $\forall \varepsilon > 0, \exists x \in A, \, s - \varepsilon < x \le s$.

Symmetrically, for $A$ non-empty and bounded below: $i = \inf A$ iff $i$ is a lower bound and $\forall \varepsilon > 0, \exists x \in A, \, i \le x < i + \varepsilon$.

Proof

We prove the equivalence for $\sup$; the case of $\inf$ is symmetric.
($\Rightarrow$) Assume $s = \sup A$. By Definition, $s$ is an upper bound of $A$, so (i) holds. For (ii), fix $\varepsilon > 0$. The real $s - \varepsilon < s$, so $s - \varepsilon$ is not an upper bound of $A$ (otherwise $s$ would not be the least upper bound). Hence there exists $x \in A$ with $x > s - \varepsilon$; combined with $x \le s$ (since $s$ is an upper bound), we get $s - \varepsilon < x \le s$.
($\Leftarrow$) Assume (i) and (ii). To show $s = \sup A$, we check that $s$ is the least upper bound. Let $M$ be any upper bound of $A$. Suppose, for contradiction, that $M < s$, and set $\varepsilon = s - M > 0$. By (ii), there exists $x \in A$ with $x > s - \varepsilon = M$, contradicting that $M$ is an upper bound. Hence $M \ge s$, so $s$ is the least upper bound, that is, $s = \sup A$.

Example

Apply the $\varepsilon$-characterization to confirm that $\sup\{1 - 1/n : n \in \mathbb{N}^*\} = 1$.

$1$ is an upper bound: $1 - 1/n < 1$ for every $n \ge 1$.
Given $\varepsilon > 0$, by the Archimedean property pick $n \in \mathbb{N}^*$ with $n > 1/\varepsilon$, that is, $1/n < \varepsilon$; then $1 - 1/n > 1 - \varepsilon$.

So $1 = \sup$.

Example

Determine $\sup, \inf, \max, \min$ of $A = [0, 1[$.

Answer

$\inf A = 0 = \min A$ ($0 \in A$). $\sup A = 1$, but $1 \notin A$, so $A$ has no maximum.

Skills to practice

Applying the $\varepsilon$-characterization