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A Construction of the Real Numbers
Dr. Holmes
November 13, 2006
This is a more complete version posted later on Monday. I’m not planning
to attach exercises; I may eventually complete the construction of the general
reals if I can see the least tangled way to do it!
We begin our construction with the system of positive rational numbers,
supposed already understood.
Axioms which are true in the positive rationals and in the reals:
Commutative laws: Addition and multiplication are commutative operations.
Associative laws: Addition and multiplication are associative operations.
Distributive law: x(y + z) = xy + xz as usual.
Identity for multiplication: 1 is the identity element for multiplication.
Inverse law for multiplication: Every number a has a multiplicative inverse a−1 . (recall that zero is not in our system).
Definition: We define x < y as ∃z(x + z = y) then >, ≤, ≥ are defined
using < as usual.
Trichotomy 1: For any x and y, either x < y, y < x, or x = y.
Trichotomy 2: For any x and y, either ¬x < y or ¬y < x.
The usual properties of order in this context are fairly easy to prove from
these axioms. We give some properties which may not be entirely obvious.
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transitivity: If x < y and y < z then there are u, v such that x + u = y and
y + v = z, so (x + u) + v = z, so x + (u + v) = z, so x < z.
additive cancellation: If a + c = b + c then a = b: suppose a + c = b + c
and a 6= b. Then either a < b or b < a; suppose the former without loss
of generality. We have k such that a + k = b. Then a + c = (a + k) + c,
whence a + c = (a + c) + k, whence a + c < a + c, which is ruled out
by Trichotomy 2.
monotonicity of addition: If a < b then there is k such that a + k = b,
whence (a + c) + k = (a + k) + c = b + c so a + c < b + c. On the other
hand, if a + c < b + c then there is k such that (a + c) + k = b + c,
whence (a + k) + c = b + c, whence a + k = b by additive cancellation,
so a < b.
monotonicity of multiplication: If a < b then there is k such that a+k =
b, whence ac + kc = (a + k)c = bc, so ac < bc. On the other hand,
if ac < bc it follows that there is k such that ac + k = bc, whence
(ac + k)c−1 = bcc−1 , whence a + kc−1 = b, so a < b.
The positive reals satisfy the Least Upper Bound Property in addition to
these axioms.
We need one more property of the positive rationals to make our proofs
work (related to the induction property of the natural numbers):
Induction Property: For any positive rational x, any nonempty set of positive rationals which is downward closed and closed under addition of
x contains all positive rationals.
This can be understood as keeping infinitesimal quantities out of our
system.
A corollary of this axiom which we also use is the following:
Unbounded Powers: For any x > 1, any nonempty set of positive rationals
which is downward closed and closed under multiplication by x contains
all positive rationals.
Proof: Let x > 1. Then x = 1 + e for some e. Let I be a nonempty
set of positive rationals which is downward closed and closed under
multiplication by 1 + e. Let a be an element of this set. Now we show
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that I is nonempty and closed under addition of ea: for any y ∈ I, we
know that y(1 + e) ∈ I. If y ≤ a then y + ea ≤ a + ea = a(1 + e) ∈ I.
If y > a, then y + ea < y + ey = y(1 + e) ∈ I.
With any positive real r a unique set Lr is associated: the set of all
positive rationals q < r.
Definition: A set A is said to be a Dedekind left set if it is a set of positive
rationals (the set of all positive rationals is denoted by Q+ ) with the
following properties:
1. A 6= ∅ and A 6= Q+
2. A is downward closed: if a is a positive rational, b is a positive
rational, b < a, and a ∈ A, then b ∈ A.
3. A has no largest element.
Observation: The Dedekind left sets are exactly the sets Lr for arbitrary
real numbers r. It is easy to verify that each Lr actually has these
properties, and further that if A is a Dedekind left set, it is equal to
Lr where r is the least upper bound of A.
Now we propose to define the positive real numbers as Dedekind left sets.
Definition: We define a (Dedekind) positive real number as a Dedekind left
set of positive rationals. If R and S are positive real numbers, we define
R + S as {r + s | r ∈ R ∧ s ∈ S} and R · S as {r · s | r ∈ R ∧ s ∈ S}.
A little attention to notation is required here: a more explicit way to
write {r + s | r ∈ R ∧ s ∈ S} is {u | ∃r∃s(r ∈ R ∧ s ∈ S ∧ r + s = u).
We need to prove that S + T and S · T are actually Dedekind positive
reals.
Suppose that S and T are Dedekind positive reals. S + T and S · T
are clearly both nonempty. Suppose s0 6∈ s and t0 6∈ t. Since s0 exceeds all
elements of S and t0 exceeds all elements of T , we see that s0 + t0 exceeds all
elements of S + T and s0 · t0 exceeds all elements of S · T . Thus neither S + T
nor S · T are the entire set of positive rationals. Now we need to show that
both sets are downward closed. Suppose s ∈ S and t ∈ T and u < s + t.
It is straightforward to show that either u2 < s or u2 < t (suppose otherwise
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and show u ≥ s + t). Suppose without loss of generality that u2 < s, so
u
+ v = s for some v. Express u as ( u2 + v) + ( u2 − v). u2 + v = s ∈ S; since
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s + ( u2 − v) = u < s + t, we have u2 − v < t so u2 − v ∈ T so u ∈ S + T .
Now suppose that u < st. u = ut−1 t < st so ut−1 < s so ut−1 ∈ S whence
u ∈ S · T . So both S + T and S · T are downward closed. Now if s ∈ S
and t ∈ T , note that if s0 > s is an element of S we have s0 + t > s + t and
s0 t > st, so if we are presented with an element of either S + T and S · T we
can produce a larger element of the same set. This completes the argument
for closure of the Dedekind positive reals under the defined addition and
multiplication operations.
We proceed through the verification that the Dedekind positive reals have
the expected properties.
Commutative properties: R + S = {r + s | r ∈ R ∧ s ∈ S} = {s + r | s ∈
S ∧ r ∈ R} = S + R by commutativity of addition of positive rationals
(and if one wants to be technical, by commutativity of ∧ as well). The
proof for commutativity of multiplication is exactly the same.
Associative properties: We show that x ∈ (R + S) + T if and only if
x ∈ R + (S + T ) for any positive rational x and Dedekind positive reals
R, S, T .
Suppose x is any positive rational number and R, S, T are any Dedekind
positive reals.
Suppose that x ∈ R + (S + T ). Then there is u ∈ R and v ∈ S + T
such that u + v = x. Further, there is a ∈ S and b ∈ T such that
a + b = v. So u + (a + b) = x, whence (u + a) + b = x. Now observe
that u + a ∈ R + S and b ∈ T , so x ∈ (R + S) + T .
Suppose that x ∈ (R + S) + T . Then there is u ∈ R + S and v ∈ T
such that u + v = x, and, further, there is a ∈ R and b ∈ S such that
a + b = U . So (a + b) + v = x, whence a + (b + v) = x, whence (because
a ∈ R and b + v ∈ S + T ) we have x ∈ R + (S + T ).
Since (R + S) + T and R + (S + T ) have the same elements, they are
the same set.
The proof of associativity of multiplication is exactly analogous.
Distributive law: Let R, S, T be Dedekind positive reals.
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Suppose that x ∈ R · (S + T ). Then there are u ∈ R and v ∈ S + T
such that u · v = x. Further, there are a ∈ S and b ∈ T such that
a + b = v. So u · (a + b) = x, so u · a + u · b = x, and, since u · a ∈ R · S
and u · b ∈ R · T , we have x ∈ R · S + R · T .
Now suppose that x ∈ R · S + R · T . It follows that there are u ∈ R · S
and v ∈ R · T such that u + v = x. Further, there is a ∈ R and b ∈ S
such that a · b = u and c ∈ R and d ∈ S such that c · d = v. Here there
is a real obstruction: we would like a to be the same number as c! This
is easily finessed, however. Let e = max(a, c). a · b = e · (b · a · e−1 ), and
further e ∈ R and b · a · e−1 ∈ S (because a · e−1 ≤ 1 and S is downward
closed). Similarly c · d = e · (d · c · e−1 ) and d · c · e−1 ∈ T . Now we have
x = e · (b · a · e−1 + d · c · e−1), from which it follows that x ∈ R · (S + T ).
This completes the proof of the distributive property.
Identity law: The identity of the Dedekind positive reals is defined as 1R =
{x ∈ Q+ | x < 1}. Clearly this is a Dedekind real. 1 · R = {x · r | x <
1 ∧ r ∈ R}. Clearly any x · r with x < 1 and r ∈ R itself belongs to R
because R is downward closed and x · r < r. Further, any element r of
R can be written in the form x · r 0 where x < 1 and r 0 ∈ r: let r 0 > r
be any element of R greater than r (R has no largest element) and let
x = r · r 0−1 < 1.
Inverse law: For any Dedekind positive real R, define R−1 as {s−1 | s ∈
Q+ − R ∧ ∃s0 (s0 < s ∧ s0 ∈ Q+ − R). Proving that this is a Dedekind
positive real is routine. We outline the proof that R·R−1 = 1R . Suppose
that x ∈ R · R−1 . It follows that there are r ∈ R and s 6∈ R such that
r · s−1 = x. Since s 6∈ R we have s > r so x = r · s−1 < 1, so x ∈ 1R .
Now suppose that x ∈ 1R , that is, x < 1. We claim that there must be
r ∈ R and s ∈ R−1 such that r · s = x, or, equivalently, there must be
r ∈ R such that r · x−1 6∈ R (and further is not the minimal element of
the complement of R, either, but this is easy to arrange by increasing
r slightly). It is sufficient to show that for any fixed x−1 > 1, any
downward closed set of positive rationals closed under multiplication
by x−1 is in fact the entire set of positive rationals (the powers of x−1
are unbounded). This is the reason we stated the Unbounded Powers
property of the positive rationals above, which says exactly this.
Definition of order: We define order on the positive reals exactly as we did
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on the positive rationals. R < S is defined as “there exists a positive
real T such that R + T = S”. (Here we do not define R ≤ S as R ⊆ S,
though this equivalence is provable as a theorem).
Definition of subtraction: For positive rationals r < s, there is (by definition of order) t such that r + t = s, and it is fairly easy to show that
this t is unique using our axioms. We define r − s as this t. We want
to extend this definition to positive reals. If R and S are reals (naively
understood) we would like to identify the set of positive reals less than
R − S. A difference of rationals r-s will be less than R − S if r < R
and s > S (not s < S)!
We define R−S as {r−s | r ∈ R∧s 6∈ S}, for all R and S for which this
set is not empty. Notice that for this set to be nonempty there must
be an element of R which is greater than some non-element of S and
so greater than all elements of S. Further, if there is any element of R
which is not an element of S (and so greater than all elements of S),
it is easy to see that R − S will be nonempty. If this set is nonempty,
it is easy to see that it is a Dedekind positive real. It does not contain
all positive rationals because it is a subset of R. Suppose r ∈ R and
s 6∈ S and t < r − s. r − t exists and is greater than s so does not
belong to S: r − (r − t) = t witnesses the fact that t ∈ R − S. r − s
is not the largest element of R − S because we can choose r 0 > r in R
and r 0 − s > r − s belongs to R − S.
We claim that (R − S) + S = R if R − S is nonempty. An element of
(R − S) + S is of the form (r − s0 ) + s, where r ∈ R, s0 6∈ S and s ∈ S.
Since s0 > s must be true, we have (r − s0 ) + s < (r − s0 ) + s0 = r,
and so (r − s0 ) + s ∈ R because R is downward closed. Now let r
be any element of R which is greater than all elements of S: we want
to show r ∈ (R − S) + S (it will follow that all elements of R are in
(R − S) + S). Let r 0 > r be an element of R. Let s0 6∈ s and s ∈ S be
such that s0 − s ≤ r 0 − r. It should be clear that s0 < r 0 . [if there were
no such s0 and s then the smallest set containing S and closed under
addition of r 0 − r would contain only elements of S, contrary to the
Induction Property]. Now we see that (r 0 − s0 ) + s ≥ r is an element of
(R − S) + S, and so is r.
Trichotomy 1: Let R and S be Dedekind positive reals. Either R = S, in
which case we are done, or either R contains an element not in S or S
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contains an element not in R. If R contains an element not in S then
R − S is not empty and S + (R − S) = R, so S < R. If S contains and
element not in R we show R < S in the same way.
Trichotomy 2: Let R and S be Dedekind positive reals. Suppose R < S
and S < R. So we have R + T = S and S + U = R for some T and U ,
whence we have R + (T + U ) = R. But this means that R is closed as a
set under the addition of elements of T + U , or indeed under addition
of any specific element of T + U , which is impossible by the Induction
Property stated for the positive rationals.
Now we need to show that the positive reals satisfy the Least Upper
Bound Property. For this we need an observation: R < S is true precisely if
S − R is nonempty, which happens precisely if R ⊆ S ∧ R 6= S. So R ≤ S is
precisely equivalent to R ⊆ S: this is the most convenient way to view it for
proving the L.U.B. property.
Suppose that A is a set of Dedekind positive reals which is nonempty
S
and bounded above by a Dedekind positive real B. We claim that A is the
least upper bound of A.
S
We need to show that A is a Dedekind positive real. It is clearly
nonempty, because each element of A is nonempty. It is not the entire set of
S
positive rationals, because we have A ⊆ B for each A ∈ A, whence A ⊆ B
as well, and B is not all of the positive rationals. It is downward closed: if
S
S
r ∈ A and q < r (both positive rationals) then there is A ∈ A such that
S
S
r ∈ A, whence q ∈ A, whence q ∈ A. Finally, if r ∈ A, we have r ∈ A
for some A ∈ A, whence we have some r 0 > r with r 0 ∈ A, whence we have
S
r 0 ∈ A as well.
S
S
S
S
A ⊆ A is obvious for A ∈ A, so A ≤ A for all A ∈ A: the
purported least upper bound is in fact an upper bound. We have already
S
seen above that A ⊆ B, where B was an arbitrarily chosen upper bound
of A, so we have the least upper bound.
What we have constructed so far is just a model of our axioms for the
positive reals. We proceed to complete our task by showing how to get a
model of all the reals, positive and negative.
We only sketch this for now: it looks like the construction of the integers
from the natural numbers on the track the book takes. The general reals
are defined as equivalence classes of ordered pairs of positive reals under the
equivalence relation (m, n) ∼ (p, q) defined as m + q = n + p. We need to
verify that this is an equivalence relation.
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Then we define [(m, n)] + [(p, q)] = [(m + p, n + q)] and [(m, n)] · [(p, q)] =
[(mp + nq, mq + np)]. We define [(m, n)] < [(p, q)] as m + q < n + p. It
is necessary to verify that these definitions do not depend on the choice of
representatives from the equivalence classes.
It is then an exercise in algebra (using additive cancellation to compensate
for the lack of subtraction) to show that this has the properties of a complete
ordered field.
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