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Set Theory
P.D.Welch
1 Introduction
The theory of sets can be regarded as prior to any other mathematical theory: any everyday
mathematical object, whether it be a group, ring or eld from algebra, or the structure of
the real line, the complex numbers etc., from analysis, or other mathematical construct,
can be constructed from sets.
The apparent simplicity of sets belies a bewildering collection of paradoxes, and logical
antinomies that plagued the early theory and led many to doubt that the theory could be
made coherent. Set theory as we are going to study it was called into being by one man:
Georg Cantor (1845-1918).
His papers on the subject appeared between 1874 to 1897. In one sense we can even
date the rst real result in set theory: it was his discovery of the uncountability of the real
numbers, which he noted on December 7'th 1873.
His ideas met with some resistance, some of it determined, but also with much support,
and his ideas won through. Chief amongst his supporters was the great German mathematician David Hilbert (1862-1943).
This course will start with the basic primitive concept of set, but will also make use
along the way of a more general notion of collection or class of objects. We shall use the
standard notation 2 for the elementhood relation: x 2 A will be read as the set x is an
element of the collection A. Only sets will occur to the left of the2. In the above A may
be a set or a class. In the beginning of the course we shall be somewhat vague as to what
objects sets are, and even more so as to what objects classes might be; we shall merely study
a growing list of principles that we feel are natural properties that a notion of set should or
could have. Only later shall we say precisely to what we are referring when we talk about
1
2
Section 1
the domain of all sets. The notion of class is not a necessary one for this development,
but we shall see that the concept arises naturally with certain formal questions, and it is
a useful shorthand to be able to talk about classes, although our theory (and this course)
is about sets, all talk about classes is fundamentally eliminable.1.1
One such basic principle is:
Principle (or Axiom) of Extensionality (for sets): For two sets a; b, we shall say
a = b i :
8x(x 2 a
! x 2 b).
Thus what is important about a set is merely its members. (There is a corresponding
Extensionality Principle for classes obtained from the above by replacing the sets a; b by
classes A; B.) It is conventional to express a collection within curly parentheses:
p
f2g =fx jx is an even prime numberg =fLargest integer less than 5 g
{Morning Star} = {Evening Star} = {x jx is the planet Venus}
{Lady Gaga} = {Stefani Joanne Angelina Germanotta}
This illustrates two points: that the description of the object(s) in the set or class
is not relevant (what philosophers would call the intension). It is only the extension of
the collection, that is what ends up in the collection, however it is specied, that counts.
Secondly we use the abstraction notation when we want to specify by a description. This
was seen at the rst line of the above and will be familiar to you as a way of specifying
collections of objects:
fy j::: y :::g where ::: y ::: is some description (often in a formal language - say the rst
order language from the Logic course), is called an abstraction term and is used to collect
together all the objects y that satisfy the description ::: y ::: into a class. We use this
notation exibly and write fy 2 z j::: y :::g to mean the class of objects y in z that satisfy
::: y :::
Axiom of Pair Set For any sets x,y there is a set z = fx; yg with elements just x and
y. We call z the (unordered) pair set of x,y.
In the above note that if x = y then we have that fx; yg=fx; xg = fxg. (This is
because fx; xg has the same members as fxg and so by the Axiom of Extensionality they
are literally the same thing.) The Axiom asserts the existence of such a pair object as a
set. It is our rst example of a set existence axiom. As is usual we say that x y if any
member of x is a member of y. We say x is a subset of y, or x is contained in y, or y
contains x. In symbols:
also:
x y ()df 8z(z 2 x ¡! z 2 y) ;
x y ()df x y ^ x =
/ y.
Denition 1.1. We let P(x) denote the class fy j y xg.
1.1. In short we do not need a formal theory of classes for mathematics.
Introduction
3
Implicit in this is the idea that we can collect together all the subsets of a given set.
Is this allowed? We adopt another set existence axiom about sets that says we can:
Axiom of Power Set For any set x P(x) is a set, the power set of x.
Notice that a set x can have only one power set (why?) which justies our use of a
special name P(x) for it.
Another axiom asserting that a certain set exists is:
Axiom of the Empty Set There is a set with no members.
Denition 1.2. The empty set, denoted by ?, is the unique set with no members.
We can dene ? as fx jx =
/ xg (since every object equals itself). Again note that there
cannot be two empty sets (Why? Appeal to the Ax. of Extensionality).
For any set (or class) A we have ? A (just by the logic of quantiers).
Example 1.3. (i) ? ?, but ? 2
/ ?; f?g 2 ff?gg but f?g ff?gg.
(ii)P(?) = f?g; P(f?g) = f?; f?gg ; P(fa; bg) = f?; fag; fbg; fa; bgg.
We are going to build out of thin air, (or rather the empty set) in essence the whole
universe of mathematical discourse. How can we do this? We shall form a hierarchy of
sets, starting o with the empty set, ?, and applying the axioms generate more and more
sets. In fact it only requires two operations to generate all the sets we need: the power set
operation, and another operation for forming unions. The picture is thus:
Figure 1.1.
4
Section 1
At the bottom is V0 =df ?; V1 =df P(V0) = P(?);V2 =df P(V1); Vn+1 =df P(Vn) ::: The
question arises as to what comes next (if there is such). Cantor developed the theory of
ordinal numbers which extends the standard natural numbers N. These new numbers also
have an arithmetic that extends that of the usual +; etc. which he developed, and which
will be part of our study here. He dened a rst innite ordinal number which comes
after all the natural numbers n and which he called !. After ! comes ! + 1, ! + 2; :::.
It is natural then to accumulate all the sets dened by the induction above, and we set
V! =df fx jx2Vn for some n 2 Ng. V!+1 will then be dened, continuing the above as P(V! ).
However this is in the future. We rst have to make sure that we have our groundwork
correct, and that this is not all just fantasy.
Exercise 1.1. List all the members of V3. Do the same for V4. How many members will Vn have for
n 2 N?
Exercise 1.2. Prove for < 3 that V+1 = V [ P(V). (This will turn out to be true for any .)
Exercise 1.3. We dene the rank of a set x (`(x)') to be the least such that x V. Compute
(ff?gg). Do the same for f?; f?g; f?; f?ggg.
1.1 Classes
We shall see that not all descriptions specify sets. This was a pitfall that the early workers
on foundations of mathematics fell into, notably Gottlob Frege (1848-1925) The
second volume of his treatise on the foundations of arithmetic (which tried to derive the
laws of arithmetic from purely logical assumptions) was not far from going to press in 1903,
when Bertrand Russell (1872-1970) informed him of a fundamental and, as it turned
out, fatal error to his programme. Frege had, in our terms, assumed that any specication
dened a set of objects. Like the Barber Paradox, Russell argued as follows.
Theorem 1.4. (Russell). The collection R = fx jx 2
/ xg does not dene a set.
Proof: Suppose this collection R was a set, z say. Then is z 2 z? If so then z 2
/ z.
However if z 2
/ z then we should have z 2 z! We thus have the contradiction z 2
/ z () z 2 z!
So there is no set z equal to fx jx 2
/ xg.
Q.E.D.
What we have is the rst example of a class of objects which do not form a set. When
we know that a class is not, or cannot be, a set, then we call it a proper class. (In general
we designate any collection of objects as a class and we reserve the term set for a class
that we know, or posit, or dene, as a set. The Russell Theorem above then proves that
the Russell class R dened there is a proper class. The problem was that we were trying
to dene a set by looking at every object in the universe of sets (which we have not yet
dened!). The moral of Russell's argument (which he took) is that we must restrict our
ways of forming sets if we are to be free of contradictions. There followed a period of
intense discussion as to how to correctly dene sets. Once the dust eventually cleared,
the following axiom scheme was seen to correctly rule out all obviously inconsistent ways
of forming sets.1.2 We hence adopt the following axiom scheme.
1.2. The word obviously is intentional: by Gödel's Second Incompleteness Theorem, we can not prove within
the theory of sets that the Axiom of Subsets will always consistently yield sets. However this is a general phenomenon
about formal systems, including formal number theory: such theories cannot prove their own consistency. Hence
this is not a phenomenon peculiar to set theory.
Introduction
5
Axiom of Subsets. Let (x) be a denite, welldened property. Let x be any set.
Then
fy 2 x j(y)g
is a set.
We call the above a scheme because there is one axiom for every property . You might
well ask what do I mean by `a welldened property ', and if we were being more formal
we should specify a language in which to express such properties1.3. This axiom rules out
the possibility of a universal set that contains all others as members.
Corollary 1.5. Let V denote the class of all sets. Then V is a proper class.
Proof: If V was a set, then we should have that R = fy 2 V j y 2
/ yg is a set by the Ax.
of Subsets. However we have just shown that R is not a set.
Q.E.D.
Note that the above argument makes sense, even if we have not yet been explicit
as to what a set is: whatever we decree them to be, if we adopt the axioms already
listed the above corollary holds. We want to generate more sets much as in the way
mathematicians take unions and intersections. We may want to take unions of innite
collections of set. For example, we know how to take the union of two sets x1 and x2:
we dene x1 [ x2 =df fz jz 2 x1 _ z 2 x2g. By mathematical induction we can dene
x1 [ x2 [ [ xk. However we may have an innite sequence of sets x1;S
x2; ; xk ; ::: (k 2 N)
all of whose members we wish to collect together. We thus dene z = X, where X = f
xk jk 2 N}, as:
S
X =df ft j9x 2 X (t 2 x)g.
This forms the collection we want. In fact we get a general exible denition. Let Z
be any set whatsoever. Then
S
S
Denition 1.6.
Z =df ft j9x 2 Z (t 2 x)g. In words: for any set Z there is a class,
Z,
which consists precisely of the members of members of Z.
We are justied in doing this by an axiom:
Axiom of Unions: For any set Z ;
S
Z is a set.
S This notation subsumes the more usual one as a special case:
fa; b; c; dg = a [ b [ c [ d.
S
fa; bg = a [ b (Check!);
S
Example 1.7. (i)
ff0; 1; 2g; f1; 2g; f2; 4; 8gg =f0; 1; 2; 4; 8g.
S
(ii)
fag = a.
1.3. It would be usual to adopt a rst order language L2_ ;= which had = plus just the single binary relation
symbol 2_ ; then well-formed formuale of this language would be deemed to express `well-dened properties.'
6
Section 1
An extension of the above is often used:
S
Notation:
If
I
is
set
used
to
index
a
family
of
sets
fa
j
j
2
I
g
we
often
write
j
j 2I A j for
S
fA j jj 2 I g.
S
Notice that this can be expressed as: x 2 j 2I A j ! (9j 2 I )(x 2 A j ) . We similarly
dene the idea of intersection:
T
Denition 1.8. If Z =
/ ? then
Z =df ft j8x 2 Z (t 2 x)g.
T
In words: for any non-empty set Z there is another set,
Z, which consists precisely
of the members of all members of Z. Using index sets we write
T
x 2 j 2I A j ! (8j 2 I )(x 2 A j )
Example 1.9.
T
ffa; bg; fa; b; cg; fb; c; dgg = fbg;
T
fa; b; cg = a \ b \ c;
T
ffagg = fag.
Suppose the set Z in the above denition were empty: then we should have that for any
t whatsoever that for any x 2 Z T
t 2 x (because there are noSx 2 Z!). However that leads
us to dene in this special case j 2? A j = V . Note that j 2? A j makes perfect sense
anyway: it is just ?.
We have a number of basic laws that
(i) I J ¡!
I J ¡!
S
T
i2I
Ai Ai S
T j 2J
Aj.
S
and
T
satisfy:
Aj
S
(ii) 8i(i 2 I ¡! Ai C) ¡! i2I Ai C.
T
8i(i 2 I ¡! Ai C) ¡! i2I Ai C.
S
S
S
(iii) i2I (Ai [ Bi) = i2I Ai [ i2I Bi.
T
T
T
i2I (Ai \ Bi) =
i2I Ai \
i2I Bi.
S
S
(iv) i2I (A \ Bi) = A \ ( i2I Bi).
T
T
i2I (A [ Bi) = A [ ( i2I Bi).
S
T
(v) Dn i2I Ai = i2I (DnAi)
T
S
D n i2I Ai = i2I (DnAi)
i2I
j 2J
(where we have written as usual for sets, X nY = fx 2 X jx 2
/ Y g). You should check that you
can justify these. Note that (iv) generalises a distributive law for unions and intersections,
and (v) is a general form of de Morgan's law.
S
S
Exercise 1.4. Give examples of sets x; y so that x =
/ y but
x = y:
S
Exercise 1.5. Show that if a 2 X then P(a) 2 P(P( X)).
S
S
Exercise 1.6. Show that for any set X: a)
P(X) = X b) X P( X); when do we have = here?
Exercise 1.7. Show that the distributive laws (iv) above are valid.
Exercise 1.8. Let I = Q \ (0; 1/2) be the set of rationals p with 0 < p < 1/2. Let Ap=R \ (1/2 ¡ p;
S
T
1/2 + p). Show that p2I Ap = (0; 1); p2I Ai = f1/2g.
Introduction
7
Exercise 1.9. Let X0 X1 and Y0 Y1 be two innite sequences of shrinking sets. Show that
T
T
T
(Xi [ Yi) = i2N Xi [ i2N Yi. If we take away the requirement that the sequences be
i2N
shrinking, does this hold in general for any innite sequences Xi and Yi?
1.2 Relations and Functions
In this subsection we shall see how the fundamental mathematical notions of relation and
function can be represented by sets. First relations, and we'll list various properties that
relations have. In general we have sets X ; Y and a relation R that holds between some of
the elements of X and of Y . If X is the set of all points in the plane, and Y the set of all
circles, the `p is the centre of the circle S' determines a relation between X and Y . We
shall be more interested in relations between elements of a single set, that is when X = Y .
We list here some properties then a relation R can have on a set X.
Type of relation
Reexive
Irreexive
Symmetric
Antisymmetric
Connected
Transitive
Dening condition
x 2 X ¡! xRx
x 2 X ¡! :xRx
(x; y 2 X ^ xRy) ¡!yRx)
(x; y 2 X ^ xRy ^ yRx) ¡!x= y)
(x; y 2 X) ¡! (x = y _ xRy _ yRx)
(x; y; z 2 X ^ xRy ^ yRz) ¡! (x Rz).
You should recall that the denition of equivalence relation is that R should satisfy
symmetry, reexivity, and be transitive.
If X = R and R = the usual ordering of the real numbers, then R is reexive,
connected, transitive, and antisymmetric. If we took R = < then we'd gain irreexivity but
so lose reexivity and anti-symmetry.
If X = P(A) for some set A and we took xRy () x y for x; y 2 X then R is reexive,
antisymmetric, and transitive. It is not connected since if both x ¡ y and y ¡ x are nonempty, then :xRy ^:yRx.
If T looks like a `tree', (think perhaps of a family tree) with an ordering aRb as `a
is a descendant of b' then we should only have irreexivity and transitivity.
1.2.1 Ordering Relations
Of particular interest are ordering relations where R is thought of as some kind of ordering
with x R y interpreted as x somehow preceding or coming before y. It is natural to
adopt some kind of notation such as or for such R. The notation of represents a
strict order: given an ordering where we want reexivity to hold, that is x x is allowed
to hold we may dene in terms of : x y () x y _ x = y. Of course we can dene
in terms of and = too, and we may want to make a choice as to which of the two
relations we think of as `prior' or more fundamental. In general (but not always) we shall
tend to form our denitions and propositions in term of the stricter ordering , dening
as and when we wish from it.
Denition 1.10. A relation on a set X is a (strict) partial ordering if it is irreexive
and transitive. That is:
(i) x 2 X ¡! :xx ;
(ii) (x; y; z 2 X ^ xy ^ y z) ¡! (x z) .
Exercise 1.10. Think about how you would frame an alternative, but equivalent denition of partial
order in terms of the non-strict ordering . Which of the dening conditions above do we need?
8
Section 1
We saw above that for any set A that P(A) with as was a partial order. We say
that an element x0 2 X is the least element of X if 8x 2 X(x0 x) and we call it a minimal
element if 8y 2 X(:y x). Note that a minimal element need not be a least element. (This
is because a partial order need not be connected: it might have many minimal elements).
Note also our implicit denition of in terms of .) Greatest element and maximal
elements are dened in the corresponding way.
Notions of least upper bound etc. carry over to partially ordered sets:
Denition 1.11. (i) If is a partial ordering of a set X, and ? =
/ Y X, then an element
z 2 X is a lower bound for Y in X if
8y(y 2 Y ¡! z y).
(ii) An element z 2 X is an inmum or greatest lower bound (glb) for Y if (a) it is lower
bound for Y, and (b) if z 0 is any lower bound for Y then z 0 z.
(iii) The concepts of upper bound and supremum or least upper bound (lub) are
dened analogously.
By their denitions if an inmum (or supremum) for Y exists, it is unique and we
write inf (Y ) (sup (Y )) for it. Note that inf (Y ), if it exists, need not be an element of Y .
Similarly for sup(Y ). If Y has a least element then in this case it is the inmum, and it
belongs to Y .
In one sense any partial order of a set X can be represented as partial order where the
ordering is , as the following shows.
Denition 1.12. (i) We say that f : (X ; 1) ¡! (Y ; 2) is an order preserving map of
the partial orders (X ; 1); (Y ; 2) i
8x; z 2 X(x 1 z ¡!f (x) 2 f (z)).
Note this implies a fortiori that f is (1-1).
(ii) Orderings (X ; 1) and (Y ; 2) are (order) isomorphic, written (X ; 1) = (Y ; 2), if
there is an order preserving map between them which is also a bijection.
(iii) There are completely analogous denitions between nonstrict orders 1 and 2.
Notice that ({Even natural numbers},<) is order isomorphic to (N; <) via the
function f (2n) = n. However (Z; <) is not order isomorphic to (N; <).
The function f (k) = k ¡ 1 is an order isomorphism of (Z; <) to itself. However as we
shall see, there are no order isomorphisms of (N; <) to itself.
If f is as in the last denition (i) then it also implies that 8x; z 2 X(x 1 z ¡
f (x) 2 f (z)) and so we have equivalence.
Theorem 1.13. (Representation Theorem for partially ordered sets) If partially orders
X, then there is a set Y of subsets of X which is such that (X ; ) is order isomorphic to
(Y ; ).
Proof: of Theorem. Given any x 2 X let X x = fz 2 X jz xg. Notice then that if x =
/y
then X x =
/ X y. So the assignment x X x is (1-1). Let Y = fX x jx 2 X g. Then we have
x y ! Xx Xy ;
x
consequently, setting f (x) = X we have an order isomorphism.
Q.E.D.
Often we deal with orderings where every element is comparable with every other - this
is strong connectivity and we call the ordering total.
Denition 1.14. A relation on X is a strict total ordering if it is a partial ordering
which is connected: 8x; y (x; y 2 X ¡! (x = y _ x y _ y x)).
If we use we call the ordering non-strict (and the ordering is then reexive). We
can then formulate the connectedness condition as: 8x; y (x; y 2 X ¡! (x y _ y x)).
Introduction
9
In a total ordering there is no longer any dierence between least and minimal
elements, but that does not imply that least elements will always exist (think of the total
ordering (Z; ).
We often drop the word strict (or non-strict) and leave it is as implicit when we
use the symbol (or ).
An extremely important notion that we shall come back to study further is that of
wellordering:
Denition 1.15. (i) (A; ) is a wellordering if (a) it is a strict total ordering and (b)
for any subset Y A, if Y =
/ ?; then Y has a -least element.
(i) A partial ordering R on a set A, (A; R) is a wellfounded relation if for any subset
Y A, if Y =
/ ?; then Y has an R-minimal element.
Then (N; <) is a wellordering, but (Z; <) is not. Cantor's greatest mathematical
contribution was perhaps generalizing this concept and recognizing its importance. The
theory of wellorderings is fundamental to the notion of ordinal number. If (A; R) is my
family tree with xRv if x is a desendant of y, then it is also wellfounded.
1.2.2 Ordered Pairs
We have talked about relations R that may hold between objects, and even used the
notation if we wanted to think of the relation as an ordering. However we shall want to
see how we can specify relations using sets. From that it is a short step to do the same for
functions. The key building block is the notion of ordered pair.
Denition 1.16. (Kuratowski) Let x; y be sets. The ordered pair set of x and y is the set
hx; yi =df ffxg; fx; ygg.
Why do we need this? Because fx; yg is by denition unordered : fx; yg = fy; xg. Hence
fx; yg = fu; vg ¡! x = u ^ y = v fails. However:
Lemma 1.17. (Uniqueness theorem for ordered pairs)
hx; yi = hu; vi ! x = u ^ y = v.
Proof: ( ¡) is trivial. So suppose hx; yi = hu; vi. Case 1 x = y. Then hx; yi = hx;
xi = ffxg; fx; xgg = ffxg; fxgg = ffxgg. If this equals hu; vi then we must have u = v
(why? otherwise hu; vi would have two elements). So hu; vi = ffugg = ffxgg. Hence, by
Extenionality fug = fxg, and so, again using Extensionality, u = x = y = v.
Case 2 x =
/ y. Then hx; yi and hu; vi have the same two elements. (Hence u =
/ v:) Hence
one of these elements has one member, and the other two. Hence we cannot have fxg = fu;
v g. So fxg = fug and x = u. But that means fx; yg = fu; yg = fu; vg. So of these last two
sets, if they are the same then y = v.
Q.E.D.
Example 1.18. We think of points in the Cartesian plane R2 as ordered pairs: hx; yi with
two coordinates, with x rst on one axis, y on the other.
Denition 1.19. We dene ordered k-tuple by induction: hx1; x2i has been dened; if hx1;
x2; ; xk i has been dened, then hx1; ; xk ; xk+1 i =df hhx1; ; xk i; xk+1 i
Thus hx1; x2; x3i = hhx1; x2i; x3i; hx1; x2; x3; x4i = hhhx1; x2;ix3i:x4i etc. Note that once
we have the uniqueness theorem for ordered pairs, we automatically have it for ordered
triples, quadruples,...
10
Section 1
This leads to:
Denition 1.20. (i) Let A; B be sets. A B =df f hx; yi jx 2 A ^ y 2 B g. If A = B this is
often written as A2.
(ii) If A1; :::Ak+1 are sets, we dene (inductively)
A1 A2 Ak+1 =df (A1 A2 Ak) Ak+1
(which equals:
fhhhhx1; x2i; x3i; ; xk i; xk+1 i j8i: 1 i k + 1 (xi 2 Ai)g ):
In general A B =
/ B A and further, the operation is not associative.
Exercise 1.11. Suppose for no sets x; u do we have x 2 u 2 x. Then if we dene hx; yi1 = fx; fx; ygg
then show hx; y i1 also satises the Uniqueness statement of Lemma 1:17.
Exercise 1.12. Let P be the class of all ordered pairs. Show that P is a proper class - that is - it is
not a set. [Hint: suppose for a contradiction it was a set; apply the axiom of union.]
Exercise 1.13. Show that if x 2 A; y 2 A then hx; yi 2 P(P(A)). Deduce that if x; y 2 Vn then hx;
yi 2 Vn+2 .
Exercise 1.14. Show that A (B [ C) = (A B) [ (A C): Show that if A B = A C and
A=
/ ?; then B = C.
S
S
Exercise 1.15. Show that A B = fA X jX 2 B g.
Denition 1.21. A (binary) relation R is a class of ordered pairs. R is thus any subset
of some A B.
Example 1.22. (i) R = fh0; 1i; h1; 2i; h1; 3i; h2; 0ig is a relation.
(ii) So are: S1 = fhx; yi 2 R2 jx y 2g;
(iii) S2 = fhx; yi 2 R2 jx2 = yg.
(iv) If x is any set, the identity relation on x is idx =df fhz; z ijz 2 xg.
(v) A partial ordering can also be considered a relation: R = fhx; yi jx yg.
Denition 1.23. If R is a relation, then dom(R) =df fx j9y hx; yi 2 Rg,
ran(R) =df fy j9x hx; yi 2 Rg.
With these denitions we can say that if R is a relation, then
R dom(R) ran(R).
Functions are just special kinds of relations.
Denition 1.24. (i) A relation F is a function ( Func(F )) if
8x 2 dom(F ) (there is a unique y with hx; yi 2 F ).
(ii) If F is a function then F is (1-1) i 8x; x 0(hx; yi 2 F ^ hx 0; yi 2 F ¡! x = x 0).
In the last Example (iii) and (iv) are functions; (i) and (ii) are not.
It is much more usual to write for functions F (x) = y for hx; yi 2 F . (ii) then
becomes the more familiar: 8x8x 0[F (x) = F (x 0) ¡! x = x 0]. We also write F : X ¡! Y instead of F X Y (with Y called the co-domain of f ). Then F is surjective, or onto
becomes 8y 2 Y (9x 2 X(F (x) = y). A function F : X ¡! Y is a bijection if it is both (1-1)
and onto (and we write F : X ! Y ).
Notation 1.25. Suppose F : X ¡! Y and A X then
(i) F A =df fy 2 Y j9x 2 A(F (x) = y)g. We call F A the range of F on A.
(ii) F A =df fhx; yi 2 F jx 2 Ag. F A is the restriction of F to A.
Introduction
11
In this terminology F A = ran(F A). As a further exercise in using this notation,
suppose T is a class of functions, with the property that that for
S any two f ; g 2 T ,
f (dom(f ) \S
dom(g)) = g (dom(f ) \ dom(g)). Then check a) F = T is a function, and
b) dom(F ) = fdom(g)j g 2 T g.
As well as considering functions as special kinds of relations, which are in turn special
kinds of sets, we shall want to be able to talk about sets of functions. Then:
Denition 1.26. If X ; Y are sets, then YX (or sometimes X Y ) =df fF jF : Y ¡! X g.
Exercise 1.16. Suppose X ; Y both have rank n ((X) = n - see Ex.1.3). Compute a) (X Y ); b) (YX). [Hint for b): show rst if X ; Y 2 Z show that YX 2 PPPP(Z):]
Denition 1.27.
/?
Q (Indexed Cartesian Products). Let I be a set, and for each i 2 I let Ai =
be a set; then i2I Ai =dfff j Func(f ), dom(f ) = I ^ 8i 2 I(f (i) 2 Ai) g
This allows us to take Cartesian products indexed by any set, not just some nite n.
Q
Example 1.28. (i) Let I = N: Each Ai = R. Then i2I Ai is the same as NR the set of
innite sequences of reals numbers.
(ii) Let Gi be a group forQeach i in some index set I; then it is possible to put a group
multiplication structure on i2I Gi to turn it into a group.
1.3 Transitive Sets
We think of a transitive set as one without any 2-holes.
Denition 1.29. A setSx is transitive, Trans(x), i 8y 2 x(y x). We also equivalently
abbreviate Trans(x) by
x x.
S
Note that easily Trans(x) ! xS
x: assume Trans(x); ifSy 2 z 2 x then, as we have
z x, we have yS2 x. WeSconclude that
x x. Conversely: if
x x then for any y 2 x
by denition of
, y x, hence y x and thus Trans(x).
Example 1.30. (i) ?; f?g; f?; f?gg are transitive. ff?gg; f?; ff?ggg are not.
Denition 1.31. (The successor function) Let x be a set. Then S(x) =df x [ fxg.
Exercise 1.17. Show the following: (i) Let Trans(Z) ^ x Z: Then Z [ fxg is transitive.
S
(ii) If x; y are transitive, then so are: S(x); x [ y, x \ y; x.
S
T
(iii) Let X be a class of transitive sets. then
X is transitive. If X =
/ ?, then
X is transitive.
(iv) Show that Trans(x) ! Trans(P(x)). Deduce that each Vn is transitive.
S
Lemma 1.32. Trans(x) ! S(x) =x.
S
S
S
S
S
Proof: First note that
S(x) = (x [ fxg) =
x[
fxg =
x [ x. For (¡!),
S
S
assume Trans(x); then
x x. Hence x S(x) x.
Q.E.D.
Exercise 1.18. Prove the ( ¡) direction of the last lemma.
Exercise 1.19. (i) What sets would you have to add to fff?ggg to make it transitive?
(ii) In general given a set x think about how a transitive y could be found with y x. (It will
turn out (below) that for any set x there is a smallest y x with Trans(y).) [Hint: consider repeated
S S0
S1
S
S2
S S1
S n+1
S Sn
applications of
:
x =df x;
x =df
x,
x =df
(
x), ,:::;
x =df
(
x)::: as in
the next denition.]
Denition 1.33. Transitive Closure TC We dene by recursion on n:
12
Section 1
S0
S S
S S
x = ( n x); TC(x) = f n x jn 2 Ng.
S
The idea is that by taking a
we are lling in 2-holes in the sets.
x = x;
S n+1
Exercise 1.20. Show that y 2
Sn
x $ 9xn ; xn¡1; ::: ; x1(y 2 xn 2 xn¡1 2 2 x1 2 x).
Note by constuction that Trans(TC(x)): y 2 TC(x) if and only if for some n y 2
S
Then y n+1 x TC(x).
Sn
x.
Lemma 1.34. (Lemma on TC) For any set x (i) x TC(x) and Trans(TC(x)) ; (ii)
If Trans(t) ^ x t ¡! TC(x) t. Hence TC(x) is the smallest transitive set t satisfying
x t. (iii) Hence Trans(x) ! TC(x) = x.
S
Proof (i) The is clear as x = 0 x TC(x), and by the comment above.
S
S
(ii): x t ¡! 0 x t TC(t). Now by induction on k, assume k x t. Now use
S
S
A B ^ Trans(B) ¡!
A B to deduce k+1 x t and it follows that TC(x) t.
However t was any arbitrary transitive set containing x. (iii): x TC(x) by (i). If Trans(x)
then substitute x for t in the above: we conclude TC(x) x.
Q.E.D
T As TC(x) is the smallest transitive set containing x we could write this as TC(x) =
ft j Trans(t) ^ x t)g (the latter is indeed transitive, see Ex. 1.17).
Exercise 1.21. (i) Show that x 2 y ¡! TC(x) TC(y).
S
(ii) TC(x) = x [ fTC(y) j y 2 xg (hence TC(fxg) = fxg [ TC(x).)
The point to note is that taking TC(x) ensures that 2TC(x) TC(x) is a partial
ordering.
Number Systems
13
2 Number Systems
We see how to extend the theory of sets to build up the natural numbers N. It was R.
dedekind (1831-1916) who was the rst to realise that notions such as innite number
system needed proper denitions, and that the claim that a function could be dened by
mathematical induction or recursion required proof. This required him to investigate the
notion of such innite systems. About the same time G. Peano (1858-1932) published a
list of axioms (derived from Dedekind's work) that the structure of the natural numbers
should satisfy.
Figure 2.1. Richard Dedekind
2.1 The natural numbers
Proceeding ahistorically, there were several suggestions as to how sets could represent the
natural numbers 0; 1; 2; :::.
E. Zermelo (1908) suggested the sequence of sets ?; f?g; ff?gg; fff?ggg; :::
Later von Neumann (1903-1957) suggested a sequence that has since become the usually
accepted one. Recall Def.1.31.
0
1
2
3
=df ? ,
=df f0g = f?g = 0 [ f0g = S(0) ,
=df {0,1}= f?; f?gg = 1 [ f1g = S(1) ,
=df {0,1,2} = f?; f?g; f?; f?ggg = 2 [ f2g = S(2).
In general n =df f0; 1; :::; n ¡ 1g. Note that with the von Neumann numbers we also have
that for any n S(n) = n + 1: 1 = S(?), 2 = S(1) etc. This latter system has the advantage
that n has exactly n members, and is the set of all its predecessors in the usual ordering.
Both Zermelo's and von Neumann's numbers have the advantage that they can be easily
generated. We shall only work with the von Neumann numbers.
Denition 2.1. A set Y is called inductive if (a) ? 2 Y (b) 8x 2 Y (S(x) 2 Y ).
14
Section 2
Notice that we have nowhere yet asserted that there are sets which are innite (not
that we have dened the term either). Intuitively though we can see that any inductive
set which has to be closed under S cannot be nite: ?; S(?); S (S(?)) are all distinct
(although we have not proved this yet). We can remedy this through:
Axiom of Innity: There exists an inductive set:
9Y (? 2 Y ^ 8x 2 Y (S(x) 2 Y )).
One should note that a picture of an inductive set would show that it consists of Schains: ?; S(?); SS(?); :::. but possibly also others of the form u; S(u); SS(u); SSS(u):::.
thus starting with other sets u. Given this axiom we can give a denition of natural number.
Denition 2.2. (i) x is a natural number if 8Y [Y is an inductive set¡!x 2 Y ].
(ii) ! is the class of natural numbers:
T
We have dened:
! = fY jY an inductive set}
by taking an intersection over (what one can show is a proper) class of all inductive sets.
But is it a set?
Proposition 2.3. ! is a set.
Proof: Let Z be any inductive set (by the Ax. of Inf. there is such a Z). By the Axiom
of Subsets: there is a set N so that:
(x 2 N !(x 2 Z ^ 8Y [Y an inductive set ¡!x 2 Y ]).
Q.E.D.
Proposition 2.4. (i) ! is an inductive set. (ii) It is thus the smallest inductive set.
Proof: We have proven in the last lemma that ! as a set. To show it is inductive, note that
by denition ? is in any inductive set Y so ? 2 !. Hence (a) of Def. 2.1 holds. Moreover,
if x 2 !, then for any inductive set Y ; we have both x and S(x) in Y . Hence S(x) 2 !. So
! is closed under the S function. So (b) of Def. 2.1 holds. (ii) is immediate.
Q.E.D.
To paraphrase the above: if we have an inductive subset of ! we know it is all of !. It
may seem odd that we dene the set of natural numbers in this way, rather than as the
single chain ?; S(?); ::: and so on. However it is the insight of Dedekind's analysis that we
obtain the powerful principle of induction, which of course is of immense utility. Note that
we may prove this principle, which is prior to dening order, addition, etc. We formally
state this as a principle about inductive sets given by some property :
Theorem 2.5. (Principle of Mathematical Induction) Suppose is a welldened denite
property of sets. Then
(0) ^ [8x 2 !((x) ¡! (S(x))] ¡! 8x 2 !(x).
Proof: We show that the set of x 2 ! for which (x) holds is inductive. Let
Y = fx 2 ! j(x)g. Then 0 2 Y . However if x 2 Y then the antecedent says that S(x) 2 Y .
Hence Y is inductive as required. Hence ! Y . And so ! = Y .
Q.E.D.
Proposition 2.6. Every natural number y is either 0 or is S(x) for some natural number.
To emphasise: this need not be true for a general inductive set: not every element can
be necessarily be reached eventually by repeated application of S to ?.
Number Systems
15
Proof: Let Z = fy 2 ! jy = 0 _ 9x 2 !(S(x) = y)g. Then 0 2 Z and if u 2 Z then u 2 !.
Hence S(u) 2 !, (as ! is inductive). Hence S(u) 2 Z. So Z is inductive and is thus !.
One should note that actually the Principle of Mathematical Induction has been
left somewhat vague: we did not really specify what a welldened property was. This we
can make precise just as we can for the Axiom of Subsets: it is any property that can be
expressed using a formal language for sets.
Exercise 2.1. Every natural number is transitive. [Hint: Use Principle of Mathematical Induction in other words, show that the set of transitive natural numbers is inductive.]
Lemma 2.7. ! is transitive.
Proof: Let X = fn 2 ! jn ! g. If X = ! then by denition Trans(!). So we show that
X is inductive. ? 2 X; assume n 2 X, then n ! and fng !, hence n [ fng !. Hence
S(n) 2 X. So X is inductive, and ! = X.
Q.E.D.
2.2 Peano's Axioms
Dedekind formulated a group of axioms could that capture the important properties of
the natural numbers. They are generally known as Peano's Axioms. We shall consider
general Dedekind systems:
A Dedekind system is a triple hN ; s; ei where
(a) N is a set with e 2 N ;
(b) Func(s) ^ s: N ¡! N and s is (1-1) ;
(c) e 2
/ ran(s) ;
(d) 8K N (e 2 K ^ sK K ¡!K = N ).
Note that sK K is another way of saying that K is closed under the s function. We
shall prove that our natural numbers form a Dedekind system; furthermore, any structure
that satises (a) - (d) will look like !.
Firstly then, let =fhk; S(k)i jk 2 !g = S ! the restriction of the successor operation
on sets in general, to the natural numbers.
Proposition 2.8. h!; ; 0i forms a Dedekind system.
Proof: We have that 0 2 !, : ! ¡! !, and that 0 =
/ (u) (? =
/ S(u)) for any u. The
axiom (d) of Dedekind system just says for h!; ; 0i that any subset A !,that is, of the
structure's domain, that contains 0 and is closed under (i.e. that is inductive) is all of
!. But ! itself is the smallest inductive set. So certainly A = !. So (a),(c)-(e) hold and
all that is left is to show that is (1-1).
S
S
Suppose S(m) = (m) = (n) = S(n). Hence
S(m) =
S(n). By the last exercise
S
S
Trans(m); Trans(n). By Lemma 1.32,
S(m) = m, and
S(n) = n;so m = n.
Q.E.D.
Remark 2.9. We shall later be showing that any two Dedekind systems are isomorphic.
2.3 The wellordering of !
Denition 2.10. For m; n 2 ! set m < n () m 2 n. Set m n () m = n _ m < n.
Lemma 2.11. (i) < (and ) are transitive; (ii) 8n 2 !8m(m < n ! S(m) < S(n)); (iii)
8m 2 ! (m<m)
16
Section 2
Proof: (i) That < is transitive follows from the fact that our natural numbers are
dened to be transitive sets: n 2 m 2 k ¡! n 2 k.
(ii): If S(m) < S(n) then we have m 2 S(m) 2 S(n) = n [ fng. If S(m) = n, then
m 2 S(m) = n, so m < n. If S(m) 2 n then as Trans(n) we have m 2 n and so m < n. We
prove the converse by the Principle of Mathematical Induction (PMI). Let (k) say:
8m(m < k ¡! S(m) < S(k)). Then (0) vacuously; and so we suppose (k), and prove
(S(k)).
Let m < S(k). Then m 2 k [ fk g. If m 2 k then, by (k) we have
S(m) < S(k) < S(S(k)). If m = k then S(m) = S(k) < S (S(k)). Either way we have
(S(k)). By PMI we have 8n(n).
(iii) Note 0< 0 since 0 2
/ 0. If k 2
/ k then S(k) 2
/ S(k) by part (ii).
So X = fk 2 ! jk 2
/ k g is inductive, i.e. all of !.
Q.E.D.
Lemma 2.12. < is a strict total ordering.
Proof: All we have left to prove is connectivity (often called Trichotomy): 8m; n 2
! (m = n _ m < n _ n < m). Notice that at most one of these three alternatives can hold
for m; n: if, say, the rst two then we should have n < n, and if the second two then m < m
(by transitivity of <) and these contradict (iii) of the last Lemma. Let
X = fn 2 ! j 8m 2 !(m = n _ m < n _ n < m)g. If X is inductive, the proof is complete.
This is an Exercise.
Q.E.D.
Exercise 2.2. Show this X is inductive.
Exercise 2.3. Show that 8m; n 2 ! (n < m
!n $ m ).
Theorem 2.13. (Wellordering Theorem for !) Let X =
/ ?; and X !. Then there is
n0 2 X so that for any m 2 X either n0 = m or n0 2 m.
Note: such an n0 can clearly be called the least element of X, since 8m 2 X(n0 m).
Thus the wellordering theorem, can be rephrased as:
Least Number Principle: any non-empty set of natural numbers has a least element.
Proof: Suppose X ! but X has no least element as above. Let
Z =df fk 2 ! j 8n < k(n 2
/ X) g
We claim that Z is inductive, hence all of ! and so X = ?. This suces. Vacuously
0 2 Z. Suppose now k 2 Z. Let n < S(k). Hence n 2 k [ fkg. If n 2 k then n 2
/ X (as k 2 Z).
But if n 2 fkg then n = k and so n 2
/ X because otherwise it would be the least element of
X and X does not have such. So S(k) 2 Z. Hence Z is inductive. Q.E.D.
Exercise 2.4. Let X =
/ ?; X !. Show that there is n 2 X ; with n \ X = ?.
Exercise 2.5. (Principle of Strong Induction for !) Let X ! and suppose is a denite
welldened property of natural numbers. Show that
8n[8k < n(k) ¡! (n)] ¡!8n(n).
[Hint: Suppose for a contradiction X = f n 2 ! j:(n)g =
/ ?. Apply the Least Number Principle.]
2.4 Denition by recursion on !
We shall now show that it is legitimate to dene functions by recursion on !.
Theorem 2.14. (Recursion theorem on !) Let A be any set, a 2 A,and f : A ¡! A,
any function. Then there exists a unique function h: ! ¡! A so that
(i)
h(0) = a
;
Number Systems
17
(ii) For any k 2 !: h(S(k)) = f (h(k)).
Proof: We shall nd h as a union of k-approximations where u is a k-approximation if
a) Func(u) ^ dom(u) = k b) If k > 0 then u(0) = a; if k > S(n) then u(S(n)) = f (u(n)).
In other words u satises the dening clauses above for our intended h - without our
requiring that dom(u) is all of !.
(1) If u is a k-approximation and v is a k 0-approximation, and l = min fk; k 0g then
u l = v l. In particular, if k k 0 then v k = u ^ u v, and if k 0 k then u k 0 = v ^ v u.
Proof: If not let 0 m < l be least with u(m) =
/ v(m). Then by b) m =
/ 0. So
m = S(m 0) and u(m 0) = v(m 0). But then again by b) u(m) = f (u(m 0)) = f (v(m 0)) = v(m).
Contradiction!
This proof also shows:
(2) (Uniqueness) If h exists, then it is unique.
Proof: Suppose h; h 0 are two functions satisfying (i) and (ii) of the theorem. Assume
that X = fn 2 ! jh(n) =
/ h 0(n)g and is non-empty. By the least number principle, (in other
words the Wellordering Theorem for !), there is a least number n0 2 X. As h(0) = h 0(0) = a
we have that 0 =
/ n0. So n0 = S(m) for some m. As m 2
/ X, then h(m) = h 0(m). But then
0
0
h(n0) = f (h(m)) = f (h (m)) = h (n0)! Contradiction! So X = ?.
QED (2).
(3) (Existence). Such an h exists.
Proof: (This is the harder part.) Let u 2 B () 9k(u is a k-approximation). We have
seen any two such approximations agree on the common partSof their domains. In other
words, for any u; v 2 B either u v or v u. So we take h = B.
(i) h is a function.
Proof: If hn; ci and hn; di are in h, with c =
/ d then there must be two dierent
approximations u with u(n) = c, and v with v(n) = d. But this is impossible!
(ii) dom(h) = !.
Proof: Let ? =
/ X =df fn 2 ! jn 2
/ dom(h)g. As fh0; aig 2 B we have that the least element
of X is not 0. Suppose it is n0 = S(m). Then m 2 dom(h), let us say h(m) = c. But then
h n0 [ fhn0; f(c)ig is a legitimate S(n0)-approximation. So h n0 [ fhn0; f(c)ig h, i.e.
n0 2
/ X! Contradiction!
Q.E.D.
Example 2.15. Let n 2 !. We can dene an add n function An(x) as follows:
An(0) =n;
An(S(k)) = S(An(k)).
We shall write from now n + 1 for S(n). Then we would more commonly write An(k)
as n + k. Assuming we have dened the addition functions An(x) for any n:
Example 2.16. (i) Mn(x) function: Mn(0) = 0; Mn(k + 1) = Mn(k) + n.
(ii) En(x): En(0) = 1; En(k + 1) = En(k) n
Again we more commonly write these as Mn(k) as n:k, and En(k) as nk.
Proposition 2.17. The following laws of arithmetic hold for our denitions:
(a) m + (n + p) = (m + n) + p
(b) m + n = n + m
(c) m (n + p) = m n + m p
18
Section 2
(d) m:(n:p) = (m:n):p
(e) m.n=n.m
(f) mn+ p = mn:m p
(g) (mn) p = mn:p .
Proof: These are all proven by induction. As a sample we do (c) (assuming (a) and
(b) proven). We do the induction on p. p = 0: then m:(n + 0) = m:n = m:n + m:0. Suppose
it holds for p. Then
m (n + (p + 1)) =m ((n + 1) + p) (by (a) and (b))
= m (n + 1) +m p (inductive hypothesis)
= (m n+m) +m p (by denition of Mm)
= m n+ (m + m p) =m n + m (p + 1)
using (a) again and nally (b) and the denition of Mm.
Q.E.D.
Exercise 2.6. Prove some of the other clauses of the last Proposition.
Now the promised isomorphism theorem on dedekind systems.
Theorem 2.18. Let hN ; s; ei be any Dedekind system. Then h!; ; 0i = hN ; s; ei.
Proof: By the Recursion Theorem on ! (2.14) there is a function f : h!; ; 0i ! hN ; s; ei
dened by: f (0) = e;
f ((k)) =f (k + 1) = s(f (k)).
The claim is that f is a bijection. (This suces since f has sent the special zero element
0 to e and preserves the successor operations ; s.)
ran(f ) = N : because ran(f ) satises (d) of Dedekind System axioms;
dom(f ) = !: because dom(f ) likewise satises the same DS(d).
f is (1-1): let X =df fn 2 ! j8m(m =
/ n ¡! f (m) =
/ f (n)g. We shall show X is inductive
and so is all of !. By DS(c) 0 2 X (because f (0) = e =
/ s(u) for any u 2 N , so if m =
/ 0,
¡
¡
¡
m = m + 1 say, and so f (m) = s(f (m )) 2 ran(s) and s(f (m )) =
/ e = f (0).) Suppose
now n 2 X. But now assume we have m with f (m) = u =df f (n + 1) 2 N (and we show
that m = n + 1), then for the same reason, namely e 2
/ ran(s) and so u = s(f (n)) =
/ e, we
¡
¡
¡
have m =
/ 0. So m = m + 1 for some m , and then we know f (m) = s(f (m )). But by
assumption on m and denition of f : f (m) = f (n + 1) = s(f (n)). We thus have shown
s(f (n)) = s(f (m¡)); s is (1-1) so f (m¡) = f (n). But n 2 X so m¡ = n. So m = n + 1.
Hence n + 1 2 X.
Thus X is inductive, which expresses that f is (1-1).
Q.E.D.
Example 2.19. Let s(k) = k + 2, let E be the set of positive even natural numbers. Then
hE ; s; 2i is a Dedekind system.
Exercise 2.7. (i) Let h: ! ¡! ! be given by: h(0) = 4 and h(n + 1) = 3 h(n). Compute h(4).
(ii) Let h: ! ¡! ! be given by h(n) = 5 n + 2. Express h(n + 1) in terms of h(n) as simply as possible.
Exercise 2.8. Assume f1 and f2 are functions from ! to A, and that G is a function on sets, so that
for every n f1 n and f2 n are in dom(G). Suppose also f1 and f2 have the property that
f1(n) = G(f1 n) and f2(n) = G(f2 n). Show that f1 = f2.
Exercise 2.9. Let h: ! ¡! ! be given by: h(k) = k ¡ 10 if k > 100; and h(k) = h(h(k + 1)) if k 100.
Give a denition of h if possible, using the standard formulation of a denition by recursion, which
involves only computing values h(k) from smaller values, or constants. If this is impossible show it so.
Exercise 2.10. Find (i) innitely many functions h: ! ¡! ! satisfying: h(k) = h(k + 1); (ii) the unique
function h: ! ¡! ! satisfying: (a) h(0) = 2; h(k) = h(k + 1)(h(k + 1) + 1) if k > 0.
Number Systems
19
Exercise 2.11. (The Ackermann function) Dene using the equations the Ackermann function:
A(0; x; y) = x:y
A(k + 1; x; 0) = 1
A(k + 1; x; y + 1) = A(k; A(k + 1; x; y); x)
Show that A(k; x; y) is dened for all x; y; k. [Hint: Use a double induction: rst on k assume that
for all x; y A(k; x; y) is dened; then assume for all y 0 <y A(k + 1; x; y 0) is dened.] What is A(1; x; y)?
Exercise 2.12. Prove that for any n; m 2 ! that n + m = 0 $ (n = 0 _ m = 0).
Exercise 2.13. Prove that for any n; m; k 2 ! (i) n < m ¡! n + k < m + k; (ii) k > 0 ^
n < m ¡! n:k < m:k.
Exercise 2.14. Prove that for any n; m 2 ! that if n m then there is a unique k 2 ! with n + k = m.
20
Section 3
3 Wellorderings and ordinals
It is possible to wellorder an innite set in many ways.
Example 3.1. Dene on N by n m () (n is even and m is odd)_(n; m are both
even or both odd, and n < m)
Then hN; i is a wellordering.
Exercise 3.1. Let < be the usual ordering on N+ =df fn 2 ! jn =
/ 0g. For n 2 N+ dene f (n) to be the
number of distinct prime factors of n. Dene a binary relation mRn () f (m) < f (n) _ (f (m) = f (n) ^
m < n). Show that R is in fact a wellordering of N+. Draw a picture of it.
Example 3.2. If hA; i is a set with a wellordering and B A then hB ; i is also a
wellordering. Note that if y 2 A is any element then y has a unique successor, namely
inf fx 2 A j y xg.
Exercise 3.2. Show that hA; i 2 WO ¡! :9fxn 2 Ajn 2 !g(8n xn+1 xn). (Is there a reason one
might hesitate to replace `¡!' by ` !' here?)
Theorem 3.3. (Principle of Transnite Induction) Let hX ; i 2 WO. Then
[8z 2 X ( ( 8y z (y) ) ¡! (z) )] ¡! 8z 2 X(z).
Proof: Suppose the antecedent holds but ? =
/ Z =df fw 2 X j:(w)g. As hX ; i 2 WO
there is a -least element w0 2 Z. But then 8y w0 (y). So (w0) by the antecedent.
Contradiction! So Z = ?.
Q.E.D.
Denition 3.4. If hX ; i 2 WO then the -initial segment Xz (or just (initial) segment)determined by some z 2 X is the set of all predecessors of z: Xz =df fu 2 X ju z g.
In Example 3.1, N1 is the set of evens, N4 = f0; 2g. We now prove some basic facts
about any wellordering.
Recall the denition of (order) isomorphism.
Lemma 3.5. If f : hX ; i ¡! hX ; i is any order preserving map of hX ; i 2 WO into
itself, then 8z 2 X(z f (z)). (NB f is not necessarily an isomorphism.)
Proof: If for some z we had f (z) z, then as hX ; i is a wellordering, there is
a least element z0 with this property. Then as f is order preserving, we should have
f (f (z0)) f (z0) z0 thereby contradicting the -leastness of z0. Q.E.D.
Note: this fails if hX ; i 2
/ WO: f : hZ; <i ¡! hZ; <i dened by f (k) = k ¡ 1 is an order
isomorphism.
Lemma 3.6. If f : hX ; i ¡! hY ; 0i is an order isomorphism with hX ; i; hY ; 0i 2 WO,
then f is unique.
Note: again this fails for general total orderings: f 0: hZ; <i ¡! hZ; <i is also an order
isomorphism where f 0(k) = k ¡ 2.
Wellorderings and ordinals
21
Proof: Suppose f ; g: hX ; i ¡! hY ; 0i are two order isomorphisms. Then h =df f ¡1 g: hX ; i ¡! hX ; i is also an order isomorphism. By Lemma 3.5 x h(x) for any x 2 X.
But f is order preserving, so f (x) 0 f (h(x)) = g(x). Applying the same argument with
h¡1 = g ¡1 f we get g(x) 0 f (x). Hence f (x) = g(x) for any arbitrary x 2 X.
Q.E.D.
Exercise 3.3. Let f : hX ; i ¡! hY ; 0i be an order isomorphism with hX ; i; hY ; 0i 2 WO as in the
hY f (z); 0i.
last Lemma 3.6. Show that for any z 2 X, f Xz: hXz ; i =
Lemma 3.7. (Cantor 1897) A wellordered set is not order isomorphic to any segment
of itself.
Proof: If f : hX ; i ¡! hXz ; i is an order isomorphism then by 3.5 we have x f (x)
for any x, and in particular z f (z). But f (z) 2 Xz ! In other words z f (z) z!
Contradiction!
Q.E.D.
Lemma 3.8. Any wellordered set hX ; i is order isomorphic to the set of its segments
ordered by (recall means proper subset:$).
Proof: Let Y = fXa ja 2 X g. Then a Xa is a (1-1) mapping onto Y the set of
segments, and since a b () Xa Xb the mapping is order preserving.
Q.E.D.
Exercise 3.4. Find an example of two totally ordered sets which are not order isomorphic, although
each is order isomorphic to a subset of the other.
3.1 Ordinal numbers
We can now introduce ordinal numbers. Recall that we generated the sequence of sets
?; f?g; f?; f?gg; f?; f?g; f?; f?gg::: calling these successively 0; 1; 2; 3; :::where each
is the set of its predecessors. Each member is the set of all those sets that have gone
before. We shall call such wellordered sets with this property ordinal numbers (or more
plainly ordinals). We thus have seen already some examples: any natural number is an
ordinal, as is !. We rst dene ordinal through another property that h!; <i had.
Denition 3.9. hX ; 2i is an ordinal i X is transitive and setting = 2 , then hX ;
i is a wellorder of X. (In which case we also set u v ! u = v _ u 2 v, for u; v 2 X:)
Example 3.10. h!; 2i is an ordinal, and we had 3 = (!)3 = fk 2 ! jk 2 3g = f0; 1; 2g.
Lemma 3.11. hX ; 2i is an ordinal implies that every element z 2 X is identical with the
2-initial segment Xz : z = Xz = fw 2 Xj w 2 z g
Proof: Suppose X is transitive and 2 wellorders X. Let z 2 X. Then w 2 Xz ()
w 2 X ^ w 2 z () w 2 z (as z X). Hence z = Xz .
Q.E.D.
So what we are doing in dening ordinals is generalising what we saw obtained for the
von Neumann natural numbers: that each was the set of its predecessors in the ordering
< that was dened as 2 in this case. Since the ordering on an ordinal is always 2 we can
drop this and simply talk about a set X being an ordinal. Note that is somehow more
natural to talk about strict total orderings when using 2 as the ordering relation.
We shall see that we can have many innite ordinals. Note that if hX ; 2i is an ordinal
then, as a = Xa (by the last lemma) for any a 2 X, we have that (a 2 b) () a $ b ()
Xa $ Xb. Hence for ordinals, the ordering is also nothing other than $= restricted to
the elements of X.
22
Section 3
Lemma 3.12. Any 2-initial segment of an ordinal is itself an ordinal.
Proof: Suppose w is an element of the segment Xu. Then t 2 w ¡! t 2 Xu. Hence
Trans(Xu). Since 2wellorders X and Xu X, 2 wellorders Xu. Hence the latter is an
ordinal.
Q.E.D.
Lemma 3.13. If Y X is a proper subset of the ordinal X,and Y is itself an ordinal, then
Y is an 2- initial segment of X.
Proof: Let Y be an ordinal which is a proper subset of the ordinal X. If a 2 Y , then
as Y is an ordinal a = Ya, and as a 2 X, a = Xa: Then Xa = Ya. As Y is not all of X, then
if we set c = inf fz 2 X jz 2
/ Y g we have that Y = Xc.
Q.E.D.
Lemma 3.14. If X ; Y are ordinals, so is X \ Y.
Proof: As 2 wellorders X, it wellorders X \ Y . Suppose w 2 X \ Y . Let t 2 w. As
Xw = w = Yw, we have t 2 X \ Y . So w X \ Y , and thus Trans(X \ Y ), and so is an
ordinal.
Q.E.D.
Exercise 3.5. Show that if hX ; 2i is an ordinal, then so is hS(X); 2i (where S(X) = X [ fX g).
Theorem 3.15. (Classication Theorem for Ordinals) Given two ordinals X ; Y either
X = Y or one is an initial segment of the other.
Proof: Suppose X =
/ Y . By the last lemma X \ Y is an ordinal. Then
Either (i) X = X \ Y or (ii) Y = X \ Y (and since X =
/ Y , (in case (i)) X \ Y is an
initial segment of Y , or (in case (ii)), an initial segment of X);
Or X \ Y is an ordinal properly contained in both X and Y . We show this is impossible.
Then by Lemma 3.13 X \ Y is simultaneously a segment Xa say of X , and a segment Yb
say of Y . But a = Xa = Yb = b in that case. Hence a = b 2 X \ Y = Xa: But then a 2 Xa
which is absurd!
Q.E.D.
Lemma 3.16. For any two ordinals X ; Y ; if X and Y are order isomorphic then X = Y.
Proof: Suppose X =
/ Y . Then by the last theorem X is an initial segment of Y (or
vice versa). However, if we had that X and Y were order isomorphic, then we should have
that the wellordered set hY ; 2i isomorphic to an inital segment of itself. This contradicts
Lemma 3.7.
Q.E.D.
By the last lemma if hA; i 2 WO then it can be isomorphic to at most one ordinal
set. (Check!) We shall show that it will be so isomorphic to at least one ordinal. We rst
give an argument for what will be the inductive step in argument to follow.
Lemma 3.17. If every segment of a wellordered set hA; i is order isomorphic to some
ordinal, then hA; i is itself order isomorphic to an ordinal.
Proof: By the last comment before the lemma, we can dene a function W which
assigns to each element b 2 A, a unique ordinal W (b) so that hAb ; i = hW (b); 2i. Let
Z = ran(W ).3.1 So
Wellorderings and ordinals
23
Z = fW (b) j 9b 2 A 9gb(gb: hAb ; i = hW (b); 2i)g.
(Note that there is only one such gb by Lemma 3.6.) Now notice that if c b, with c;
b 2 A then Ac = (Ab)c. Hence gb Ac: hAc ; i = h(W (b)) gb(c); 2i (see Ex.3.3). So in fact
gc = gb Ac (again by uniqueness of the isomorphism given by Lemma 3.6, we can say gb Ac
is the isomorphism gc). This can only mean that
W (c) = (W (b)) gb(c).
(1)
(So: (W (b)) gb(c) is the ordinal assigned by F to Ac.) In other words:
c b=)W (c) = (W (b)) gb(c) 2 W (b).
(We can write this 2 here since W (b) is an ordinal, and so its initial segments are its
members.) But since only one of c b and b c can occur we have then
c b () W (c) 2 W (b).
We thus have that W : hA; i = hZ ; 2i, i.e. that W is an order isomorphism. We'd be
done if we knew hZ ; 2i was an ordinal. This is the case: because is a wellorder, (and W
is an isomorphism) Z is also wellordered by 2. All we have to check is that Trans(Z). But
this is easy: let X = W (b) 2 Z be arbitrary. If u 2 X, then u = gb(c) for some c b, Then
u = Xu = W (b) gb(c) = W (c) (the rst equality holds for any ordinal X, and element u 2 X,
the last holds as in (1) above). Hence u 2 Z. Thus X Z and thus Trans(Z).
Q.E.D.
Theorem 3.18. (Representation Theorem for Wellorderings, Mirimano 1917)
Every wellordering hX ; i is order isomorphic to one and only one ordinal.
Proof: Uniqueness follows from the comment after Lemma 3.16. Existence will
follow from the last lemma by the Principle of Transnite Induction, Theorem 3.3: the
wellordering hX ; i will be order isomorphic to an ordinal, if all its initial segments
are. Suppose
Z =df fv 2 X jXv is isomorphic to an ordinalg.
If Z = X then by the last Lemma we have achieved our task. Clearly if v0 is the least element of X then hXv0; i = ? = 0: But if v 2 X and Xv Z, again by the Lemma
applied to hXv ; i; v 2 Z. Hence Z = X.
Q.E.D.
Denition 3.19. If hX ; i 2 WO then the order type of hX ; i is the unique ordinal
order isomorphic to it. We write it ot(hX ; i).
Corollary 3.20. (Classication Theorem for Wellorderings, Cantor 1897) Given
two wellorderings hA; i and hB ; 0i exactly one of the following holds:
0
(i) hA; i = hB ; i
0
(ii) 9b 2 B hA; i = hBb ; i
0
(iii) 9a 2 A hAa ; i = hB ; i.
Proof: If hX ; 2i and hY ; 2i are the unique ordinals isomorphic to hA; i , hB ; 0i
respectively, then by Theorem 3.15, either hX ; 2i = hY ; 2i (in which case (i) holds); or
hX ; 2i is isomorphic to an initial segment of hY ; 2i (in which case we have (ii)), or vice
versa, and we have (iii).
Q.E.D.
3.1. Why does this set Z exist? We shall discuss later the Axiom of Replacement that justies this.
24
Section 3
Denition 3.21. Let On denote the class of ordinals.
For ; 2 On; we write < =df 2 . =df < _ = .
We shall summarise below some of the basic properties of ordinals. In the sequel, as in
the last denition we follow the convention of using lower case greek letters to implicitly
denote ordinals.
3.2 Properties of Ordinals
Basic properties of ordinals:
(1) 2 On ¡! is a transitive set, Trans(); 2 wellorders .
(2) 2 2 ^ ; ; 2 On ¡! 2 .
(3) 2 On ^ 2 ¡! 2 On ^ = .
(4) ; 2 On ^ h; 2i = h; 2i ¡! = .
(5) ; 2 On ¡! exactly one of (i) = (ii) 2 ; (iii) 2 holds.
Lemma 3.22. (6) Principle of Transnite Induction for On
If C =
/ ?; C On then 9 2 C8 2 C[ 2 _ = ].
(1) here is Def. 3.9; (2) simply expresses again the transitivity of the 2-ordering; (3) is
3.11 and 3.12, and (4) is 3.16. (5) follows from (3) and (4) and 3.20 (notice that we cannot
have 2 as this would contradict the strictness of the wellorder 2).
Proof of (6): let 0 2 C as C is non-empty. If for no 2 C do we have 2 0 then
0 was the2-minimal element of C. Otherwise we have that C \ 0 =
/ ?. As 0 2 On, by
denition2wellorders 0. Hence C \ 0 is non-empty and so has an2minimal element 1;
and then 1 is the minimal element of C.
Q.E.D.
Note: This last argument seems a little unnecessary, but it is not: we know any individual ordinal is wellordered: (6) claims the whole class On is wellordered. Note also that
we did not require C to be a set, it could be a proper class.
Exercise 3.6. Let C be as in (6) above. Let 2 C. Check that is a minimal element of C i
\ C = ?. Deduce from (6) that < wellorders On.
The following was originally noted as a paradox by Burali-Forti. This was the rst
of the set theoretical paradoxes to appear in print. Burali-Forti noted (as in the argument
below) that On itself formed a transitive class of objects well-ordered by 2. Hence, as On
consists of all such transitive classes, (On; 2) is isomorphic to a member of itself! A plain
contradiction! The reaction to this contradiction was messy: Burali-Forti thought he had
shown that the class of ordinals was merely partially ordered. Russell thought that the
class of ordinals was linearly ordered only (although two years later he saw the need for the
distinction between sets and classes, and reasoned that On had to be a proper class). Again
we must distinguish between sets as objects of study, and proper classes as collections of
sets brought together by an arbitrary description. Burali-Forti's argument when properly
dressed in its modern clothes is the following.
Lemma 3.23. (Burali-Forti 1897) On is a proper class.
Wellorderings and ordinals
25
Proof. Suppose x is a set and x = On. Then as we have seen we can wellorder x by
the ordering < on On. But then hx; 2i is itself a wellordering (as 2 is a strict total ordering
and every non-empty subclass C x has a least element by (6) above) and furthermore
Trans(x). Hence x 2 On. But then x 2 x, and x becomes an ordinal that is a member of
itself. This is nonsense as 2, is a strict ordering on any ordinal!
QED
Denition 3.24. For A a set of ordinals, sup A is the least ordinal 2 On so that
8 2 A( ).
This conforms entirely to our notion of supremum as the lub of a set. In particular:
(i) If A has a largest element then sup A = .
(ii) Suppose A =
/ ? has no largest element; then sup A is the smallest ordinal strictly
greater than all those in A.
Example 3.25. sup 3 = 2 =sup {0,2}; sup f3g = 3; sup fEvensg = ! =sup ! = sup f!g;
sup f0; 3; ! + 1g = ! + 1.
Many texts simply dene sup (A) as
S
A. This makes sense:
Lemma 3.26. Let A be a set of ordinals then sup A is properly dened, and equals
S
A.
Proof: First note that sup (A) is properly dened: there is an ordinal which is an upper
bound for A. Suppose not, then we have that for every 2 On
S there is 2 A with
S < .
By the axiom of union: as A is assumed to be a set, so is
A. But On =
A! This
contradicts Lemma 3.23. Hence A has an upper bound, and sup (A) exists.
S
Claim: sup A = A.
S
Proof: Let = sup A. Suppose
2 A. Then for some 2 A we have: 2 2 A. So
S
< and so 2 . Hence
A . Conversely suppose 2S. Then < =
Ssup A and
so there is 2 A with < . Hence 2 2 A and so 2 A. Thus A.
Q.E.D.
Observe also that if X Y are sets of ordinals, then by denition, sup X sup Y .
S
Exercise 3.7. (i) Compute sup ( + 1) and verify that it equals
( + 1). Show that is a limit
S
ordinal i = .
(ii) Prove that if X is a transitive set of ordinals, then X is an ordinal.
Denition 3.27. Succ() () 9 ( = S() = [ f g( = + 1))
Lim() () 2 On ^ =
/ 0 ^ :Succ().
We thus have ordinals are divided into three types: (i) 0; (ii) those of the form + 1,
i.e. those that have an immediate predecessor, and (iii) the rest, the limit ordinals which
have no immediate predecessors. Notice we have written S() as ` + 1', that is because
we shall dene our ocial `+1' operation to coincide with S (see Lemma 3.37 below) as we
did for natural number addition. So we are getting slightly ahead of ourselves. Note that
A=
/ ? has no largest element; then sup A is a limit ordinal.
Example 3.28. Successors are: 2, n; ! + 1; ( ! + 1) + 1; :::
Limits: ! is the rst limit ordinal; the next will be ! + !, then (! + !) + !; ::: !:! ; :::
when we come to dene these arithmetic operations, which we shall now turn to.
Denition 3.29. Let hA; Ri; hB ; S i be total orderings, with A \ B = ?. We dene the
sum of hA; Ri; hB ; S i to be the ordering hC ; T i where C = A [ B and we set
26
Section 3
xTy !(x 2 A ^ y 2 B) _ (x; y 2 A ^ xRy) _ (x; y 2 B ^ xSy)
The picture here is that we take a copy of hA; Ri and place all of it before a copy of
hB ; S i.
Exercise 3.8. Show that if hA; Ri; hB ; S i 2 WO,then the sum hC ; T i 2 WO.
Note that the denition required that A; B be disjoint (so that the orderings did not
become confused. We should like to use ordinals themselves for A; B but they are not
disjoint. Hence it is convenient to use a simple disjointing device as follows. If ; 2 On;
then f0g and f1g are disjoint copies of and . We could now dene the sum
of and as
+ 0 =df ot(h f0g [ f1g; T i where h ; iiT h; j i
! (i = j ^ < ) _ i < j.
The operation + 0 is pretty clearly associative, but it is not commutative as the following
examples will show.
Example 3.30. 2 + 0 3; 2 + 0 !; ! + 0 2; ! + 0 !; (! + 0 !) + 2; (! + 0 !) + 0 ! :::
sup f!; ! + 0 !; (! + 0 !) + 0 !:::g =!: 0! = sup f!: 0n jn 2 !g.
Denition 3.31. Let hA; Ri; hB ; S i be total orderings.We dene the product of hA; Ri;
hB ; S i to be the ordering hC ; U i where C = A B and we set U to be the anti-lexicographic
ordering on C:
hx; yi U hx 0; y 0i !(ySy 0) _ (y = y 0 ^ xRx 0).
This is dierent: here we imagine taking a copy of hB ; S i and replacing each element
y 2 B with a copy of all of hA; Ri.
Exercise 3.9. Show that if hA; Ri; hB ; S i 2 WO, then the product hC ; U i 2 WO.
Again we could dene ordinal products : 0 by setting : 0 to be:
: 0 =df ot(h ; U i) where h ; i U h 0; 0i !( < 0) _ ( = 0 ^ < 0).
Again : 0 will turn out to be associative (after some thought) but non-commutative.
Example 3.32. 2: 03; 2: 0! ; !: 02 ; !: 0! ; (!: 0!):2 ; (!: 0!): 0! :::
Exercise 3.10. (i) Express (! + 0 !) + 0 ! using the multiplication symbol : 0 only.
(ii) Informally express !: 0! using the addition symbol + 0 only.
Exercise 3.11. Show that the distributive law ( + 0 ): 0 = : 0 + 0 : 0 is not valid. On the other
hand, convince yourself that : 0( + 0 ) = : 0 + 0 : 0 will be true.
The reason we have put primes above our arithmetical operations is that we shall
soon dene them in another way, extending our everyday denition of +,. for natural
numbers. As we dened them by recursion for ! so we shall dene addition, multiplication,
and exponentiation for ordinals by recursion. However rst we have to prove a recursion
theorem that shows we can do this (just as we proved the Recursion Theorem on ! (Thm
2.14).
Theorem 3.33. (Recursion Theorem on On; von Neumann1923) Let F : V ¡! V
be any function. Then there exists a unique function H: On ¡! V so that:
8 ( H() = F (H ) ).
Wellorderings and ordinals
27
Proof: As before for the Recursion Theorem on ! we shall dene H as a union of
approximations to H where u is a -approximation if Func(u); dom(u) = , and 8 < (
u() = F (u ) ). So u satises the dening clauses of H throughout its domain up to .
Notice how this works: (i) if > 0 then u(0) = F (u ?), but u ? = ?; hence u(0) = F (?)
for any -approximation. There is a single 1-approximation: it is v = fh0; F (0)ig. (In
fact u = ? is a 0-approximation! This is because the empty set counts as a function with
empty domain, hence it can be considered a 0-approximation); (ii) further if u is a approximation, then u is a -approximation for any . We let
B = fuj 9(u is a -approximation)g
(1) If u is a -approximation and v a -approximation, with , then u = v .
Proof: As usual, look for a point of least dierence for a contradiction: suppose is
least with u( ) =
/ v( ). Then the two functions agree up to ; i.e. u = v ; but then
u( ) = F (u ) = F (v ) = v( )! Contradiction.
This same argument will establish:
(2) (Uniqueness) If H exists then it is unique.
(Replace u; v in the above with H ; H 0 any two functions that satisfy the conditions of
the theorem.)
(3) (Existence). Such an H exists.
Proof: As any two
S approximations agree on the common part of their domains, we may
sensibly dene H = B. Just as for the proof of recursion on !:
(i) H is a function.
(ii) dom(H) = On.
Proof: Let C be the class of ordinals for which there is no -approximation. So if C
is non-empty, by the Principle of Transnite Induction for On, then it will have a least
element . By Note (i) above, > 0. Consider the set A = f w 2 Bj dom(w) < g. Notice that
all the members of A agree with each other on their respective domains by the reasoning
at (1). (Also by the assumption there can be no -approximation for any again by
remark (ii) above.)
S
If were a limit ordinal then u = A would itself be a -approximation, (check that
u is a function with dom(u) = !) contradicting that 2 C.
If = + 1 then A has an element v with largest possible domain . We may dene a
-approximation by extending v to a longer u by setting u() = F (v); i.e., set u = v [ fh;
F (v)ig. However this contradicts the fact that 2 C. Hence C = ?.
Q.E.D.
Remark 3.34. As we have stated it, we have used proper classes - the function F for
example is such, and On being a proper class will entail that H is too. This is not as risky
as might be thought at rst, since we may eliminate talk of proper classes by their dening
formulae if we are careful. We have chosen to be a little relaxed about this, for the sake
of the exposition.
Remark 3.35. Although this is the common form of the Recursion Theorem for On in text
books, it is often more useful in the following form, which tends to unpack the function
F into two dierent subfunctions and a constant depending on the type of ordinal just
occurring in the denition of H. It essentially contains no more than the rst theorem:
one should think of it as a version of the rst theorem where F is dened by cases.
28
Section 3
Theorem 3.36. (Recursion Theorem on On, Second Form) Let a 2 V.
F0; F1: V ¡! V be functions. Then there is a unique function H: On ¡! V so that:
(i) H(0) = a ;
(ii) If Succ() then H() = F0(H()) where = + 1 ;
(iii) If Lim() then H() = F1(H ).
Let
Proof: Dene F : V ¡! V by:
F (x) = a if x = ?,
F (u) = F0(u) if Funct(u) ^ dom(u) is a successor ordinal,
F (u) = F1(u) if Funct(u) ^ dom(u) is a limit ordinal,
F (u) = ? in all other cases.
Now apply the previous theorem to the single function F Q.E.D.
In practise we shall be a little informal as in the following denitions of the ordinal
arithmetic operations.
Denition 3.37. We dene by transnite recursion on On:
(Ordinal Addition) A() = + :
A(0) = ;
A( + 1) = S(A()) =A() + 1;
A() = sup fA() j < g if Lim().
We write + for A().
(Ordinal Multiplication) M() = :
M(0) = 0;
M( + 1) = M() + ;
M() = sup fM() j < g if Lim() .
(Ordinal Exponentiation) E() = :
E(0) = 1;
E( + 1) = E():
E() = sup fE()j < g if Lim():
We write for M():
We write for E():
Compare these denitions with those for the usual arithmetic operations on the natural
numbers. Note that denition of multiplication (and exponentiation) assumes that addition
(respectively multiplication) has been dened for all . They are obtained in each case
by adding a third clause to cater for limit ordinals. Hence we know immediately that the
ordinal arithmetic operations agree with standard ones on !, the set of natural numbers.
Note we have immediately gone to the more informal but usual notation: the second line
of the above could have been stated as A( + 1) = S(A()) etc.
Lemma 3.38. The functions A are strictly increasing and hence (1-1). That is, for any
:
() < ¡! + < + .
Proof: This is formally a proof by induction on , but really given the denitions of
the arithmetical operations are intuitively true. For suppose as an inductive hypothesis
that () holds for all . Then we show it is true for + 1. Let < + 1. If = then
+ 2 S( + ) = + ( + 1) and hence + < + ( + 1). But if < then by IH we
know + < + and the latter we have just shown less than + ( + 1).
Suppose that () holds for all < for some limit ordinal . We show it holds for .
Suppose < . Note < + 1 < . So + < + ( + 1) sup f + j < g = + (the rst < holding by inductive hypothesis).
Q.E.D.
Wellorderings and ordinals
29
Lemma 3.39. Similarly both M ; E are also strictly increasing and hence (1-1): suppose
; ; 2 On are such that < . If > 0 then (i) < ; (ii) < .
Exercise 3.12. Prove this last lemma.
Lemma 3.40. Let ; 2 On. Then
(i) + = [ f + j < g;
(ii) = f + j < ^ < g
Proof: (i) By induction on : if = 0 then + 0 = [ ? = . Suppose (i) is true for
. Then + ( + 1) = ( + ) + 1 = S( + ) = + [ f + g =
= [ f + j < g [ f + g (by Ind Hyp.)
= [ f + j < + 1g.
It is thus true for + 1.
Now suppose Lim() and that (i) is true for < . Then
+ = sup f + j < g (by Def. of +)
S
= f [ f + j < g j < g (by Lemma 3.26 and the Ind. Hyp.)
= [ f + j < g (as Lim() implies that any + for < is also trivially
+ for < for a < ).
It is thus true for Lim() also.
(ii) Again by induction on . For = 0 then :0 = 0 = ? = f + j < 0 ^ < g.
Suppose it is true for . ( + 1) = + (by Def. of Multiplication)
= [ f + j < g (by (i) of the Lemma)
= f + j < ^ < g [ f + j < g (by the Inductive Hypothesis)
= f + j < + 1 ^ < g.
Now suppose Lim() and it is true for < , we ask the reader to complete the proof
as an exercise.
Exercise 3.13. Complete the proof of (ii) of the Lemma.
Corollary 3.41. Suppose ; 2 On and 0 < < . Then (i) there is a unique ordinal so that + = ; (ii) there is a unique pair of ordinals ; so that < ^ = + .
Proof: By the last Corollary the function M() is strictly increasing. So M() <
M( + 1) for example. By (ii) of the last lemma 2 ( + 1) = f + j < + 1^
< g. So at least one pair ; satisfying these requirements exists. Suppose 0; 0 is
another. If = 0 then = 0; but then = + = + 0. But by (i) of the last
Corollary this implies = 0.
However if, say, < 0 then + 1 0 and so
= + < + = ( + 1) 0 0 + 0 = which is absurd. So this case cannot occur.
Q.E.D.
Note: By the last corollary we could make a sensible denition of subtraction and
division for ordinals, although one rarely has any use for it, so we shall not investigate this
further here.
It is easy to see that sup f + 2n jn 2 !g = + ! (=dfsup f + n jn 2 !g). This is an
elementary example of (i) of the next exercise where we have taken X as the set of even
natural numbers.
Exercise 3.14. Let X be a set of ordinals without a largest element. Show
(i) + sup X = sup f + j 2 X g;
(ii) sup X = sup f j 2 X g ;
(iii) supX = sup f j 2 X g.
30
Section 3
Lemma 3.42. The following laws of arithmetic hold for our denitions:
(a) + ( + ) = ( + ) + (b) ( + ) = + (c) ( ) = ( ) (d) + = .
Proof: These are all proven by transnite induction. Again we do (b) as a sample. We
perform the induction on . For = 0 we have ( + 0) = + 0= + 0. Suppose
it is true for . Then ( + ( + 1)) = (( + ) + 1) = ( + ) + =( + ) + = + ( + ) = + ( + 1). So it holds for + 1. Suppose now Lim()
and it holds for < .
Then ( + ) = sup f + j < g
= sup f ( + ) j < g (by (ii) of the last Exercise
=sup f + j < g (by the Ind. Hyp.)
= + sup f j < g (by (i) of the last Exercise
= + (by Def. of Multiplication).
Lemma 3.43. (Cantor's Normal Form Theorem). Let 1 < ! . Then there exists
a unique k 2 ! and unique 0; :::; k ¡1 with 0 > :::>k¡1 and d0; ::: ; dk ¡1 2 ! so that:
= ! 0 d0 + ! 1 d1 + + ! k ¡1 dk¡1.
The Theorem says that any ordinal ! can be expressed to base !. There is nothing
special about ! here: if we could still nd nitely many decreasing ordinals i, and
0 < di < and have = 0:d0 + 1:d1 + + n¡1:dn¡1. Thus could be expressed to
base .
Proof: Let 0 = sup fj ! g. If ! 0 < then there is a largest d0 2 ! so that
! 0 d0 (thus with ! 0 (d0 + 1) > ). If ! 0 d0 = we are done. Otherwise there is a
unique 1 so that ! 0 d0 + 1 = . Note that in this case 1 < . Now repeat the argument:
let 1 = sup fj ! 1g; by virtue of our construction and the denition of 0 and d0,
we must have 1 < 0. If ! 1 < 1 then dene d1 2 ! as the largest natural number with
! 1 d1 1. If we have equality here, again we are done. Otherwise there is 2 dened
to be the unique ordinal so that ! 1 + 2 = 1. Since we have > 1 > 2 there must be
some k with k = 0, that is with ! k¡1 dk¡1 = k¡1. then has the form required for the
theorem.
Q.E.D.
Exercise 3.15. Convince yourself that a Cantor Normal Form theorem could be proven for other bases
as indicated above: if we may nd nitely many decreasing ordinals i, and 0 < di < with
= 0 d0 + 1 d1 + + n¡1 dn¡1.
Exercise 3.16. Find subsets of Q with order types ! 2; ! ! ; and ! ! + ! 3 + 17 under the natural <
ordering.
Cardinality
31
4 Cardinality
Je le vois, mais je ne le crois pas! G.Cantor 29.vii.1877.
Letter to Dedekind, after discovering that R R R.
We now turn to Cantor's other major contribution to the foundations of set theory:
the theory of cardinal size or cardinality of sets. Informally we seek a way of assigning
a number to represent the size or magnitude of a set - any set whether nite or innite.
(And we have yet to dene what those two words mean.) We extrapolate from our experience with nite sets when we say that two such sets have the same size when we can pair
o the members one with another - just as children do arranging blocks and apples.
4.1 Equinumerosity
Denition 4.1. Two sets A; B are equinumerous if there is a bijection f : A
write then A B and f : A B.
! B. We
The idea is that f is both (1-1) and onto, and thus we can use A to count B (more
usually we have A is a natural number or perhaps is N itself). An alternative word for
equinumerous here is equipollent. Notice that:
Lemma 4.2. is an equivalence relation:
(i) A A; (ii) A B ¡! B A ; (iii) A B ^ B C ¡! A C.
Denition 4.3. (i) A set B is nite if it is equinumerous with a natural number:
9n 2 !9f (f : n B).
(ii) If a set is not nite then it is called innite.
Notice that this denition makes use of the fact that our denition of natural number
has built into it the fact that a natural number is the (nite) set of its predecessors, so the
above denition makes sense.
Could a set be equinumerous to two dierent natural numbers? Well, of course not if
our denitions are going to make any sense, but this is something to verify.
Lemma 4.4. (Pidgeon-Hole Principle) No natural number is equinumerous to a proper
subset of itself.
Proof: Let Z = fn 2 ! j 8f (If f : n ¡! n and f is (1-1), then ran(f ) = n)g. (Thus
members of Z cannot be mapped in a (1-1) way to proper subsets of themselves.) Trivially
0 2 Z. Suppose n 2 Z, and prove that n + 1 2 Z. Let f be (1-1) and f : n + 1 ¡! n + 1.
Case 1 f n: n ¡! n.
Then by Inductive hypothesis, ran(f n) = n. Then we can only have f (n) = n and
thus ran(f ) = n + 1.
Case 2 f (m) = n for some m 2 n.
32
Section 4
As f is (1-1) we must have then f (n) = k for some k 2 n. We dene g to be just like
f but we swap around the action on n ; m: dene g by g(m) = k; g(n) = n and g(l) = f (l)
for all l =
/ m; n. Now g: n + 1 ¡! n + 1 and g n: n ¡! n. By Case 1 ran(g) equals n + 1,
but in that case so does ran(f ).
Q.E.D.
Corollary 4.5. No nite set is equinumerous to a proper subset of itself.
Exercise 4.1. Prove this.
Corollary 4.6. Any nite set is equinumerous to a unique natural number.
The next corollary is just the contrapositive of Cor. 4.5.
Corollary 4.7. Any set equinumerous to a proper subset of itself is innite.
Corollary 4.8. ! is innite.
Exercise 4.2. Prove the corollaries 4.6 & 4.8.
Exercise 4.3. Show that if A $ n 2 ! then A m for some m < n. Deduce that any subset of a nite
set is nite.
Exercise 4.4. Suppose A is nite and f : A ¡! A. Show that f is (1-1) i ran(f ) = A.
Exercise 4.5. Let A; B be nite. Without using any arithmetic, show that A [ B and A B is nite.
Theorem 4.9. (Cantor, Dec. 7'th 1873)
The natural numbers are not equinumerous to the real numbers: ! R.
Proof: Suppose f : ! ¡! R is (1-1). We show that ran(f ) =
/ R so such an f can never
be a bijection. This is the famous diagonal argument that constructs a number that is
not on the list. We assume that the real numbers in ran(f ) are written out in decimal
notation. f (0) = 3.1415926:::
f (1) = ¡2.4245:::
f (2) = 176.011:::
We let x be the number 0.212::: obtained by letting x have 0 integer part, and putting
at the n + 1'st decimal place a 1 if the n + 1st decimal place of f (n) is even, and a 2 if it is
odd. The argument concludes by noting that x cannot be f (n) for any n as it is deliberately
made to dier from f (n) at the n + 1'st decimal place.
Q.E.D.
Theorem 4.10. (Cantor)
No set is equinumerous to its power set: 8X ( X P(X) ).
Proof: Similar to the argument of the Russell Paradox: suppose for a contradiction
that f : X P(X). Let Z = fu 2 X ju 2
/ f (u)g. Argue that although Z 2 P(X) it cannot
be f (u) for any u 2 X. Q.E.D.
Denition 4.11. We write: (i) X Y if there is a (1-1) f : X ¡! Y.
(ii) X Y i X Y ^ Y X.
Note that then X Y ¡! X Y ^ Y X. The next theorem shows the converse is true.
Cardinality
33
Theorem 4.12. (Cantor-Schröder-Bernstein) X Y ^ Y X ¡! X Y.
Proof: Suppose we have the (1-1) functions f : X ¡! Y and g: Y ¡! X. We need a
bijection between X and Y and we piece one together from the actions of f and g.
We dene by recursion:
C0 = X ¡ ran(g)
Cn+1 = gf Cn.
Thus C0 is that part of X that stops g from being a bijection. We then dene
h(v) =
f
f (v)
if v 2 Cn for some n Case 1
g ¡1(v) otherwise.
Case 2
Note that the second case makes sense: if v 2 X but v 2
/ Cn for any n,then in particular
it is not in C0, that is v 2 ran(g).
We now dene Dn =df f Cn. (Note that this makes Cn+1 = gDn.) We claim that h is
our required bijection.
h is (1-1): Let u; v 2SX; as both f and g ¡1 are (1-1) the only problem is if, for some
m say, u 2 Cm and v 2
/ n2! Cn (or vice versa). However then:
h(u) = f (u) 2 Dm;
h(v) = g ¡1(v) 2
/ Dm (it is not in Dm because otherwise we should have v 2 Cm+1 a
contradiction).
Hence h(u) =
/ h(v).
S
h is onto Y : 8nDn ran(h). So consider u 2 Y ¡ n Dn. g(u) 2
/ C0 = X ¡ ran(g) and
g(u) 2
/ Cn+1 for any n either: this is because Cn+1 = gDn and u 2
/ Dn and g is (1-1). So
g(u) cannot end up in Cn+1 . Therefore Case 2 applies and
h(g(u)) = g ¡1(g(u)) = u.
Q.E.D.
The proof of this theorem has a chequered history: Cantor proved it in 1897 but his
proof used the Axiom of Choice (to be discussed later) which the above proof eschews.
Schröder announced that he had a proof of the theorem in 1896 but in 1898 published
an incorrect proof! He published a correction in 1911. The rst fully satisfactory proof was
due to Bernstein, but was published in a book by Borel, also in 1898.
Exercise 4.6. Show that (i) (¡1; 1) R (ii) (0; 1) [0; 1] by nding suitable bijections, without using
Cantor-Schröder-Bernstein.
Denition 4.13. Let X be any set, we dene the characteristic function of Y X to be
the function Y : X ¡! 2 so that Y (a) = 1 if a 2 Y and Y (a) = 0 otherwise.
Exercise 4.7. Show that P(!) R !2. [Hint: First show that P(!) (0; 1). May be easier to
show that 9f : P(!) (0; 1) (by using characteristic functions of X ! and mapping them to binary
expansions). Then show that 9g: (0; 1) P(!) using a similar device. Then appeal to Cantor-SchröderBernstein to obtain the rst P(!) (0; 1). Now note that P(!) !2 is easy: subsets X ! are in (11) correspondence with their characterisitc functions X . ]
Exercise 4.8. Show directly (without using that P(!) !2 or the CSB Theorem) that X X 2.
Denition 4.14. A set X is denumerably innite or countably innite if X !. It is
countable if X !.
34
Section 4
Note that nite sets are countable according to this denition. Trivially from this:
Lemma 4.15. Any subset of a countably innite set is countable.
Exercise 4.9. (i) Show that ? =
/ X is countable i there is f : ! ¡! X which is onto. [Hint for (():
Construct a (1-1) map from f , demonstrating X !.]
(ii) Prove that X is countable and innite () X is countably innite.
Lemma 4.16. Let X be an innite set, and suppose R is a wellordering of X. Then X
has a countably innite subset.
Proof: Let x0 be the R-least element of X. Dene by recursion xn+1 = R-least element
of X ¡ fx0; :::; xn g. The latter is non-empty, because X was assumed innite. Hence for
every n < !, xn+1 is dened. Then X0 = fxn jn < !g is a countably innite subset of X.
Q.E.D.
Without the supposition of the existence of a wellordering on X we could not run this
argument. We therefore adopt the following
Wellordering Principle (WP):
Let X be any set, then there is a wellordering R of X.
It will turn out that the Wellordering Principle is equivalent to the Axiom of Choice.
Lemma 4.17. Let X and Y be countably innite sets. Then X [ Y is countably innite.
Exercise 4.10. Show that ! ! !. [Hint: consider the function f (m; n) = 2m(2n + 1) ¡ 1. For future
reference we let (u)0 and (v)1 be the (1-1) unpairing inverse functions from ! to ! so that f ((u)0;
(u)1) = u.]
Exercise 4.11. Prove this last lemma.
By induction we could
S then prove for any n that if X0;:::; Xn are all countably innite
then so is their union in Xi.
Lemma 4.18. Assume the Wellordering
Principle. Then if X0; :::; Xn ; ::::(n < !) are all
S
countably innite then so is i<! Xi.
Remark 4.19. Remarkably, it can be proven that without WP we are unable to prove this.
Proof: The problem is that although we are told that each Xi is bijective with ! we
are not given the requisite functions - we are just told they exist. We must choose them,
and this is where
S ! WP is involved. Let Z = fg j9i < ! (g: ! Xi)g. Then Z isSa set (it is
a subset of
f Xi ji < ! g). Let R be a wellordering of Z. Dene g: ! ¡! i<! Xi by
g(f (i; n)) = gi(n) where f is the function of Exercise 4.10, satisfying f : ! ! ! and gi
is the R-least function g: ! Xi. Then Check that g is onto and apply Ex.4.9. Q.E.D.
Exercise 4.12. Let X ; Y ; Z be sets. Either by providing suitable bijections, or by establishing
injections in each directions and using Cantor-Schröder-Bernstein, in each case show that:
(i) X (Y Z) (X Y ) Z) and X (Y [ Z) (X Y ) [(X Z) (Assume Y \ Z = ?)
(ii) X [YZ X Z YZ ; (assume X \ Y = ?)
(iii) X (Y Z) X Y XZ ;
(iv) X (YZ) (X Y )Z .
Cardinality
35
4.2 Cardinal numbers
We shall assume the Wellordering Principle from now on. This means that for any set
X we can nd R, a wellordering of it. However if hX ; Ri 2 WO then it is isomorphic to
an ordinal. If f : hX ; Ri = h; 2i is such an isomorphism, then in particular f : X is
a bijection. In general for a set X there will be many bijections between it and dierent
ordinals (indeed many bijections between it and a single ordinal), but that allows for the
following denition.
Denition 4.20. Let X be any set, the cardinality of X, written jX j, is the least ordinal
with X .
This corresponds again with notion of nite cardinality. Note that if X is nite then
there is just one ordinal with X (namely that 2 ! with which it is bijective). This
just follows from the Pidgeon-Hole Principle.
However in general a set may be bijective with dierent ordinals: ! ! + 1 ! + !
for example. Still for an innite set X, jX j also makes sense.
Lemma 4.21. For any sets X ; Y
(iii) X Y () jX j < jY j.
(i) X Y () jX j =jY j; (ii) X Y () jX j jY j;
Proof: These are really just chasing the denitions: let = jX j ; = jY j. Let g: X ,
h: Y . For (i) (=)) Let f : X ¡! Y be any bijection. Then < since otherwise h f
is a bijection of X with < - a contradiction. Similarly < since otherwise g f ¡1:
Y < contradicting the denition of . ((=) Suppose = and just look at h¡1 g.
Complete (ii) and (iii) is an exercise.
Q.E.D.
Exercise 4.13. Complete (ii) and (iii) of this lemma.
This last lemma (together with WP) shows that we can choose suitable ordinals as cardinal numbers to compare the sizes of sets. Cantor's theorems in this notation are that
jNj < jRj and in general jX j < jP(X)j. In general when we are dealing with the abstract
properties of cardinality, the lemma also shows that we might as well restrict ourselves to
a discussion of the cardinalities of the ordinals themselves. All in all we end up with the
following denition of cardinal number .
Denition 4.22. An ordinal is a cardinal or cardinal number, if = jj.
Notice that we could have obtained an equivalent denition if we had said that an
ordinal number is a cardinal if there is some set X with = jX j.
We tend to reserve middle of the greek alphabet letters for cardinals: ; ; ; :::
Check that this means is not a cardinal i there is < with .
For any 2 On jj = j jj j. (Check!)
Exercise 4.14. Check: each n 2 ! is a cardinal, ! itself is a cardinal. [Hint: just consult the denition
and some previous lemmas and corollaries.]
Exercise 4.15. Suppose !. (i) Show + 1. (ii) Show that + n is not a cardinal, nor is + !.
[Hint: try it with = ! rst; nd a (1-1) map f from + n (or + ! respectively) into .]
Note: The last Exercise shows that innite cardinals are limit ordinals.
36
Section 4
Lemma 4.23. If jj then j j =j j.
Proof: By denition there is f : jj. Hence by Lemma 4.21(i) j jj j =jj. Now
( ! ), hence f : ¡! jj (1-1). Hence jj. But jj so trivially
jj . By CSB jj. Hence, again by Lemma 4.21(i): j j = j jj j =jj.
Q.E.D.
4.3 Cardinal arithmetic
We now proceed to dene arithmetic operations on cardinals. Note that these, other than
their restrictions to nite cardinals, are very dierent from their ordinal counterparts.
Denition 4.24. Let ; be cardinals. We dene
(i) = jK [ Lj where K ; L are any two disjoint sets of cardinality ; respectively.
(ii) = jK Lj where K ; L are any two sets of cardinality ; respectively.
Notes: (1) It really does not matter which sets K ; L one takes: if K 0; L 0 are two
others satisfying the same conditions, then there are bijections between K ; K 0 and L;
L 0 and thus between K [ L and K 0 [ L 0 (and K L and K 0 L 0) so simply as far as
size goes it is immaterial which underlying sets we consider. (ii) can be paraphrased as
jX Y j =jX j jY j for any sets X ; Y .
(2) Unlike ordinal operations, and are commutative. This is simply because in
their denitions, [ is trivially commutative, and K L L K.
(3) = j f0g [ f1gj = j 2j = 2 by denition.
(4) For any ordinal : j j = jj jj follows directly from the denition of .
Lemma 4.25. For n; m 2 ! m + n = m n < ! and m n = m n < !.
Proof: We already know that m + n; m n < !. One can prove directly that m + n =
m n (or by induction on n), and m n = m n similarly.
Q.E.D.
Exercise 4.16. Complete the details of the last lemma.
Exercise 4.17. For any ordinals ; :
hold for ordinal + and ).
j + 0 j=jjj j ; j 0 j=jj
j j (and so the same will
The next theorem shows how dierent cardinal multiplication is from ordinal multiplication. We shall use the following exercise in its proof.
Exercise 4.18. Suppose hA; Ri 2 WO and there is a cardinal with jAj , but so that for every
a 2 A the initial segment hAa ; Ri = h; 2i for a < . Show that ot(hA; Ri = .
Theorem 4.26. (Hessenberg) Let be an innite cardinal. The = .
Proof: By transnite induction on . As ! ! ! (Ex.4.10), we already know that
! ! = j! !j = !. Thus we assume the theorem holds for all smaller innite cardinals
< and prove it for . We consider the following ordering on :
h; i C h ; i ()df max f; g < max f ; g _
[ max f; g = max f ; g ^ ( < _ ( = ^ < )) ].
(Note the last conjunct here is just the lexicographic ordering on .)
(1) C is a wellorder of .
Cardinality
37
Proof: Let ? =
/ X . Let X0 = f j9 h; i 2 X g Let 0 = min X0. Then consider
0 =df min f j h0; 0i 2 X g.
(1)
The ordering starts out:
h0; 0i C h0; 1i C h1; 0i C h1; 1i C h0; 2i C h1; 2i C h2; 0i C h2; 1i C h0; !i C h1; !i C h2;
!i C h!; 0i C h!; 1i C h!; ! i
(2) Each h; i 2 has no more than jmax (; ) + 1max (; ) + 1 j < many
C-predecessors.
Proof: By looking at the square pattern that occurs, the predecessors of h; i t inside
a cartesian product box of this size. But by Remark (4) following on Def.4.24, if we set
= max f; g + 1 then j j = j j j j. As < ; then j j < and so by the inductive
hypothesis we have j j j j < as required.
(2)
By (2) it follows that C has the property that every initial segment has cardinality less
than . The whole ordering certainly has size since for every < h; 0i is in the eld
of the ordering! That means (by Exercise 4.18) that ot(h ; Ci) = . But that means
we have an order isomorphism between h ; Ci and h; 2i. But such an isomorphism is
a bijection. Hence we deduce , which translated is = .
Q.E.D.
Corollary 4.27. Let ; be innite cardinals. Then = = max f; g.
Proof: Assume , so = max f; g. Then let X ; Y be disjoint with jX j =; jY j =.
(Then Y X X f1g.) Thus we have:
X X [ Y X f0g [ X f1g = X 2 X X.
In terms of cardinal numbers this expresses:
= 2 .
However Hessenberg shows that = so we have equality everywhere above, and in
particular = = max f; g.
Further: X X Y X X. Again in terms of cardinals, and quoting Hessenberg:
and so = max f; g.
Q.E.D.
S
Denition 4.28. Let A be any set. Then <!A = nnA is the set of all functions f : n ¡! A
for some n < !.
=
Exercise 4.19. Show that nA A A (the n-fold cartesian product of A).
S
Exercise 4.20. () Assume WP. Let jXn j = ! for n < !. Show that that j n Xn j = . (This is
()
the generalisation of Lemma 4.18 for uncountable sets Xn.) [Hint: The
means it is supposed to be
slightly harder. Follow closely the format of Lemma 4.18; use the fact that we now know ! to
replace ! ! = ! in that argument.]
Corollary 4.29. Let be an innite cardinal. Then j<!j = .
Proof: It us enough to show that Xn =df n has cardinality and then use Exercise
4.20. However n (the former by Exercise 4.19, the latter by repeated
use of the Hessenberg Theorem).
Q.E.D.
Denition 4.30. (WP) Let ; be cardinals, then =df jj.
(Recall that =df ff j f : ¡! g.) We need WP here (unlike the denitions of the
other cardinal arithmetic operations) since we need to know that the set of all possible
functions can be bijective with some ordinal.
38
Section 4
Exercise 4.21. Show that jXY j =j X jY j j =jjX jjY j j =jY j jX j. Deduce that we could have dened to be the cardinality of XY for any sets X ; Y with jX j =, jY j =.
Lemma 4.31. If ! and 2 , then 2 P(). Hence 2 = = (=
jP()j).
Proof: We can establish 2 P() by identifying characteristic functions of subsets
of with those subsets themselves. Now see that: 2 P( ) P() 2
(using Hessenberg's Theorem on the rst). Hence we have throughout.
Q.E.D.
Lemma 4.32. (WP) If ; ; are cardinals, then
(i) = ; (ii) () = .
Proof: (i) This is Exercise 4.12 (ii) with, for example, X = f0g; Y = f1g, and
Z = .
=dfjj = j X [Yj = jX Y j ( the second equality by Ex. 4.21, the last by
Ex 4.12 (ii))
X
Y
= j j j j ( def. of )
= .
(ii) () =df j ()j = j() j (the latter equality by Ex. 4.21)
= j j
(by Exercise 4.12 (iv))
=j
j = (since jAj =jAj - Ex.4.21 - for any set A).
Q.E.D.
Theorem 4.33. (Hartogs' Theorem). For any ordinal there is a cardinal > .
Remark: The observant may wonder why we prove this: after all Cantor's Theorem
showed that for any , P() and so jP()j > . This is true, but this required the WP
(to argue that P() is bijective with an ordinal, and so has a cardinality). Hartogs' theorem
does not require WP - although it does require the Axiom of Replacement - which we have
not yet discussed. It shows that there are arbitrarily large cardinals without appealing to
Cantor's theorem.
Proof: For nite this is trivial. Let ! be arbitrary. Let S =df fR j h; Ri 2 WOg.
Then S is a set - it is a subset of P( ) and so exists by Power Set and Subset Axioms.
Let S~ = fot(h; Ri) jR2S g. Then to argue that S~ is a set we need to know that the range
of the function that takes a wellordering to its order type, when restricted to a set of
wellorderings yields a set of ordinals. To do this we appeal to the Axiom of Replacement
that says that any denable function F : V ¡! V when restricted to a set has a set as its
range: (8x2V )(F x 2 V ) (see next Chapter).
Then, knowing that S~ is a set, we form sup S~ which is then an ordinal > . As S~ has
no largest element (Exercise), is a limit ordinal (Lemma 3.31). Hence 2
/ S~. Hence there
is no onto map f : ¡! (for if so we could dene a wellordering R by R ! f () < f ();
R is a wellordering as h ; <i is such.) Hence . But then for any < , since
for such there is an onto map showing that (because < 0 for some 0 2 S~ - in
fact < ¡! 2 S~). So j j =.
Q.E.D.
Corollary 4.34. Card =df f 2 On j a cardinal} is also a proper class.
Proof: If there were only a set of cardinals, then sup(Card) 2 On. By Hartogs' (or Cantor's)
Theorem there is nevertheless a cardinal >sup (Card)!
Q.E.D.
Corollary 4.35. For any set x there is an ordinal so that x.
Cardinality
39
Exercise 4.22. (Without WP) Prove the last corollary. [Hint: this is really Hartogs' theorem, with
the set x substituted for throughout.]
Denition 4.36. We dene by transnite recursion on the ordinals:
!0 = !; !+1= least cardinal number >!; Lim() ¡! ! = sup f! j < g.
Denition 4.37. An innite cardinal ! with > 0,is called an uncountable cardinal;
it is also called a successor or a limit cardinal, depending on whether is a successor or
limit ordinal.
We are thus dening by transnite recursion a function F : On ¡! On which enumerates
all the innite cardinals starting with F (0) = !0 = !. This function is strictly increasing
( < ¡! F () = ! < ! = F ()) and it is continuous at limits (meaning that
F () = sup fF () j < g for Lim()).
Technically we should also call nite cardinals > 0 successor cardinals as well.
(Innite) successor cardinals are however of the form ! +1 . Given any ordinal then,
the least cardinal > must then be a successor cardinal, and is written +.
Exercise 4.23. Let S be a set of cardinals without a largest element. Show that sup S is a cardinal
and a limit cardinal. Deduce that for the limit case of Denition 4.37 that ! must indeed be a
cardinal (as was implicitly assumed).
Exercise 4.24. Are there ordinals so that = !? If so nd one. (Such would be a xed point of
the cardinal enumeration function F , we should have F () = .)
Denition 4.38. (Cantor) Continuum Hypothesis CH:
2!0 = !1;
The Generalised Continuum Hypothesis GCH: 8 2! = !+1.
As N ! and P(!) ! 2, we can express CH as jP(N)j =!1, and again as R P(!),
that jRj =!1. The GCH says that 8 jP(!)j =!+1 , the exponential function 7! 2 thus
always takes the very least possible value it could.
Again CH could be expressed as saying that for any innite set X R, either X N
or X R.
Cantor believed that CH was true but was unable to prove it. We now know why
he could not: the framework within which he worked, was prior to any formalisation of
axioms for sets, but even once those axioms were written down and accepted, (the ZFC
axioms which we have introduced above) we have the following contrasting (and startling)
theorems:
Theorem (Gödel 1939) In ZFC set theory we cannot prove :CH : it is consisttent
that jRj be !1.
Theorem (Cohen 1963) In ZFC set theory we cannot prove CH: it is consistent that
jRj be !2 (or !17; !!+5; :::).
CH on the basis of the ZFC axioms is thus an undecidable statement. Set theorists have
searched subsequently for axioms to supplement ZFC that would decide CH but to date,
in vain. We simply do not know the answer, or indeed any simple way of even trying to
answer it.
Indeed the cardinal exponentiation function in general is problematic in set theory, little
can denitely be said about in general, but work on this function for so-called singular
limit cardinals (and < ) has resulted in a lot of information about V the universe of
sets.
40
Section 4
Denition 4.39. (The beth numbers)We dene by transnite recursion on the ordinals:
i0 = !; i+1 =2i ; Lim() ¡! i = sup fi j < g.
Note that if the GCH holds, then 8 (i = @).
Exercise 4.25. Prove that that there is with = i.
Exercise 4.26. Place in correct order the following cardinals using =; <; :
!1
?
@13 , @!2, ?, @@
!1 , sup f@n j n < !g, @!1 @! , @! , @!1 @!1, @! @!1, 2 , @!1.
You should give your reasons; apart from the `! 2' in the second cardinal, the arithmetic is all
cardinal arithmetic.
Exercise 4.27. Simplify where possible: 2@0; @! @!1; (2@0)@1; (@!)3 (@5)2.
You should do this twice: the rst time without assuming the Generalised Continuum Hypothesis,
and the second time assuming it. (The operations are all cardinal arithmetic.)
Exercise 4.28. A real number is said to be algebraic if it is a root of a polynomial anxn + an¡1xn¡1 +
+a1x + a0 = 0 where each ai 2 Q. Show that there are only countably many algebraic numbers. A
real number that is not algebraic is called transcendental . Deduce that almost all real numbers are
transcendental, in that the set of such is equinumerous with R:
Exercise 4.29. A word in an alphabet is a string of symbols from of nite length. Show that the
number of possible words made up from the roman alphabet is countable. If we enlarge the alphabet
to be now countably innite, is the answer dierent?
Exercise 4.30. What is the cardinality of (i) the set of all order isomorphisms f : Q ! Q; (ii) the set
of all continuous functions f : R ¡! R? ; (iii) the set of all convergent sequences 1
n=0 an?
¡n
Exercise 4.31. The Cantor set C is the set of all real numbers of the form 1
with
n=0 an 3
an 2 f0; 2g. Show that C R.
Axioms of Replacement and Choice
41
5 Axioms of Replacement and Choice
We consider in this chapter the Axiom of Choice (AC) and its various equivalents, one
of which we have already mentioned: the Wellordering Principle (WP). However we rst
look more closely at another axiom which delimits the existence of sets.
5.1 Axiom of Replacement
This axiom (which we have already used in one or two places) asserts that the action of a
function on a set produces a set.
Axiom of Replacement Let F : V ¡! V be any function, and let x be any set. Then
Fx =df fz j9u 2 x ( F (u) = z )g is a set.
The import of the axiom is one of delimitation of size: it says that a function applied
to a set cannot produce a proper class, i.e. something that is not too large. It thus appear
prima facie to be dierent from those of the other axioms, which assert simple set existence.
The `replacement' is that of taking a set X and `replacing' each element u 2 X by some
dierent a; and which a it is is specied by F . If this is done for each u 2 X the resulting
X 0 should still be considered a set.
Examples: (i) Let F (x) = fxg for any set x. Then the Axiom of Replacement ensures
that F ! = ff0g; f1g; f2g; :::; fng; :::g is a set.
(ii) Likewise Replacement is needed to justify that f@0; @1; @2; @3; :::g is a set which we
can think of as F@ ! where F@() = @ for 2 On. Without Replacement we cannot say
the supremum of this set exists, and this supremum is @!.
S
(iii) Similarly V!+! , which will be dened below as fVj < ! + !g, requires the use
of the function FV where FS
V () = V, in order to justify FV ! + !=fVj < ! + !g to be
a set, before we can apply
to it.
A slightly less trivial example occurs in the proof of Hartogs' Theorem (Thm. 4.33).
There we had a set of wellorders S. Consider the function F that takes x to 0 unless x = hA;
Ri where R wellorders the set A, in which case F (hA; Ri) returns the ordinal ot(hA; Ri).
Then F : V ¡! V is a legitimate function, and the Axiom of Replacement then asserts that
S~ = fot(h; Ri) jR 2S g = F S is a set of ordinals.
The axiom was introduced in a paper by Zermelo who attributed it to Fraenkel
(although it had been considered by several others before in various versions). In Zermelo's earlier paper there was no mention of any principle such as Replacement (in German
Ersetzung) and thus in his axiomatic system (which was, and is, called Z for Zermelo)
the set of nite numbered alephs in Example (ii) above, did not exist as a set (and nor
did V!+! ). Since the set of nite numbered alephs did not exist, @! did not exist.
42
Section 5
Figure 5.1. Abraham Fraenkel 1891-1965
Other important examples are aorded by proofs of transnite recursion theorems such
as Theorem 3.32 (although we brushed these details under the carpet at the time). In
axiomatic set theory it us usual to think of the function F as given to us dened by some
formula '(u; v) where we have proven that 8u 2 x9!v'(u; v) (recall that 9!v is read there
exists a unique v ). The conclusion then can be expressed as 9w8u 2 x9v 2 w'(u; v) and
then w in eect has been dened as a set containing F x. (Then if we want a set that is
precisely F x we may use the Axiom of Subsets to pick out from w just the set of elements
in the desired range.)
5.2 Axiom of Choice
This is an axiom that is ubiquitous in mathematics. It appears in many forms: analysts use
it to form sequences of real numbers, or to justify that the countable union of countable
sets is countable. Algebraists use it to form maximal prime ideals in rings, and functional
analysts to justify the existence of bases for innite dimensional vector spaces. We have
adopted as a basic axiom the Wellordering Principle that every set can be wellordered: for
any A we may nd R so that hA; Ri 2 WO. In particular this meant that hA; Ri = h; <i
for some ordinal , and then we could further dene jAj the cardinality of A. Without
WP we could not have done this. A very common form in set theory text books of AC the Axiom of Choice - which is the following:
Axiom of Choice - AC Let G be a set of non-empty sets. Then there is a choice
function F so that 8X 2 G (F (X) 2 X).
The reason for the name choice function is obvious: F (X) picks out for us, or chooses
for us, a unique element of the set X (which is why we specify that X =
/ ?). AC turns out
to equivalent to WP. We shall prove this.
Theorem 5.1. (Zermelo 1908) AC () WP.
Proof: (=)) Assume AC. Let Y be any set. We may assume that Y =
/ ? (otherwise
the result is trivial). We seek a wellordering R of Y . Let G = fX Y jX =
/ ?g. By AC
let F be a choice function for G. Let u be any set not in Y . We dene by recursion H:
Axioms of Replacement and Choice
43
¡! Y a (1-1) onto function with domain some 2 On. If we succeed here then we can
dene easily a wellordering R: put xRy ! H ¡1(x) < H ¡1(y) (this makes sense as H is
a bijection). Dene:
H0() = F (Y ¡ fH0() j < g) if the latter is non-empty.
= u otherwise.
Note that this denition implies that H0(0) = F (Y ¡ ?) = F (Y ) 2 Y . Then by the
Theorem on Transnite Recursion 3.32 there is a function H0: On ¡! Y [ fug.
Claim There is 2 On with H0() = u.
Proof: (The Claim says that sooner or later we exhaust Y .) Suppose not. Then H0 is
a (1-1) function sending all of On into the set Y . But then H0¡1 is a function. Look at
H0¡1Y . By the Axiom of Replacement this is a set. But it is On itself, and by the BuraliForti Lemma On is a proper class! This is a absurd.
QED Claim
Let be least with H0() = u and let H = H0 . By the above comment
this
suces.
S
((=) Suppose WP. Let G be any set of non-empty sets. Let A =df G = fu j 9X 2
G(u 2 X)g. By WP suppose hA; Ri 2 WO. We need a choice function F for G. Let X 2 G
and dene F (X) to be the R-least element of X. Check that this works!
Q.E.D.
A collection G of sets is called a chain if 8X ; Y 2 G(X Y _ Y X).
Zorn's Lemma (ZL) Let F be a set so that for every chain G F then
F contains a maximal element Y, that is 8Z 2 F(Y =
/ Z ¡! Y Z).
S
G 2 F: Then
Theorem 5.2. WP () AC () ZL.
Proof: (ZL ) AC) Let G be a set of nonempty sets. We dene F to be the set of all
choice functions that exist on subsets of G. That is we put f 2 F if (a) dom(f ) G (b)
8x 2 dom(f ) f (x) 2 x. Such an f thus acts as a choice function on its domain, and it may
only fail to be a choice function for all of G if dom(f ) =
/ G. Consider a chain H F. H is
thus a collection of partial choice functions of theSkind f ; g 2 F with the property that
either f g orSg f . However then if we set h = H we have that h is itself a function
and dom(h) = fdom(f ) j f 2 Hg. That is h is a partial choice function, so h 2 F. Now
by ZL there is a maximal m 2 F.
Claim m is a choice function for G.
Proof: m is a partial choice function for G: it satises (a) and (b) above. Suppose it
failed to be a choice function. Then there is some x 2 G with x 2
/ dom(m). As G consists
of non-empty sets, pick u 2 x. However then m [ fhx; uig 2 F as it is still a partial choice
function, but now we see that it is not maximal. Contradiction!
S
(WP ) ZL) Let F be a set so that for every chain G F then G 2 F: By WP F can
be wellordered, and a fortiori there is a bijection k: ! F for some 2 On. We dene
by transnite recursion on a maximal chain by inspecting the members of F in turn.
For < dene F () =1 if 8 < ( F () = 1 ¡! k() k())
=0 otherwise.
F can then be thought of as a characteristic function of a subset of F namely C = fX 2
F jF (k ¡1(X)) = 1g. We claim that C is a chain. This is obvious as we only add X = k()
say toSC, if it contains as subsets all the previous elements fk() j < S
g. We further claim
S
that C is a maximal element of F. By our dening property of F ; C 2 F. If Y C
then Y contains every element of C as a subset.
S However Y = k() for some , and so by
the denition
of F , F () = 1. So Y 2 C. So Y C. This suces since we have now shown
S
Y = C.
QED
44
Section 5
There are many equivalents of AC. We state without proof some more.
5.1
Uniformisation Principle (UP) If R X Y is any relation, then there is a function
f : X ¡! Y with (i) dom(f ) = dom(r) =df fx j 9y(hx; yi 2 Rg and (ii) f R.
Inverse Function Principle (IFP) For any onto function H: X ¡! Y between sets
X ; Y, there is a (1-1) function G: Y ¡! X with 8u 2 X(G(H(u)) = u).
Cardinal Comparison For any two sets X ; Y either X Y or Y X.
Hessenberg's Principle For any set X X X.
Vector Space Bases Every vector space has a basis.
Tychono Property Let G be any set of non-empty sets. then the direct product
X 2GX =
/ ? [Here i2IXi =df ff j dom(f ) = I ^ 8i 2 I (f (i) 2 Xi g.]
Tychono-Kelley Property Let Xi (for i 2 I) be any sequence of compact topological
spaces. Then the direct product space i2IXi is a compact topological space.
It can also be shown that GCH =) AC but this is not an equivalence.
Exercise 5.1. Show that AC , UP
Exercise 5.2. Show that AC ) IFP.
In general with the above exercises the converse implications are harder.
Exercise 5.3. Show that WP , Cardinal Comparison. [Hint: for (() use the Cor. 4.35.]
Exercise 5.4. Show that AC , Tychono Property.
Exercise 5.5. Show that WP ) Vector Space property. [Hint: use the argument for nite dimensional
vector spaces, but transnitely; use WP to wellorder the space, to be able to keep choosing the `next'
linearly independent element.]
Exercise 5.6. Show that if C is any proper class and F any (1-1) function, then F C is a proper class.
Figure 5.2. Ernst Zermelo 1871-1953
5.3 Weaker versions of the Axiom of Choice.
5.1. There is whole book devoted to listing and proving such equivalents: Equivalents of the Axiom of Choice
by H.Rubin & J.Rubin, Studies in Logic Series, North-Holland Publishing, 1963.
Axioms of Replacement and Choice
45
Clearly AC implies the following:
Denition 5.3. (AC! ¡ the countable axiom of choice) Every countable family of nonempty sets has a choice function.
But we cannot assume AC! and hope that it implies the general AC:
Theorem 5.4. Assume AC!. Then (i) the union of countably many countable sets is
countable. (ii) (Russell & Whitehead 1912) Every innite set has a countably innite
subset.
Denition 5.5. (i) DC!. Let R be a relation on a set A with the property that for any
u 2 A there is b 2 A with bRa. Then there is a sequence of elements fuij i 2 !g of A with
ui+1Rui for all i 2 !:
It can be shown that DC! ) AC! (Bernays 1952) but not conversely (Jensen 1966).
46
Section 6
6 The Wellfounded Universe of Sets
At the very start of this course we introduced a picture of the universe of sets of
mathematical discourse, which we dubbed V . The idea was that we could start with the
empty set and build up a hierarchy of sets that would be sucient for all of mathematics.
We dened V0 = ?, and then Vn+1 = P(Vn). The suggestion was that this idea would
be continued into the transnite. Now that we have a theory of ordinals, and theorems
concerning the possibility of denitions along all the ordinals by transnite recursion, we
can make complete this picture.
Denition 6.1. (The Wellfounded hierarchy of sets) We dene the V function by
transnite recursion as:
S
S
V0 = ?; V+1 = P(V); Lim() ¡! V = < V; we set V = 2On V.
Denition 6.2. (The rank function) For any x 2 V we let:
(x) = the least so that x 2 V +1.
Note that by the denition of V +1 we could just as easily have dened rank by setting
(x) to be the least so that x V . Notice also that if y 2 x then (y) < (x). As the
ordinals are wellordered, this means that the 2-relation is a wellfounded relation on V .
Examples If x; y 2 V then: fxg; fx; yg 2 P(V) = V+1. Hence hx; yi = ffxg;
fx; ygg 2 P(V+1) = V+2. Hence if (x) = (y) = then (fx; yg) = + 1, and
(hx; yi) = + 2.
Hence V V V+2 and so V V 2 V+3. As any ordering R on V is a subset of
V V we have R V+2 as well, and so is also in V+3. So (R) = + 2.
Exercise 6.1. Suppose (x) = (y) = and f : x ! y: Compute (S(x)); (
(f ); ( xy ) ; ( x).
S
x); (hx; y; xi);
Exercise 6.2. What if in the above example is a limit ordinal? Can we improve the bounds on
ranks? If h!; Ri is an ordering, what is (R)? [Hint: compute (! !), and ((! + 1) (! + 1)).]
It is so useful to have sets organised in this hierarchical fashion that we adopt from
now on one last axiom:
Axiom of Foundation: Every set is wellfounded, that is, 8x9y 2 x(y \ x=?).,
We thus rule out by at the existence of sets such as x and y with the properties that
x 2 x, or x 2 y 2 x, because for such a set, whatever 2 means it is not a wellfounded
relation on x. Consequently since we adopt this axiom, we have that 2- is a wellfounded
relation on every set, and every set appears somewhere in the V-hierarchy. Some texts
write WF for the class of wellfounded sets in the V-hierarchy, prove a lemma such as 6.3
for WF, and then later introduce the Axiom of Foundation.
Exercise 6.3. Show the equivalence of the following: (i) The Axiom of Foundation; (ii) 8x9(x 2 V);
(iii) 8x9(x V).
Is the Axiom of Foundation justied? Perhaps there are mathematical objects that
cannot be represented by sets or stuctures in V ? If so this would destroy our claim that
set theory provides a sucient foundation for all of mathematics. In fact this turns out not
to be the case: if we assume AC we can prove that every structure that mathematicians
invent can be seen to have an isomorphic copy in V - and since mathematicians only worry
about truths in mathematical structures up to isomorphism this will do for us.6.1
The Wellfounded Universe of Sets
47
Exercise 6.4. Let G = hG; ; e; ¡1i be a group. Assume WP, but not the Axiom of Foundation. Show
that there is a group G 2 V with G = G . [Hint: By WP nd R so that hG; Ri 2 WO. Then copy G
onto = ot(hG; Ri).]
Exercise 6.5. a) There is no sequence of sets xi for i 2 !, with xi+1 2 xi. b) WP implies that
a) itself implies the Axiom of Foundation:
Lemma 6.3. For any : (i) Trans(V)
(ii) < ¡! V V;
(iii) V = f x2V j (x) <g;
(iv) If x 2 V then 8y 2 x(y 2 V ^ (y) < (x));
(v) If x 2 V , then (x) = sup f(y) +1j y 2 xg = sup+ f(y) j y 2 xg;
(vi) () = ;
(vii) On \ V = .
Note that in (v), the denition is equivalent to writing (x) = sup+ f(y) j y 2 xg ,
where sup+X is dened to the the strict sup of a set of ordinals X: that is the least ordinal
strictly greater than any in X.
Proof: For (i) and (ii) use transnite induction on : = 0 is trivial; if = + 1 then
Trans(X) ¡! Trans(P(X)) (see Exercise 1.17) and we conclude by the inductive hypothesis
that Trans(V). That V V then is immediate by transitivity of V . If Lim() then as
a union of transitive sets
S is transitive, (Exercise 1.17(iii)) so Trans(V) is immediate from
the denition of V as < V , as is V V .
For (iii): If x 2 V , then (x) < ()df 9 < (x 2 V +1)) () x 2 V.
For (iv): Let = (x). Then x 2 V+1 = P(V). So if y 2 x then y 2 V and so (y) < by (iii).
For (v): Let = sup f(y) + 1 j y 2 xg. By (iv) (x). But each y 2 x has
(y) < %(y) + 1 which implies y 2 V; so x V and hence x 2 V+1, i.e. (x) .
For (vi): By transnite induction on : assume true for all < . Then < ¡! 2
V +1 V. Hence V and 2 V+1 . Then by (v) () = sup+ f j < g = .
For (vii): this is immediate from (vi) and (iii).
QED
So we have a picture of sets, V , in which as an object x lives at a certain rank on the
V-hierarchy, and its members y 2 x live below that at lesser levels, and in turn whose
members u 2 y live below (y) and so forth.
We can thus think of a set x is given by a graph or picture of nodes in a certain kind
of tree where we go downwards in the 2-relation as we descend the tree. The tree will most
likely have innitely many nodes, and any one node may have innitely many members
immediately below it, but what it does not have is any innitely long downwards growing
branches: this is because every level of a node comes with an ordinal denoting the rank of
the set attached at that point, and we can have no innite descending chains through the
ordinals. This idea provides us with a new way of dening functions or proving properties
about sets: since 2 is wellfounded we have:
Lemma 6.4. Principle of 2-induction Let (v) be any welldened and denite
property of sets.
(i) (Set Form) Let Trans(X) and suppose Then
8y 2 X( (8x 2 y (x)) ¡! (y) ) ¡! 8y 2 X (y):
6.1. There should be a slight caveat here: some category theorists deal with proper class sized objects because
they wish to work with the category of all groups, or the category of all sets, but there are ways of dealing also
with these notions, so the spirit of the claim is true.
48
Section 6
(ii) (V or Class form)
8y ( (8x 2 y (x)) ¡! (y) )
¡!
8y(y) .
Proof: (i) Let Z =df fy 2 X j:(y)g. Suppose Z =
/ ? and we shall show that the antecedent
of the induction scheme fails. Let y0 2 Z have (y0) least amongst all ranks of members
of Z. Then for any x 2 y0 we have (x) < (y0), x 2 X (since Trans(X)), and hence (x)
holds for such x. Suppose 8y( (8x 2 y (x)) ¡! (y) ) were true (for a contradiction).
However we have just argued that 8x 2 y0 (x). If this were true we'd conclude (y0) - a
contradiction! This nishes (i).
(ii) Notice this is exactly the same, thinking of X as the transitive class V ! Instead
now take Z =df fy j:(y)g. The rest of the argument makes perfect sense.
QED
Theorem 6.5. 2-Recursion Theorem Let G: V ¡! V be any function. Then there is
exactly one function H: V ¡! V so that
8xH(x) =G(H x) [ =G(fhy; H(y)i j y 2 xg) ].
Proof This is done in just the same format as Theorem 3.32 - the Recursion Theorem
for On. As before we shall dene H as a union of approximations where now u is an
approximation if (a) Func(u), Trans(dom(u)), and (b) 8w 2 dom(u)( u(w) = G(u w) ). We
call it an x-approximation, if additionally x 2 dom(u). So u satises the dening clauses of
H throughout its domain. Note for later that TC(fxg) dom(u) for any x-approximation
u.
(1) If u and v are approximations, and we set y = dom(u) \ dom(v) then u y = v y is
an approximation.
Proof: Note that Trans(y) as the intersection of any two transitive sets is transitive.
Suppose we have shown that for some x 2 y that 8z 2 x(u(z) = v(z)). Then u x = v x;
but then u(x) =df G(u x) = G(v x) =df v(x)! We thus have shown
8x 2 y(8z 2 x(u(z) = v(z) ¡! u(x) = v(x)))
By the (set form of the) Principle of 2-induction applied to the transitive set X = y
we conclude that 8x 2 y(u(x) = v(x)), and we are done.
(2) (Uniqueness) If H exists then it is unique.
Proof: This is really the same as before but we repeat the detail: if H, H 0 were two
such functions dened on all of V , there would be an 2-least set z on which they disagreed.
Note that z cannot be ?: Let x = TC(fz g). So then H x =
/ H 0 x are two dierent xapproximations, which is impossible by (i).
(3) (Existence). Such an H exists.
Proof: Let u 2 B () fuj u is an approximationg. B is in general a proper class of
approximations, but this does not matter as long as we are careful. As any two Ssuch
approximations agree on the common transitive part of their domain, we dene H = B.
Just as for the proof of recursion on !:
(i) H is a function
(ii) dom(H) = V .
Proof: We use the principle of 2-induction. It suces to show then that 8z(8y 2 z(y 2
dom(H)) ¡! z 2 dom(H)).
Let C be the class of sets z for which there is no z-approximation. So if we suppose
for a contradiction that C is non-empty, by the Principle of 2-Induction, then it will have
an 2-minimal element z such that 8y 2 z9u(u is a y-approximation). But now we let h be
the function
The Wellfounded Universe of Sets
49
S
fh y jh y is a y-approximation ^ y 2 z ^ dom(h y ) = TC(fyg)g.
By the above these functions h y all agree on the parts of their domains they have in
common. Note that the domain of h is transitive, being the union of transitive sets dom(h y )
for y 2 z. Hence z dom(h) and thus fz g [ dom(h) is transitive. We can extend h to
h 0 = h [ fhz; G(h z)ig and h 0 is then a z-approximation. However we assumed that z 2 C!
A contradiction. Hence C = ? and (ii) holds.
Q.E.D.
Exercise 6.6. Show for any x that (x) = (TC(x)).
Exercise 6.7. Let X be any set. Show that Trans(X) ! X 2 On:
Exercise 6.8. Does Trans(X) ^ X =
/ ? imply that ? 2 X?
Exercise 6.9. () We say that a function j: V ¡! V is an elementary embedding if it preserves the
truth about objects. In other words if '(v0; :::; vn) is a formula and x0; :::; xn are sets; then
'(x0; :::; xn) ! '(j(x0); :::; j(xn)).
If we assume the axioms of set theory (but not AC) our current state of knowledge allows the
possibility that such a class function j could exist which is not the identity (so j(x) =
/ x for some x 2 V ).
Show if there is such a j then for some ordinal , j() =
/ . [Hint: Consider the formula u = rk(v).]
(It is known by a result of K. Kunen that AC rules out the existence of such a j.)