DEFINITION. Let x and y be elements of F.
If either x or y is positive, define the product x × y to be the equivalence class
of the cut (A, B)(C, D), where x is the equivalence class of (A, B) and y is the
equivalence class of (C, D).
If either x or y is 0, define x × y to be 0.
If both x and y are negative, i.e., both -x and -y are positive, define x
× y =
(-x) × (-y).
The next exercise is tedious. It amounts to checking a bunch of cases.
bf Exercise A.6. (a) Prove that multiplication in F is commutative.
(b) Prove that multiplication in F is associative.
(c) Prove that multiplication in F is distributive over addition .
(d) Prove that the product of two positive elements of F is again positive.
We define the element 1 of F to be the equivalence class of the cut (A1,B1),
where A1 = {r : r ≤ 1} and B1 = {r : r > 1}
bf Exercise A.7. (a) Prove that the elements 0 and 1 of F are not equal.
(b) Prove that x × 1 = x for every element x
∈ F.
(c) Use the associative law and part (b) to prove that if xy = 1 and xz = 1,
then
y = z.
THEOREM A.3. With respect to the operations of addition and multiplication
defined above, together with the definition of positive elements, F is an ordered
eld.
PROOF. The first five axioms for a field, given in Chapter I, have been established
for F in the preceding exercises , so that we need only verify axiom 6 to
complete
the proof that F is a field. Thus, let x ∈ F be a non zero element . We must show
the existence of an element y of F for which x × y = 1. Suppose
first that x is a
positive element of F. Then x is the equivalence class of a positive cut (A, B),
and
therefore A contains some positive rational numbers . Let a0 be a positive number
that is contained in A. It fol lows then that every element of B is greater than
or
equal to a0 and hence is positive. Define
to be the set of all rational
numbers r
for which r ≥ 1/b for every b ∈ B. Then define
to be the set of all rationals r
for
which
for every
It follows directly that the pair
is a Dedekind
cut.
Let
, and note that every element d
∈ D is of the form
and hence is greater than or equal to 1. We claim that (C, D) is
equivalent
to the cut (A1,B1) that de termines the element 1 of F. To see this we must
verify
that D contains every rational number r that is greater than 1. Thus, let r > 1
be
given, and set ε = a0(r-1). From Condition (3) of Theorem A.1, choose an a'
∈ A
and a b' ∈ B such that b' - a' < ε . Without loss of generality, we may as sume
that
a ' ≥ a0. Finally, set
Clearly
for all b ∈ B, so that
Also
and
implying that r ∈ D. Therefore, (C, D) is equivalent to the cut (A1,B1), implying
that
is equivalent to the cut (A1,B1). Therefore, if y is the
element
of F that is the equivalence class of the cut
then x
× y = 1, as
desired.
If x is negative, then -x is positive. If we write z for the multiplicative
inverse
of the positive element -x, then -z is the multiplicative inverse of the element
x.
Indeed, by the definition of the product of two negative elements of F, x
× (-z) =
(-x) × z = 1.
The properties that guarantee that F is an ordered field also have been
established
in the preceding exercises, so that the proof of this theorem is complete.
So, the Dedekind field is an ordered field, but we have left to prove that it is
complete. This means we must examine upper bounds of sets, and that requires us
to understand when one cut is less than another one. We say that a cut (A,B) is
less than or equal to a cut C, D) if a ≤ d for every a ∈ A and d
∈ D. We say that
an element x in the ordered field F is less than or equal to an element y if y - x
is
either positive or 0.
THEOREM A.4. Let x and y be elements of F, and suppose x is the equivalence
class of the cut (A, B) and y is the equivalence class of the cut (C, D). Then x
≤
y
if and only if (A, B) ≤ (C, D).
PROOF. We have that x ≤ y if and only if the element y - x = y + -x is positive
or 0. Writing, as before, (A', B') for the cut -(A, B), we have that y - x is the
equivalence class of the cut (C, d) - (A, B) = (C, D) + (A', B'), so we need to
determine when the cut (G, H) = (C, D) + (A', B') is a positive cut or the 0 cut,
which is the case when the set H only contains nonnegative numbers. By definition
of addition, the set H contains all numbers of the form h = d + b' for some d
∈
D
and some b' ∈ B'. Since B' = -A, this means that H consists of all elements of
the
form h = d - a for some d ∈ D and a ∈ A. Now these numbers h are all greater
than or equal to 0 if and only if each a ∈ A is less than or equal to each d
∈
D, i.e.,
if and only if (A, B) ≤ (C, D). This proves the theorem
We are now ready to present the first of the two main theorems of this appendix,
that is Theorem 1.1 in Chapter I.
THEOREM A.5. There exists a complete ordered field. Indeed, the Dedekind
field F is a complete ordered field.
PROOF. Let S be a nonempty subset of F, and suppose that there exists an upper
bound for S, i.e., an element M of F such that x ≤ M for all x ∈ S. Write (A, B)
for a cut such that M is the equivalence class of (A, B). We must show that there
exists a least upper bound for S.
For each x ∈ S, let (Ax, Bx) be a Dedekind cut for which x is the equivalence
class of (Ax, Bx), and note that ax ≤ b for all ax ∈Ax and all b
∈ B. Let A0 be
the
union of all the sets Ax for x ∈ S. Let B0 be the set of all rational numbers r
for
which r ≥ a0 for every a0 ∈ A0. we claim first that the pair (A0,B0) is a Dedekind
cut. Both sets are nonempty, A0 because it is the union of nonempty sets, and B0
because it contains all the elements of the nonempty set B. Clearly Condition
(2)
for a cut holds from the very definition of this pair. To see Condition (1), let
r be
a rational number that is not in B0. We must show that it is in A0. Now, since r
is
not in B0, there must exist some a0 ∈ A0 for which r < a0. But
so
that there must exist an x ∈S such that a0 ∈Ax, and hence r is also in Ax. But
then r ∈A0, and this proves that (A0,B0) is a Dedekind cut.
Let M0 be the equivalence class determined by the cut (A0,B0). Since each
Ax ⊆ A0, we see that ax ≤ b0 for every ax
∈Ax and every b0 ∈ B0. Hence,
(Ax, Bx) ≤ (A0,B0) for every x ∈S, and therefore, by Theorem A.4, x
≤ M0 for
all x ∈S. This shows that M0 is an upper bound for S.
Finally, suppose M' is another upper bound for S, and let (A',B') be a cut for
which M' is the equivalence class of (A',B'). Then ax ≤ b' for every ax
∈Ax and
every b' ∈B', implying that a0 ≤ b' for every a0
∈A0 and every b' ∈ B'.
Therefore,
(A0,B0) ≤ (A',B'), implying that M0 ≤ M'. This shows that M0 is the least upper
bound for S, and the theorem is proved.
We come now to the second major theorem of this appendix, i.e., Theorem 1.2 of
Chapter I. This one asserts the uniqueness, up to isomorphism, of complete
ordered
fields.
THEOREM A.6. Let
be a complete ordered
field. Then there exists an isomorphism
of
onto the Dedekind
field F. That is, there exists a one-to-one function
J .
→ F that is onto all of F, and that satisfies
(1) J(x + y) = J(x) + J(y).
(2) J(xy) = J(x)J(y).
(3) If x > 0, then J(x) > 0.
PROOF. We know from Chapter I that, inside any ordered field, there is a subset
that is isomorphic to the field Q of rational numbers. We will therefore identify
this special subset of
with Q.
If x is an element of
, let Ax ={r
∈ Q : r ≤ x} and let Bx = {r ∈ Q : r > x}.
We claim first that the pair (Ax, Bx) is a Dedekind cut. Indeed, from the definition
of Ax and Bx, we see that Condition (2), i.e., that each ax ∈ Ax is less than or
equal to each bx ∈ Bx, holds. To see that Condition (1) also holds, let r be a
rational number in
. Then, because
is an ordered
field, either r ≤ x or r >
x,
i.e., r ∈ Ax or r ∈ Bx. Hence, (Ax, Bx) is a Dedekind cut.
We define a function J from
into F by setting J(x) equal to the equivalence
class determined by the cut (Ax, Bx). We must check several things.
First of all, J is one-to-one. Indeed, let x and y be elements of
that are
not
equal. Assume without loss of generality that x < y. Then, according to Theorem
1.8, which is a theorem about complete ordered fields and hence applicable to
,,
there exist two rational numbers r1 and r2 such that x < r1 < r2 < y, which
implies
that r1 ∈ Bx and r2 ∈ Ay. Since r2 > r1, the cut (Ay, By) is not equivalent to
the
cut (Ax, Bx), and therefore J(x) ≠ J(y).
Next, we claim that the function J is onto all of the Dedekind
field F. Indeed,
let
z be an element of F, and let (A, B) be a Dedekind cut for which z is the
equivalence
class determined by (A, B). Think of A as a subset of the complete ordered field
. Then A is nonempty and is bounded above. In fact, every element of B is an
upper bound of A. Let x = sup A. (Here is another place where we are using the
completeness of the field
.) We claim that the cut (A, B) is equivalent to the
cut
(Ax, Bx), which will imply that J(x) = z. Thus, if ax ∈ Ax, then ax
≤ x, and x ≤ b
for every b ∈B, because x is the least upper bound of A. Similarly, if a
∈ A,
then
a ≤ x, and x < bx for every bx ∈ Bx. This proves that the cuts (A, B) and (Ax, Bx)
are equivalent, as desired.
If x and y are elements of
, and bx
∈ Bx and by ∈ By, then bx > x and by > y,
so that bx + by > x + y, and therefore bx + by ∈ Bx+y for every bx
∈Bx and
by ∈ By. On the other hand, if r ∈Bx+y, then r > x + y. Therefore, r - x > y,
implying, again by Theorem 1.8, that there exists an element by ∈ By such that
y < by < r - x. But then r - by > x, which means that r - by = bx for some
bx ∈ Bx. So, r = bx + by, and this shows that Bx+y = bx + By. It follows from
this that the cuts (Ax+y, Bx+y) and (Ax, Bx) + (Ay, By) are equal, and therefore
J(x+y) = J(x)+J(y). A consequence of this is that J(-x) = -J(x) for all x
∈
.
If x and y are two positive elements of
, then an argument just like the one
in the
preceding para graph shows that J(xy) = J(x)J(y). Then, since J(-x) = -J(x),
the fact that J(xy) = J(x)J(y) for all x, y ∈
follows.
Finally, if x is a positive element of
, then the set Ax must contain some
positive
rationals, and hence the cut (Ax, Bx) is a positive cut, implying that J(x) > 0.
We have verified all the requirements for an isomorphism between the two
fields
and F, and the theorem is proved.
