Chapter 5
Construction of the real numbers
The construction of Z from ω and of Q from Z can be found in the book. Here we detail the
construction of R from Q.
Unlike the previous constructions we do not define R as the quotient space of an equivalence
relation on Q × Q. Rather we take it to be a subset of P(Q). In other words a real number is
taken to be a set of rational numbers.
In effect we will take x ∈ R to be the set of rational numbers strictly smaller than x. Note
however that this is very far from a definition. We assume x exists and describe it in terms of
rational numbers. To construct R from Q, we must define a real number as a set of rationals
that satisfies some specific properties.
1
Dedekind cuts
For historical reasons a real number is often called a Dedekind cut. The distinction is sound
for everyday use, as Dedekind cuts are sets of rationals and real numbers are points on the
real line. For our purposes, though the distinction has no practical use, as we will define a real
number to be a Dedekind cuts!
An important notational remark: as the building set is Q, the binary operations +, − and ·
will be assumed to be addition, subtraction and multiplication in Q; and < will be assumed to
be the linear ordering in Q.
Definition. A real number or Dedekind cut x is a subset of Q that satisfies the following three
properties:
(i) ∅ =
6 x ( Q.
(ii) x is closed downward, i.e.,
q ∈ x ∧ p < q =⇒ p ∈ x .
(iii) x has no largest element, i.e., if q ∈ x, then there exists r ∈ x that satisfies q < r.
The set of real numbers is denoted by R.
1
Unlike ω, Z and Q we begin by putting a linear ordering in R and leave binary operations for
later.
2
Ordering the real numbers
Let us see what ordering do we expect to have in R. Like we mentioned above the informal
way to think about a real numbers x is {q ∈ Q : q < x} Therefore if x is a strictly smaller real
number than y, then {q ∈ Q : q < x} ⊂ {q ∈ Q : q < y}.
So the following definition does not contradict out intuition.
Definition. Let x and y be real numbers. Then
x <R y ⇐⇒ x ⊂ y.
So <R : R × R is the relation
<R = {hx, yi ∈ R × R : x ⊂ y}.
For brevity we write
x ≤R y ⇐⇒ (x = y ∨ x <R y) .
Theorem 5RA. <R is a linear ordering on R.
Proof. The relation is transitive as x ⊂ y ⊂ z, implies x ⊂ z.
We must show it satisfies trichotomy on R. Let x and y be real numbers. It is clear that at
most one of
x⊂y
x=y
y⊂x
is true. We must show that at least one holds.
Suppose the first two fail. Then there exists q ∈ x \ y.
Every p ∈ y satisfies p ≤ q, for otherwise there exists q < p ∈ y and, by downward closedeness
of y, q ∈ y. It follows that p ∈ x, as x is closed downward, and consequently that y ⊆ x.
Now that we have an ordering in R, we prove one of the most fundamental results about real
numbers: that every bounded set of real numbers has a least upper bound. Some definitions
first.
2
Definitions. Let A ⊂ R be a set of real numbers. A real number x ∈ R is a upper bound of
A, if a ≤R x for all a ∈ A. x is a least upper bound if it less than any other upper bound. A is
bounded if there exists a bound of A.
Theorem 5RB. Any bounded non-empty subset of R has a least upper bound.
Proof. Let ∅ 6= A ⊂ R be a non-empty subset of R and b be an upper bound for A. We will
S
show that A is a least upper bound for A.
S
The first step of the proof is to show that A is a real number.
S
- (i) A ⊆ Q as every a ∈ A is a subset of Q. A is non-empty. So there exists a ∈ A. a ∈ R,
S
S
S
so there exists q ∈ a. Hence q ∈ A and so A 6= ∅, Furthermore A ⊆ b as every a ∈ A is
S
a subset of b. b ∈ R and so b 6= Q, and so A 6= Q.
S
S
- (ii) A is closed downward. Let q ∈ A, then q ∈ a for some a ∈ A. If p < q, then p ∈ a (as
S
a is closed downward) and so p ∈ A.
S
S
- (iii) A has no greatest element. Suppose that x ∈ A is the greatest element. Then x ∈ a
S
for some a ∈ A and x must be the greatest element of a (as a ⊆ A). This is a contradiction.
S
The second part of the proof is to show that A is an upper bound for A. Every a ∈ A is a
S
S
subset of A and so a ≤R A.
S
The third and final is to show that if z is an upper bound for A, then A ≤R z. We simply
S
note that every a ∈ A is a subset of z and so A ⊆ z.
3
Addition in the real numbers
The next step is to define the addition operation +R for the real numbers. Naturally we use as
a building block +, addition in the rationals.
Definition. Let x, y ∈ R. Their sum x +R y is defined as
x +R y = {q + r : q ∈ x ∧ r ∈ y}.
Our first task is to show that the above definition does indeed produce a real number.
Lemma 5RC. Let x and y be real numbers. Their sum x +R y is a real number.
Proof. We check the three properties.
(i) x +R y ⊆ Q as every element of the set is a rational.
3
x, y ∈ R so x, y are non-empty. Let q ∈ x and r ∈ y. Their sum q + r lies in x +R y and so
x + Ry is non-empty.
It is furthermore not the whole of Q. x, y ∈ R so there are a ∈ Q \ x and b ∈ Q \ y. Let q ∈ x.
Then a 6= q and a 6< q, as x is closed downward and a < q would imply a ∈ x. By trichotomy
we must have that a > q for all q ∈ x. Similarly b > r for all r ∈ y. Adding the two inequalities
gives: a + b > q + r. So a + b 6= q + r for all q ∈ x and r ∈ y and so a + b ∈
/ x +R y.
(ii) x +R y is closed downward. Let z = q + r ∈ x +R y (q ∈ x, r ∈ y) and suppose p < z. Then
p − r < q and as x is closed downward we have p − r ∈ x. Therefore, p = (p − r) + r ∈ x +R y.
(iii) x +R y has no largest element. Let z = q + r ∈ x +R y (q ∈ x, r ∈ y). x ∈ R and so q is not
largest in x. So there is q < q 0 ∈ x. Similarly there is r < r0 ∈ y. Adding the two inequalities
gives
z = q + r < q 0 + r0 ∈ x +R y.
The commutativity and associativity of addition in R follows from the corresponding property
of addition in Q and is left as an exercise. We next define zero, which is the additive identity
in R.
Definition. The zero element in R is defined to be the set of negative rational numbers
0R = {q ∈ Q : q < 0}.
Let us know prove that the definition is sensible.
Theorem 5RE. (a) 0R is a real number.
(b) For every x ∈ R, we have x +R 0R = x.
Proof. (a) We check the three properties.
(i) By definition 0R ⊆ Q. −1 ∈)R and so 0R is non-empty. 1 ∈
/ 0R and so 0R 6= Q.
(ii) 0R is closed downward. Let q ∈ 0R and p < q. Then p < q < 0 and so p ∈ 0R .
(iii) 0R has no greatest element. Let q ∈ 0R . Then q < 0 and, by the density of the rationals,
there exists q < r < 0 (e.g. r = q/2, c.f. Q.5 in HWK 7). So q < r and r ∈ 0R .
(b) We have to prove x +R 0R ⊆ x and x +R 0R ⊇ x.
⊆: Let z = r + q ∈ x +R 0R (r ∈ x and q ∈ 0R ), we show r + q ∈ x. Note that q < 0. So adding
r to both sides gives r + q < r. x is closed downward and r ∈ x, so r + q ∈ x, as claimed.
4
⊇: Let r ∈ x, we show r ∈ x +R 0R by expressing it as a sum of another element of x and an
element of 0R . r is not the largest element of x, so there is r < s ∈ x such that s ∈ x. Then
r = s + (r − s)
with s ∈ x, and r − s < 0 and so r − s ∈ 0R . So r ∈ x +R 0R , as claimed.
4
Negatives of real numbers
We next define the negative of a real number. An intuitive definition of −x is {q ∈ Q : −q ∈
/ x}.
The problem is that such a set may have a largest element. Indeed 0 is the largest element of
{r ∈ Q : −r ∈
/ 0R } = {r ∈ Q : −r ≥ 0}. To fix this we utilise a standard trick: r ∈ −x if there
exists an other rational s, that is bigger than r, and whose negative does not lie in x.
Definition. Let x ∈ R. its negative or additive inverse is
−x = {r ∈ Q : ∃s > r (−s ∈
/ x)}.
For example,
−0R = {r ∈ Q : ∃s > r (−s ∈
/ 0R )}
= {r ∈ Q : ∃s > r (−s ≥ 0)}
= {r ∈ Q : ∃s > r (s ≤ 0)}
= 0R .
We must prove that the negative has the desired properties.
Theorem 5RF. Let x ∈ R.
(a) −x ∈ R.
(b) x +R (−x) = 0R .
Proof. (a) We must check the three properties.
(i) −x ⊆ Q by definition.
x ∈ R, so there exists t ∈ Q \ x. Then −t − 1 ∈ −x (we may take s = −t in the definition of
−x). Hence −x 6= ∅.
x ∈ R, so there exists q ∈ x. We show that −q ∈
/ −x, which implies −x 6= Q. Suppose that
s > −q. Then −s < q and so, as x is closed downward, −s ∈ x.
5
(ii) −x is closed downward. Let r ∈ −x and q < r. Then there exists s > r > q such that
−s ∈
/ x. So q ∈ −x (by “using the same s as r”).
(iii) −x has no greatest element. Let r ∈ −x. There exists s > r such that −s ∈
/ x. By the
density of the rationals, there exists r < p < s (e.g. p = (r + s)/2, c.f. Q.5 in HWK 7). Hence
p ∈ −x (“using the same s as r”) and so r is not greatest in −x.
(b) ** Non-examinable ** Left as an exercise (Q.3 and Q.4 in HWK 8).
We now deduce the cancelation law for addition in R.
Corollary 5RG. Let x, y, z ∈ R.
x +R z = y +R z =⇒ x = y.
Proof.
x +R z = y +R z
=⇒
(x +R z) +R (−z) = (y +R z) +R (−z)
=⇒
x +R (z +R (−z)) = y +R (z +R (−z))
=⇒
x +R 0R = y +R 0R
=⇒
x = y.
It takes a little more work to show the following.
Theorem 5RH. Let x, y, z ∈ R.
(x <R y) ⇐⇒ (x +R z < y +R z).
** Proof **. Can be found in p. 118 of the book.
5
Absolute value and multiplication
Multiplication is trickier to define as the sign of the real numbers whose product we want to
define plays a role.
Let us first note that the absolute value of x ∈ R is the least upper bound of {x, −x} and so
equals
|x| = x ∪ (−x).
6
Let x, y ∈ R. Their product x · y is defined as follows
(i) If both x and y are nonnegative, then
x ·R y = 0R ∪ {0 ≤ r · s : r ∈ x ∧ 0 ≤ s ∈ y}.
(ii) If both x and y are negative, then
x ·R y = |x| ·R |y|.
(iii) If one of x, y is nonnegative and the other is negative, then
x ·R y = −(|x| ·R |y|).
The binary operation of multiplication as defined above satisfies all the expected properties.
We do not prove them here however.
6
Q as a subset of R
Our final task is to find a subset of R that can safely be identified with Q. Just like we found
a subset of Q that may identified with Z and a subset of Z that may be identified with ω.
Let q ∈ Q and define
E(q) = {p ∈ Q :< q}
be the set consisting of smaller rationals.
It is left as an exercise to check that E(q) is a real number.
We may think of E : Q 7→ R as a function.
Theorem 5RJ. Let E : Q 7→ R be the function defined above. Then
(i) E is an injection.
(ii) E(r + s) = E(r) +R E(s).
(iii) E(r · s) = E(r) ·R E(s).
(iv) E(0) = 0R .
(v) (r < s) ⇐⇒ (E(r) <R E(s)).
7
Proof. (i) Let r 6= s ∈ Q. By trichotomy we may without loss of generality assume that r < s.
The transitivity of < implies, E(r) ⊆ E(s). Moreover E(r) 6= E(s) as r ∈ E(s) \ E(r). Hence
E(r) ⊂ E(s) and the claim follows.
(ii) ** Non examinable ** We first establish the ⊆ inclusion: Let t ∈ E(r + s) and set ε :=
(r + s − t)/2 noting that ε > 0. Observe that
t = (r − ε) + (s − ε),
with r − ε < r and so r − ε ∈ E(s); and similarly s − ε ∈ E(s). So t ∈ E(r) +R E(s), as claimed.
Conversely,
E(r) +R E(s) = {p + q : p ∈ E(r) ∧ q ∈ E(s)}
= {p + q : p < r ∧ q < s}
= {p + q : p + q < r + s}
⊆ E(r + s).
(iii) Omitted.
(iv) By definition of 0R and E.
(v) =⇒ : proved in (i) above.
⇐= : E(r) <R E(s) means E(r) ⊂ E(s). It follows immediately that r =
6 s. Furthermore,
s∈
/ E(r) (otherwise E(s) ⊆ E(r) by downward closedness), which means s ≥ r. Trichotomy
implies r < s.
Another useful fact is that E(1) is the multiplicative identity in R. In other words, x·R E(1) = x,
for every x ∈ R.
8