§ 1. Definition of real numbers

1. Definition of real numbers

1. The ordered group of rational numbers

We have defined the ordering $x \leq y$ on the set $\mathbf{Q}$ of rational numbers; we have seen that this ordering makes $\mathbf{Q}$ a linearly ordered set, and that it is compatible with the additive group structure of $\mathbf{Q}$ , i.e. for each $z \in \mathbf{Q}$ the relation $x \leq y$ is equivalent to $x + z \leq y + z$ (that is, the ordering is invariant under translations). We recall the notation (which is used in any linearly ordered group)

x^+ = \sup (x, 0), \\ x^- = \sup (-x, 0) = (-x)^+, \\ |x| = \sup (x, -x);

$|x|$ is called the absolute value of $x$ , and we have

x = x^+ - x^-, \qquad |x| = x^+ + x^-

and the triangle inequality

(1)

|x + y| \leq |x| + |y|,

together with the inequality (2)

||x| - |y|| \leq |x - y|

which is an immediate consequence of (1); moreover (3)

|x^+ - y^+| \leq |x - y|.

The relations $x \geq 0,\ x = x^+,\ x^- = 0,\ |x| = x$ (resp. $x \leq 0,\ x = -x^-,\ x^+ = 0,\ |x| = -x$ ) are equivalent. The relation $|x| = 0$ is equivalent to $x = 0$ ; if $a \geq 0$ , the relation $|x| \leq a$ is equivalent to $-a \leq x \leq a$ , and the relation $|x| \geq a$ is equivalent to " $x \geq a$ or $x \leq -a$ ". For all $x, y$ in $\mathbf{Q}$ , we have

(4)

\sup (x, y) + z = \sup (x + z, y + z),

(5)

\inf (x, y) = - \sup (-x, -y),

and, as particular cases,

(6)

\sup (x, y) = x + (y - x)^+ = x + (x - y)^-,

(7)

\inf (x, y) = x - (y - x)^- = x - (x - y)^+.

Finally, let $\mathbf{Q}_+$ denote the set of rational numbers $\geq 0$ ; we have then

(8)

\mathbf{Q}_+ + \mathbf{Q}_+ \subset \mathbf{Q}_+,

(9)

\mathbf{Q}_+ \cap (- \mathbf{Q}_+) = \{ 0 \},

(10)

\mathbf{Q}_+ \cup (- \mathbf{Q}_+) = \mathbf{Q}.

The relation $x \leq y$ is equivalent to $y - x \in \mathbf{Q}_+$ .

We shall use this ordering to define a topology on $\mathbf{Q}$ compatible with its additive group structure.

2. The rational line

Consider the set $\mathfrak{F}$ of symmetric open intervals $]-a, +a[$ , where $a$ runs through the set of rational numbers $> 0$ ; we shall show that $\mathfrak{F}$ is a fundamental system of neighbourhoods of $0$ in a topology compatible with the additive group structure of $\mathbf{Q}$ .

The group $\mathbf{Q}$ is commutative, and axiom $(\mathrm{GV}'_{\mathrm{II}})$ is clearly satisfied; it is therefore enough to show that axiom $(\mathrm{GV}'_{\mathrm{I}})$ is also satisfied, in other words, that for each $a > 0$ there exists $b > 0$ such that the conditions $|x| < b$ and $|y| < b$ together imply $|x + y| < a$ . The triangle inequality shows that we may take $b = a/2$ .

Definition 1

The rational line is the topological space consisting of the set $\mathbf{Q}$ together with the additive group topology for which the symmetric open intervals $]-a, +a[$ $(a > 0)$ form a fundamental system of neighbourhoods of $0$ .

The topological group $\mathbf{Q}$ thus defined is called the additive group of the rational line.

If $a$ is any rational number $> 0$ , there is an integer $n > 0$ such that $1/n < a$ ; hence the open intervals $]-\frac{1}{n}, +\frac{1}{n}[$ $(n = 1, 2, \ldots)$ form a fundamental system of neighbourhoods of $0$ on the rational line.

We obtain a fundamental system of neighbourhoods of any point $x \in \mathbf{Q}$ by taking the open intervals $]x - a, x + a[$ , where $a$ runs through the set of rational numbers $> 0$ (or the set of numbers $1/n$ ).

Definition 1 is therefore equivalent to that given in Chapter I, § 1, no. 2.

For each pair of rational numbers $(a, b)$ such that $a < b$ , there exists $c \in \mathbf{Q}$ such that $a < c < b$ [for example $c = (a + b)/2$ ]; it follows that the rational line is a non-discrete Hausdorff space.

For each $a > 0$ , let $U_a$ be the set of pairs $(x, y)$ in $\mathbf{Q} \times \mathbf{Q}$ such that $|x - y| < a$ . As $a$ runs through the set of rational numbers $> 0$ (or just the set of numbers $1/n$ ), the sets $U_a$ form a fundamental system of entourages of the uniformity of the additive group $\mathbf{Q}$ of the rational line. Relations (2) and (3) show that $|x|, x^+$ and $x^-$ are uniformly continuous on $\mathbf{Q}$ . It follows that the functions $\sup(x, y)$ and $\inf(x, y)$ are uniformly continuous on $\mathbf{Q} \times \mathbf{Q}$ .

3. The real line and real numbers

Definition 2

Let $\mathbf{R}$ denote the topological group which is the completion of the additive group $\mathbf{Q}$ of the rational line. The elements of $\mathbf{R}$ are called real numbers; as a topological space, $\mathbf{R}$ is called the real line; as a topological group, $\mathbf{R}$ is called the additive group of the real line.

We shall always identify $\mathbf{Q}$ with the dense subgroup of $\mathbf{R}$ to which it is canonically isomorphic. With this convention, every rational number is a real number. Every real number which is not rational is said to be irrational; we have seen in Chapter II, § 3, no. 3 that such numbers exist (we shall show this in another way in § 3, no. 3 of this chapter; see also Exercise 2 to § 2); hence (Chapter III, § 2, no. 1) the set $\mathbf{CQ}$ of irrational numbers is dense in $\mathbf{R}$ .

We shall show that the order structure of $\mathbf{Q}$ can be extended to $\mathbf{R}$ in such a way that the extended ordering is still compatible with the additive group structure of $\mathbf{R}$ :

Proposition 1

The relation $y - x \in \overline{\mathbf{Q}}_+$ is an ordering on $\mathbf{R}$ which makes $\mathbf{R}$ into a linearly ordered set; is compatible with the additive group structure on $\mathbf{R}$ , and induces the ordering $x \leq y$ on $\mathbf{Q}$ .

We begin by showing that the relations $y - x \in \overline{\mathbf{Q}}_+$ and $z - y \in \overline{\mathbf{Q}}_+$ imply $z - x \in \overline{\mathbf{Q}}_+$ . Indeed, the function $x + y$ is continuous on $\mathbf{R} \times \mathbf{R}$ , and therefore by (8) we have $\overline{\mathbf{Q}}_+ + \overline{\mathbf{Q}}_+ \subset \overline{\mathbf{Q}}_+$ (Chapter I, § 2, no. 1, Theorem 1). Next, we shall show that the relations $y - x \in \overline{\mathbf{Q}}_+$ and $x - y \in \overline{\mathbf{Q}}_+$ imply $x = y$ ; this will establish that $y - x \in \overline{\mathbf{Q}}_+$ is an ordering on $\mathbf{R}$ . It is enough to show that $\overline{\mathbf{Q}}_+ \cap (-\overline{\mathbf{Q}}_+) = \{0\}$ . Now the functions $x \to x^+$ and $x \to x^-$ are uniformly continuous on $\mathbf{Q}$ and can therefore be extended by continuity to $\mathbf{R}$ (Chapter II, , no. 6, Theorem 2); let $f$ and $g$ be their respective extensions. By extension we have $x = f(x) - g(x)$ for all $x \in \mathbf{R}$ ; if $x \in \overline{\mathbf{Q}}_+$ then $g(x) = 0$ , and since $-\overline{\mathbf{Q}}_+$ is the closure of $-\mathbf{Q}_+$ by the continuity of $-x$ , it follows that if $x \in -\overline{\mathbf{Q}}_+$ then $f(x) = 0$ . Hence if $x \in \overline{\mathbf{Q}}_+ \cap (-\overline{\mathbf{Q}}_+)$ we have $f(x) = g(x) = 0$ and therefore $x = 0$

By (10), we have $\overline{\mathbf{Q}}_+ \cup (-\overline{\mathbf{Q}}_+) = \mathbf{R}$ , and hence $\mathbf{R}$ is linearly ordered by the ordering $y - x \in \overline{\mathbf{Q}}_+$ .

Furthermore, since the relations $y - x \in \overline{\mathbf{Q}}_+$ and $(y + z) - (x + z) \in \overline{\mathbf{Q}}_+$ are equivalent, the ordering $y - x \in \overline{\mathbf{Q}}_+$ is compatible with the additive group structure of $\mathbf{R}$ .

Finally, if $x$ and $y$ belong to $\mathbf{Q}$ the relations $y - x \in \overline{\mathbf{Q}}_+$ and $y - x \in \mathbf{Q}_+$ are equivalent, and therefore the relation $y - x \in \overline{\mathbf{Q}}_+$ induces the relation $x \leq y$ on $\mathbf{Q}$ . This completes the proof.

The relation $y - x \in \overline{\mathbf{Q}}_+$ is again denoted by $x \leq y$ . The set $\overline{\mathbf{Q}}_+$ is the set of all $x \geq 0$ in $\mathbf{R}$ and is denoted by $\mathbf{R}_+$ ; it is a closed set. The set of all $x > 0$ is denoted by $\mathbf{R}_+^*$ ; it is the complement of $-\mathbf{R}_+$ and is therefore open in $\mathbf{R}$ .

4. Properties of intervals in $\mathbf{R}$

Proposition 2

Every closed (resp. open) interval in $\mathbf{R}$ is a closed (resp. open) set in $\mathbf{R}$ .

The sets $[a, \to[ = a + \mathbf{R}_+$ and $]\leftarrow, a] = a - \mathbf{R}_+$ are obtained by translation from $\mathbf{R}_+$ and $-\mathbf{R}_+$ respectively and are therefore closed (Chapter III, § 1, no. 1); the sets $]\leftarrow, a[$ and $]a, \to[,$ which are their complements, are open; finally, the closed interval $[a, b]$ (resp. the open interval $]a, b[$ ) is the intersection of $[a, \to[$ and $]\leftarrow, b]$ (resp. of $]a, \to[$ and $]\leftarrow, b[$ ) and is therefore a closed (resp. open) set.

The closed intervals $[-a, +a]$ ( $a > 0$ ) in $\mathbf{R}$ are therefore neighbourhoods of 0. Let us show that they form a fundamental system of neighbourhoods of 0 as $a$ runs through $\mathbf{R}_+^*$ . For this it is enough to establish the following proposition:

Proposition 3

As $r$ runs through the set of rational numbers $> 0$ , the intervals $s_r = [-r, +r]$ in $\mathbf{R}$ form a fundamental system of neighbourhoods of 0.

By Proposition 7 of Chapter III, § 3, no. 4 we obtain a fundamental system of neighbourhoods of 0 in $\mathbf{R}$ by taking the closures in $\mathbf{R}$ of the intervals

S_r \cap \mathbf{Q} = [-r, +r] \textit{of} \mathbf{Q}. The proof will be complete if we show that S_r is the closure of S_r \cap \mathbf{Q}. Now S_r is closed in \mathbf{R}, and we need therefore only prove that, if x is a real number such that -r < x < r, then x is in the closure of S_r \cap \mathbf{Q}. The interval ]-r, +r[ is an open set in \mathbf{R} and therefore for all sufficiently small neighbourhoods V of o in \mathbf{R} we have x + V \subset ]-r, +r[; but \mathbf{Q} being dense in \mathbf{R}, there is a rational number r' \in x + V, so that -r < r' < r and therefore r' \in S_r \cap \mathbf{Q}.

Corollary

Every point of the real line has a countable fundamental system of neighbourhoods.

Proposition 4

If (x, y) is any pair of real numbers such that x < y, there is a rational number r such that x < r < y.

Since \mathbf{Q} is dense in \mathbf{R}, it is enough to show that ]x, y[ is not empty; by translation we may assume x = 0 and y > 0. Now \mathbf{R} is a Hausdorff space and therefore, by Proposition 3, there is a rational number r > 0 such that y \notin ]-r, +r[, and this implies that 0 < r < y.

Proposition 5

Let I be any interval in \mathbf{R}. Then the topology induced on I by the topology of \mathbf{R} is generated by the open intervals of I (where I is considered as linearly ordered by the relation x \leq y).

Every open interval of I is the trace on I of an open interval of \mathbf{R}. This is clear for a bounded interval, and the unbounded interval ]a, \to[ of I is the trace of the unbounded interval ]a, \to[ of \mathbf{R}. We may therefore restrict ourselves to the case I = \mathbf{R}; but in this case the result follows from Proposition 3, since every neighbourhood of a point x \in \mathbf{R} contains an open interval ]x - a, x + a[.

Remark

If A is a dense subset of \mathbf{R}, the topology of \mathbf{R} is generated by the open intervals whose end-points belong to A. For if

]x - a, x + a[

is an open interval containing x, there exist two points y, z of A such that x - a < y < x and x < z < x + a; hence ]y, z[ contains x and is contained in ]x - a, x + a[. This proof shows, moreover, that the intervals under consideration form a base (Chapter I, § 1, no. 3) of the topology of \mathbf{R}. In particular, if we take A = \mathbf{Q}, we see that the topology of \mathbf{R} has a countable base.

5. Length of an interval

Definition 3

The length of a bounded interval with end-points a and b (a \leq b) is defined to be b - a.

Every bounded interval which contains more than one point therefore has length > 0. If $a \leq b$ , the four intervals $[a, b], ]a, b[, [a, b[$ and $]a, b[$ all have the same length. An interval with end-points $a + c$ and $b + c$ has the same length as an interval with end-points $a$ and $b$ ; in other words, the length of an interval is invariant under translation.

If $a \leq c \leq d \leq b$ we have $d - c \leq b - a$ . Hence if a bounded interval $I$ is contained in a bounded interval $I'$ , the length of $I$ is less than or equal to the length of $I'$ .

If $n$ mutually disjoint open intervals $I_1, I_2, \ldots, I_n$ are contained in the interval $[a, b]$ ( $a < b$ ) it is easily seen by induction on $n$ that, if $I_k = ]c_k, d_k[$ , there is a permutation $\sigma$ of the indices $k$ ( $1 \leq k \leq n$ ) such that $d_{\sigma(k)} \leq c_{\sigma(k+1)}$ for $1 \leq k \leq n - 1$ . It follows immediately that the sum of the lengths of the intervals $I_k$ is at most equal to the length of $[a, b]$ , and that equality holds if and only if $c_{\sigma(1)} = a, d_{\sigma(n)} = b$ and $d_{\sigma(k)} = c_{\sigma(k+1)}$ for $1 \leq k \leq n - 1$ .

6. Additive uniformity of $\mathbf{R}$

Since the group $\mathbf{R}$ is linearly ordered, the functions $x^+, x^-$ and $|x|$ are defined on $\mathbf{R}$ in the same way as on $\mathbf{Q}$ and satisfy all the relations listed above for $\mathbf{Q}$ , notably relations (1) to (7). Let $a$ be a real number > 0 and let $U_a$ be the set of all pairs $(x, y) \in \mathbf{R} \times \mathbf{R}$ such that $|x - y| < a$ ; as $a$ runs through the set of real numbers > 0 (or the set of numbers $1/n$ ), the sets $U_a$ form a fundamental system of entourages of the uniformity of the additive group $\mathbf{R}$ of the real line (called the additive uniformity of the real line).

The functions $|x|$ , $x^+$ and $x^-$ are uniformly continuous on $\mathbf{R}$ , and the functions $\sup(x, y)$ and $\inf(x, y)$ are uniformly continuous on $\mathbf{R} \times \mathbf{R}$ ; these functions therefore coincide with those obtained by extending by continuity the corresponding functions defined on $\mathbf{Q}$ and $\mathbf{Q} \times \mathbf{Q}$ , respectively.

Exercises

See the exercises for § 1.