B. 1
Proof of the Well-ordering Theorem
We follow the pattern outlined in Exercises 2-7 on pp. 72-73 of the text.
Then
h: J-YE.
be well-ordered sets; let
ar;d E
J
Let
Theorem B.1.
the following are equivalent:
of
or a section
E
equals
h(J)
is order preserving and
(i) h
E.
smallest [E - h(S4 )] for each
=
hj(,)
(ii)
J;
be an arbitrary element of
Let
holds.
(i)
Suppose
Proof.
·
let
e0.
)
h(
and suppose that
smallest [E - h(SB )],
=
e0
h(J)
Now we show that
and
</3 , then
On the other hand, if
e0 .
This contradicts the fact that h(J)
(ii)
Suppose
does not.
The set
t< .
, "h(
Since
h(A );
h(, )
If
h(p )
e.
And
that
e
belongsto
is injective; this follows
h(S, ),
wlhile by
Suppose
is order preserving:
s:nce the statement
h(P ),
h(J).
)
h(S& )
h
h
is
is injective.
Suppose that
te proof is complete.
h(
E.
or a section of
would contradict the fact that
h(J) cannot contain any
then the fact that
but does not contain
is the smallest element not in h(S<), we have
be the smallest element not in
than
h
equality cannot hold because
h(J) = E,
h
lies in
h(ol)
does not contain
) for some $<c ,,
) = h(
injective.
then
We then show that
h(S )
E
equals
show that
TWecfirst
holds.
from the fact that if o
(ii)> h(P )
eO,
contains an element greater than
h(J)
Thus
so that
h(Sp ),
belongs to
h(c)
)> eC.
h(o
we have
0
e0 .
by definition of
/ e0
h(di)
Kz_,
h(l )>e
Since
eO .
does not contain
is order preserving, then for all
h
e0.
contains an element greater than
h(J)
Thus
e@.
h(A )
Hence
is injective.
h
would contradict the fact that
tlhis in turn
;
for some A/S
)
h(
=
h()
so that
hS6 ),
lies in
)
h(
h(p) c eO , for that would imply that
It cannot be true that
Then
h(J) X E; let
h(J) contains every element less
element greater than
is the smallest element not in
ar;d hence to
e
h(J).
e)
for if
h(A) e,
h(Se) would imply
We conclude that
h(J) = Se.
B.2
map
of
or a section
E
thlat is order preserving and whose image is
E
h:J-
There is at most one
be well-ordered sets.
E
and
J
Let
B.2.
Corollary
E.
nor can two different sections of
J;
the order type of
has
J
is a well-ordered set, no section of
J
If
Ccrollary B.3.
have the same order
J
type.
If
Proof.
Sd is a section of
the conditions specified for the map
J, then inclusion
i : S -
4i-3 , then inclusion
S
in
e0
Coose
Proof.
h(oC)
(*)
whenever
E - h(S_)
Now, given p,
(i)
E- h(S
)
h(
(iii)
to
h(o<) = e0
h(
otherwise.
for all o(c.
is not empty.
)
k(
).
for 4<<
Given (i), we have the inequalities
, which imply that
k((3)
does not belong
It then follows from the definition of
h(( ) is the smallest element of
that, since
E.
consider the following conditions:
Thus (ii) holds.
h(Sn ).
If there is an
smallest[E - h(S4 )]
We show that (i) implies (ii) and (iii).
h(dL)4k(&)- k(o)
·
the principle of recursive definition, we
is nonempty, and
h(4) f k(4)
(ii)
=
E
not in
h
h(SA ), we have
).
) c k(
The fact that (i) implies (iii) shows, by induction, that
The fact that (i) implies (ii) then shows that
for all d,.
for all
Similarly,
by setting
E
h: J
may define a function
B
E.
h: S--> S
be well-ordered sets.
or a section of
E
-whose image is
h: S --~ J.
then there is an order -preserving map
k : J-- E,
order-preserving map
h : J-- E
E
and
J
TLct
B.4.
satisfies
satisfies these same conditions, so
there is no surjective order -preserving map
Theorem
J
of the preceding corollary.
h
Hence there is no surjective order-preserving map
if
i:S,-
.
We then apply
Theorem B.l.
l
h
h(id) < k(o)
satisifes (8)
B.3
Theorem B.5 (Comparability theorem). Let
sets.
A
and
B
be well-ordered
Eactly one of the following conditions holds:
(i)
A
(ii)
has the order type of
A
(iii)
B.
has the order type of a section of
B
B.
has the order type of a section of
A.
A
Assume without loss of generality that
Proof.
Order the set
C = A
declaring that
a<b
and
by using the order relations on
B
for
a
in
A
b in
and
are disjoint.
B
A
B; and by
and on
It is easy to see that
B.
C
is well-ordered.
Let
by
be the smallest element of
b0
theorem that there is an order-preserving map
or a section of
of
by
C
A
C
A.
b> b
of
type of a section of
If
whose image is
C
h(B)
And if
A.
B,
or if
C
by an element of
C
the section of
equals
by an element
has the order
C
the section of
equals
h : B-
has the order type of
then B
b0,
the section of
then
h(B)
If
has the order type of a section of
B
then
C.
A
Then
is order preserving; it follows from the preceding
i : B--C
b0 .
Inclusion
B.
A.
equals the section
equals
h(B)
equals all of
h(B)
C,
B.
The preceding corollary implies that only one of the conditions (i)-(iii)
can hold .n
Lenma B.6.
Let
is a subset of
A
(A,< )r, where
bej the collection of all pairs
be a set; let
X
X
and
is a well-ordering of
<
A.
Define
(A,<) <
if
(A,<)
order on
a
equals a section of
.
If
3
fr all
B,
union of the relations
Proof.
(A', <').
Then
is a strict partial
-
is a simply ordered subcollection of
the union of the sets
upper bound fr
(A',<
C')
<
in
,
fcr all
(B,< )
(B,--)
; and let
in
in
.
Then
,
let
< C
C
ecual,
equal the
(C,<C)
is an
A .
We, check the conditions for a.strict partial order. Nonreflexivity
is immediate, for A
cannot equal a section of itself.
immediate, since if
A!
Al
A 3.
is a section of
is a section of
A2
and
A2
Transitivity is also
is a section of
A,
then
Now consider the set
of
C.
of them (because
b0
and
bl
contains both
B
such that
D
One of these elements is less
is simply ordered by -< ).
ib
than the other under C<,
of
of
(B,<)
there is an element
C,
Given two distinct elements
arid which relation holds is independent of the choice
(B,< ), again because
is simply ordered.
Hence one is less than the
other under < C
Since the relation
we cannot have
b<b
cannot hold in
b0 ,
hold in
b0 <C
all belong to
b-
b
Therefore
D
C.
intersects
B1,
then one of
(B0, <C0
a section of
(C,<
'.
-
of
DLt
co
B
contains no element
B
equals. the section of
So suppose
such that
of
B
because
because
B
C
B contains
B
C
by
o
c0
that
d
element is independent of
such that
D
equals a section
DOB
and
D
B
are
C.
(C,<C) or it equals
(B,< ) equals
is contained in <C.
and that <
c be the smallest element of C
C by
is greater than
cO
And
and
B
c.
c;
c. We show that
this implies that
(B0, O)
B0 cannot be a section of
cannot be a section of
but not the smaller element
contradiction to the fact that
B0,
c. As before, there is an element
contains both
cO
be
D
c.
B does not contain. c .
contains
Let
iS well-ordered.
contains the section of
of
b2
is an upper bound for ' ; that is,
BcC
Suppose that equality does not hold.
B
b1
and
is the smallest element of
d
either
,
We know that
tlat is not in B. Then
such that
sco that we have
(B1, <1)
and
(C,< C)
Finally, we must show that
(B,<)
43
bl
is another element of
A similar argument shows that
given an element
in
)
b0
B,
This
(Bo
0 ,< 0).
(Bl', 1 )
For if
B.
such that
C
intersects soma set
D
Then
of the other, so that the smallest elements of
the same.
,
. Let us take the smallest element
belongs to-
BO
in the well-ordered set
0
the choice of
C
is simply ordered; we show
C
(B0, <0)
where
holds in
b2
bo0
an arbitrary nonempty subset of
of
(B,
anid the relations
B
Then the relation
B.
in
C b2
bl
is simply ordered, there is an element
and
b
are elements of
b,
b 1, and
be 0C
Because
(B,)
b<C b.
Finally, suppose
b0
B, for any
B0
c. Th-,us we reach a
is simply ordered. ­
B.5
The maximum principle is equivalent to the well-ordering
Theorem B.7.
theorem.
We have sketched in the text (p.70) how one can use the principle
Proof.
of recursive definition to show that the well-ordering theorem implies the
maximum principle.
a set
Given
preceding lemma provides a proof of tiereverse implication.
Thl
The maximum principle gives one
one proceeds as in the lemma.
X,
x
X, for if
must equal all of
x
to be larger than every element of
D
by
C
Ten
C.
ordered subcollection of LA
to
x
C
one
C,
and declaring
would equal the section of
wculd give us a simply
to the collection
D
Adjoining
by adjoining
D
C
Its upper bound
not in
X
were an element of
could form a larger well'ordered set
x.
.
that is simply ordered by
a maximal subcollection
that properly contains
d ,
contradicting
maximality. The choice axiom is equivalent to the well-ordering
B 8.
Tl-eorem
theorem.
It is immediate that the well-ordering theorem implies the choice
Proof.
axiom.
We prove the converse.
Given
If
T
i
let
X,
be a choice function for the nonempty subsets of
c
is a tower in
X
=
S (T)
Step 1.
T,
wsay that
and if for each x
T
is a well-ordering of
if <
x
where
is a relation on
ar;d <
X
a subset of
X.
(T,< )
in
T,
c(X - S(T)),
is the section of
Given two towers
T
x.
by
(T1,'
1)
and
(T,
< 2)
in
X,
either they
are equal or one equals a section of the other.
Switching indices if necessary, the comparability theorem tells us there is
an order-preserving map
h: T
whose image is either
T2
T 2
or a section of
T 2.
Theorem
B.1
tells us that
h
B.6
rmst be given y the formula
( )
h(x)
This in turn implies that
=
sm:allest[T2 - h(Sx(T
1))].
for all
h(x) = x
We proceed by transfinite induction.
for all
h(x) = x
that
We show
x<y.
Consider the restricted function
image must be a section of
contain
h(y).)
x
in
as we now show:
Suppose that
y
is in
T]
and
h(y) = y.
h: Sy(T 1 )-
(It cannot equal
T 2.
T 1,
Because (*) holds, the
T2 .
T2,
because it does not
This section is of course the section by the element
smallest [T2 - h(Sy(Tl))],
which by (*) is just
h(y).
Thbls
=
h(Sy(T 1 ))
Sh(y)(T 2)
It follows that
h(y)
Thus
T1
h(x) = x
equals either
Step 2.
=
c(X- Sh(y)(T 2 )) by definition of a tower,
=
c(X - h(Sy(T 1))
=
c(X - Sy(TI))
=
y
for all
x
(T,
h(x) = x
in
Ti
)
T1.
It follows that
(T,< )
ard let
of Lemma B.6.
This is in fact easy.
x
T
TI
equals
T1
is a tower,
(T)
1,
T
X.
X
is simply
It follows from this lemma that
in
x
T,
w
must show that
c(X - Sx(T)).
such that
in
X
or a section of
T.
Therefore,
x
ALt
be the union of all the relations
<
l1
Now there is a tower
X.
We show that it is actually a tower.
Given
=
so that
T 2.
is a tower in
is a well-ordered set.
x<y,
h(T 1) = T1 ,
We showed in Step 1 that the collection of all towers in
ordered by the relation -<
for
be the collection of all towers in
(T.<i)
be the union of all the sets
< i . We show that
because
byt definition of. a tower.
T 2 or a section of
LE!t
as just noted,
=
our desired result follows.
c(X - Sx(T1));
Tg
contains
Sx(T 1 ) = Sx(T).
x.
By Lemma B.6,
Because
B.7
If
T = X.
Step 3. We show that
T
is not all of
X
we can set
y = c(X - T),
and make the set
x
in
T uhy} into a well-ordered set by declaring
y
x
for every
X.
Then not only is this set well-ordered, it is also a tower in
T.
This contradicts the fact that
towers in
X
is obtained by, taking the union of all
T
a
EXERCISES
Suppose we alter the statement of Lemma B.6 by declaring that
1.
(A,<)
•
(A', <')
if
A
is contained in
Show that the resulting set
A'
and
is simply ordered.
C
<
'.
is contained in
Give an example to show that
it need not be well-ordered.
2. LIet
be a collection of sets.
Let us define two sets to be equivalent
if there is a bijection between them; the equivalence classes are called
cardinal numbers. Let us denote the equivalence class of the set
c(A); and let us define
but no injection of
B
c(A)< c(B)
into
A
if there is an injection
A
i : A-
by
B
Show that this is a well-defined relation,
and that this collection of cardinal numbers is well-ordered by this relation.
The cardinal number of the positive integers is commonly denoted
(Read "aleph naught.")
(surprise!) (i1.
10
The next cardinal number after this one is denoted
Thle cardinal number of the reals is denoted
cardinality of the continuum")'
c
("the
The continuum hypothesis is the statement that
=c.
[It is tempting to try to construct the collection of all cardinal numbers
by beginning with the collection-of all sets and introducing the above equivalence
relation.
notion.
Te problem is that the collection of all sets is a contradictory
See Exercise 6 of §9.
Logicians have formulated a way around this
difficulty, so that they can consider arbitrarily large car dinal numbers.
.]