Set Theory Lecture Notes
Introduction.
Set Theory
(1) Set Theory as the foundations of mathematics.
• Zermelo Frankael Axioms with the Axiom of Choice (ZFC) widely
accepted as the “foundations” of all of modern mathematics.
• All basic intuitive objects of mathematical study, integers and their
properties (Peano axioms), rational numbers, real numbers, functions,etc...
can be formally and rigourously defined (as sets) in this
√ framework.
E.g., the natural number 0 is a set. The real number 2 is a set as
are functions, functionals, measures, algebraic operations etc....
• And every possible proposition formulated in mathematics can, in principle, be written out using nothing but the symbols ∈, logical symbols
(∃, ¬, etc...), and variables (x, y, a, b, ....).
• So, questions about what is a proof, notion of consistency and independence of mathematical statements wrt the axioms can be precisely
formulated and rigorously studied.
(2) Set Theory as topic of study in its own right.
• Initiated by Cantor (and Dedekind ) in late 1800s
• Flourished in the late 20th c to the present after Cohen’s proof of the
independence of Cantor’s Continuum Hypothesis.
• Set-theoretic results and techniques have wide ranging applications
especially towards proving consistency and independence results in
other areas of mathematics: topology, functional analysis, algebra,
measure theory, real analysis....
We’ll do a bit of (1) but mainly focus on (2).
Prerequisites In addition to the usual fundamentals regarding functions, relations
(equivalence, order, etc...) and basic algebra of sets, some background in topology
of the real line (and Euclidean space) is needed. For example, the following results
should be familiar:
(1) Compact subsets of Euclidean spaces (covering definition and characterizations, e.g., Heine-Borel and closed and bounded subsets of Euclidean
space)
(2) Baire Category Theorem: Given a countable family of dense open subsets
of R, its intersection is dense. True for any completely metrizable space,
and for any compact space.
(3) Lebesgue measure and the algebra of measure zero sets.
(4) Basic topology of R, continuous functions, closures, etc...
The Axioms
If all mathematical objects are sets, then there should be clear rules for defining
new sets from established existing sets. We are interested in the simplest, most
intuitive assumptions needed to prove the existence of N and from that the set of
reals R and prove its properties, algebraic, topological etc... one of the first interesting objects constructed in real analysis, is Lebesgue measure on the collection of
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2
subsets of the reals. One is forced to refer to
{X : X is a subset of R}
the collection of all subsets of R. Or even before we get there, other more basic
objects like the set of ordered pairs of real numbers, etc....
It is instructive to (once in your life) go through the development of how this is
done formally.
If this were a more rigorous couse in mathematical logic, we would spend some
time talking about the language of set theory (with two relational symbols = and
∈) and the distinctions from syntax (formulas, formal proofs, rules of inference)
and semantics (models of set theory, satisfaction) and formalizing what is meant
by ‘consistency’ ‘truth’ etc...
Instead we take an informal approach, assuming that all mathematical objects
are sets, and the universe of all mathematical objects (while not a set can be
thought of as a collection, or more rigorously a “class”) satisfy some basic self
evident properties (axioms). Some are more intuitive and self-evident than others.
Axiom of Existence The class of all sets is not empty:
∃x(x = x)
Axiom of Extensionality A set is determined by its elements: If x and y are sets
and every element of x is an element of y and every element of y is an element of
x then x = y
∀x∀y (∀z(z ∈ x → z ∈ y) ∧ ∀z(z ∈ y → z ∈ x)) → x = y
The next axiom is a natural way to avoid Russell’s paradox which underlines
that there are some natural collections that can’t be formal objects in our sets.
E.g., the collection of all sets that do not contain themselves as elements:
{x : x 6∈ x}
If this were a set, then call it a and ask is a ∈ a? If not then a 6∈ a and so satisfies
the defining clause of itself so indeed it is an element of itself. If, on the other hand,
it is the case that a ∈ a, then by definition of a must satisfy a 6∈ a.
However, if instead we restrict formations of such sets (the set of all elements
satisfying some property) by bounding them as subsets of already obtained sets,
then we get a natural (and non-contradictory) way of forming new sets from old:
Axiom Scheme of Comprehension For any property of sets φ (i.e., any formula
in the language of set theory with one free variable) and for any set x, the set of
elements of x satisfying φ is a set:
∀x∃y(z ∈ x ∧ φ(z) ⇐⇒ z ∈ y)
Remark It is called a scheme since it is not a single axiom. But for every formula
φ we obtain an axiom.
Proposition 1. There is a unique set with no elements.
AKA ∅, the empty set.
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Proof. By the axiom of existence, there is a set, call it A. By Comprehension with
the formula φ being x 6= x, there is a set B with the property that x ∈ B if and
only if x ∈ A and x 6= x. By definition, since every set is equal to itself, B has no
elements. By Existentionality, any other set with no elements is equal to B.
In general, given a set A and property φ, the set B asserted to exist by Comprehension, is, by Existentionality, unique, so it is handy to have notation to refer to
this set, namely B can be denoted
{x ∈ A : φ(x)}
And, since the unique set with no elements comes up frequently, we call it “the
empty set” and denote it ∅.
How about the collection of all subsets of a fixed set A? Being a subset is a property (X is a subset of A if ∀x(x ∈ X → x ∈ A)), so does Axiom of Comprehension
give the collection of all subsets of A as a set?
{x : x ⊆ A}?
Of course, if we have a set that contains all subsets of A as elements, then we
can use Comprehension, but that is not provable to exist from our axioms so far,
so we need another one:
Power Set Axiom The collection of all subsets of a given set is itself a set.
∀x∃y∀a(a ⊂ x → a ∈ y)
And so, we denote the unique set of all subsets of a set A by P(A).
Other natural constructions cannot be proven to exist from the previous axioms:
Axiom of Pairing Given x and y there is another set that has x and y as elements.
∀x∀y∃z(∀w(w ∈ z ⇐⇒ w = x ∨ w = y))
And this set is denoted {x, y}.
Axiom of Union Given a collection of sets, its union is a set.
∀x∃y(∀z(z ∈ y ⇐⇒ ∃w(w ∈ x ∧ z ∈ w))
S
x denotes the union of the sets in x.
Exercises A big chunk of Chapters 1 and 2 of [HJ] show how basic operations of
sets are shown to exist from the axioms. E.g., justify the existence of each of the
following:
(1) Given two sets A and B, the sets corresponding to A ∪ B, AT
∩ B and A \ B?
(2) Given a collection of sets A, the intersection of the family A
How about ordered pairs and A × B?
(a, b) = {{a}, {a, b}} which exists by two applications of Pairing. and has the
property that (a, b) = (c, d) if and only if a = c and b = d (prove this if it is not
obvious).
What about A × B, by Comprehension
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A × B = {(a, b) : a ∈ A ∧ b ∈ B}
This is an unbounded (hence unjustified) application of Comprehension....??
Note: {a}, {a, b} ∈ P(A ∪ B), and so each (a, b) ∈ P(P(A ∪ B))....
Relations and functions. Linear orders, partial orders, functions, equivalence
relations etc... are all examples of relations with particular properties. Now that
we have established how to define A×B for arbitrary sets A and B, we can consider
subsets with particular properties keeping in mind that, formally, a relation on a
set A is merely a subset of A × A and a relation between A and B is a subset of
A × B.
E.g., A relation F ⊆ A × B is a function from A to B if
(1) ∀a ∈ A∃b ∈ B(a, b) ∈ F
(2) ∀a ∈ A∀b, c ∈ B((a, b) ∈ F ∧ (a, c) ∈ F → b = c)
Basic notions such as the range of a function, injectivity, surjectivity etc...
Properties of relations (transitivity, reflexivity, symmetry, etc...) should be familiar as well as special types of relation: Orders (including partial, linear and well)
and equivalence relations.
N and the Peano Axioms Z, Q, R
Given (N, 0, S), satisfying the Peano Axioms, namely:
(PA1)
(PA2)
(PA3)
(PA4)
0∈N
∀n ∈ NS(n) ∈ N \ {0}
∀n, m(n 6= m ⇒ S(n) 6= S(m))
∀φ in the language of arithmetic, if φ(0) is true and for all n (φ(n) ⇒
φ(S(n)) then ∀nφ(n) is true.
From these, one can define 1 = S(0), and addition, multiplication can be defined
recursively using
(1)
(2)
(3)
(4)
(5)
n + 1 = S(n)
n+0=n
n + (m + 1) = (n + m) + 1
n·0=0
n · (m + 1) = n · m + n
Then one can reformulate induction by
(6) (N, 0, 1, +) is inductive, i.e., for every formula φ in the language of arithmetic, if φ(0) is true and for all n φ(n) implies φ(n + 1), then φ(n) is true
for all n ∈ N.
And prove by induction that
(7) If n 6= 0 then n = k + 1 for some k.
(8) + and · satisfy commutative, associative and distributive laws.
And after that, given N satisfying the Peano Axioms, the collection of axioms
fixed so far suffices to define Z, Q and R and establish their defining properties
(e.g., Q dense in R, Z unbounded in R, completeness of R etc....)
[Say something here about how....]
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But, what are the elements of N, E.g., what is 0? or 1? or 17 for that matter.
And how to construct the collection of all natural numbers into a single set N?
Need 0 to start and some definition of the successor function S to define 1 and
n + 1 for every n.
The idea, mainly due to von Neumann, is to let 0 = ∅ and 1 = {0} etc... Note:
0 ∈ 1.
How about 2, 3, or given n how to define n + 1 etc... How does one deal with
“etc...”?
2 = 1 ∪ {1}
3 = 2 ∪ {2}.
So, indeed, 1 = {0, {0}} and 2 = {0, 1} = {0, {0, {0}}} Etc...
In general, give a set n, let S(n) = n ∪ {n}. We will also henceforth denote
S(n) = n + 1
Now define a set A to be inductive if ∅ ∈ A and S(x) ∈ A for all x ∈ A. Is there
an inductive set? If so then it turns out there is a minimal inductive set, and that
minimal inductive set we call ω (to use the modern set-theoretic notation) and can
check it satisfies the Peano Axioms.
Axiom of infinity There is an inductive set.
∃x (∅ ∈ x ∧ ∀y ∈ x(y ∪ {y} ∈ x))
Let A be any inductive set and let
ω = {x ∈ A : x ∈ y for every inductive set y}
I.e., ω is the intersection of all inductive sets and so has the property that ω ⊆ A
for any inductive set A. Note, 0, 1, 2, 3... ∈ ω (as defined above) and indeed ω along
with the successor function S(n) = n ∪ {n} satisfies the Peano Axioims: To see this
we need to prove:
Proposition 2. For any formula φ, suppose that φ(∅) and that for all n if φ(n)
holds then φ(S(n)) holds. Then φ(n) holds for every n ∈ ω
Proof. Let Z = {n ∈ ω : φ(n)}. Then by properties of φ we have that ∅ ∈ Z and
n ∪ {n} ∈ Z for all n ∈ Z. So Z is inductive. By minimality of ω it follows that
ω ⊆ Z and so φ(n) holds for all n ∈ ω.
Definition 1. A set x is transitive if every element of x is also a subset of x
Examples: ∅, 1, 2, 3....
EXERCISE 1. Prove that every n ∈ ω is a transitive set. Prove also that ω is
transitive.
EXERCISE 2. Prove that n 6∈ n for any n ∈ ω
EXERCISE 3. Prove that for every n, m ∈ ω either n 6∈ m or m 6∈ n
Proposition 3. Peano axiom P A(3) holds
Proof. By induction on n we show that φ(n): for all m if n 6= m then n ∪ {n} =
6
m ∪ {m}.
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Note this holds for n = 0. Since if m 6= 0, then m ∪ {m} has an element which
is not empty, while {0} has only one element which is the emptyset.
Suppose that φ(n) holds for some n. We need to show that φ(S(n)) holds. To
see this consider m 6= S(n) = n ∪ {n}. Case 1. m = n: then since n 6∈ n we have
by transitivity of the elements of ω that n ∪ {n} 6∈ n = m. And since n ∪ {n} =
6 m
it follows that n ∪ {n} 6∈ m ∪ {m} = S(m). Hence n ∪ {n} ∈ S(n ∪ {n}) but
n ∪ {n} 6∈ S(m). So S(n + 1) 6= S(m).
Case 2. m 6= n. In which case either m 6∈ n or n 6∈ m. In the first case, since m 6=
n and m 6= n∪{n} we have m 6∈ n∪{n}∪{{n∪{n}} So m∪{m} =
6 n∪{n}∪{{n∪{n}}.
In the case that n 6∈ m we have, by transitivity, n∪{n} 6∈ m. And since n∪{n} =
6 m
we have that n ∪ {n} 6∈ m ∪ {m} and again m ∪ {m} =
6 n ∪ {n} ∪ {{n ∪ {n}}.
Proposition 4. The relation ∈ on ω is a strict linear order (i.e., transitive, asymmetric and total).
Proof. Transitivity of ∈ is already proven since every element of ω is a transitive
set.
Linear is proven by showing that for all n 6= m either n ∈ m or m ∈ n.
And strict is proven since we have that n 6∈ n for all n.
asymmetry similarly because for all n, m either n 6∈ m or m 6∈ n.
EXERCISE 4. Prove that for every n ∈ ω \ {0} there is a k ∈ ω such that
n = k + 1. (Hint: if n is not of this form, consider n as a set of its predecessors
{m ∈ ω : m ∈ n}).
Notation: If n = k + 1 we denote k by k = n − 1
Proposition 5. Prove that ∈ is a well-ordering of ω. I.e., every nonempty subset
of ω has an ∈-minimal element. (Hint: if A has no minimum, consider {x ∈ ω :
x < a for all a ∈ A}).
Proof. Suppose that A ⊆ ω is nonempty, and as in the hint, let B = {x ∈ ω : x <
a for all a ∈ A}. Since A has no minimum and 0 is the minimum of ω, we have
that 0 ∈ B. If n ∈ B then n + 1 ∈ B since if not, then n + 1 6< m for some m ∈ A.
So either n + 1 = m in which case m would be the minimum of A, or m ∈ n + 1
in which case either m ∈ n or m = n in both cases this is impossible since n ∈ B
means that n ∈ m.
EXERCISE 5. Find another inductive set (distinct from ω).
So, we can define + and · on ω recursively in such away that (ω, 0, 1, +, ·, ∈)
satisfies the Peano Axioms.
Principle of Recursion Intuitive, but imprecise, version. Suppose that A is a
set, a ∈ A and for each n, having defined f n : n → A, we have a rule Φ by which
to define f (n) ∈ A (which depends on f n, then there is a functions f : ω → A
such that f (0) = a and for all n f (n) = Φ(f n).
This is more or less correct, but is a bit circular since it presupposes some
sequence f and we are indeed trying to construct f some clarification needed.
Firstly, let A<ω denote all finite sequences of elements of A. I.e., the set of all
functions s such that there is an n ∈ ω such that s : n → A.
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Does A<ω provably exist from our Axioms?
A<ω = {s ⊆ ω × A : s is a function and ∃n ∈ ω(dom(s) = n)}
Principle of recursion as stated above is a bit circular, since “having defined fn ..”
is not clear. What we really mean is, no matter how we end up defining fn we have
a way of extending it. Hence our consideration of all possible functions from n into
A.
Theorem 1. Principle of Recursion. Suppose that Φ : A<ω → A is given and
suppose that a0 ∈ A. Then there is a f : ω → A such that f (0) = a and f (n) =
Φ(f n) for all n 6= 0.
Proof. Let
F = {s ∈ A<ω : s(0) = a0 and for all n ∈ dom(s)(s(n) = Φ(s n)}
S
Let f = F . Claim: f is a function satisfying the conclusion of the Theorem.
As an illustration:
Theorem 2. A non-empty linearly ordered set (A, ≤) is a *not* a well-ordering if
and only if there is a strictly decreasing function f : ω → A. I.e., f is such that
f (m) < f (n) for all n ∈ m.
Proof. Suppose that A is non-empty and not well-ordered. Fix a0 ∈ A. For each
s ∈ A<ω , let φ(s) ∈ A be given such that φ(s) < s(n − 1) where n = dom(s) (so
n − 1 is the largest element in the domain of s.
Then, by the principle of induction, there is f : ω → A such that f (0) = a0 and
for all n > 0 f (n) < f (n − 1). By induction it is easy to show that f (n) < f (m)
whenever m ≤ n.
Cardinality and cardinal arithmetic
Definition 2. (Cantor, 1870) Given two sets, we say they are equipotent or of the
same cardinality and we write |A| = |B| if there is a bijection f : A → B.
Definition 3. We say that |A| ≤ |B| if there is an injection from A to B. And
|A| < |B| if |A| ≤ |B| but |A| =
6 |B|.
Note: This does not mean that we associate some value or meaning to |A|
(though, later we will).
Definition 4. A set A is finite if |A| = |n| for some n ∈ ω and in this case we do
just write |A| = n.
Definition 5. A is countable or denumerable if either A is finite or |A| = |ω|. If
A is countable and not finite we say it is countably infinite.
Definition 6. A is uncountable if A is not countable.
.
Some familiar results you may be familiar with, we will prove things more general
once we have discussed ordinals, but please recall:
Theorem 3. The following types of sets are all countable:
(1) a product of any finite family of countable sets
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(2) the union of a countable family of countable sets is countable.
(3) any subset of a countable set.
(4) the range of a function with countable domain.
So, Q is countable. And indeed the set of algebraic numbers (roots polynomials
with rational coefficients) is countable.
However,
Theorem 4. (Cantor) For any set X, |X| < |P(X)|
Corollary 1. (Cantor) R is uncountable.
Basic results:
• If |A| ≤ |B| and |B| ≤ |A| then |A| = |B|.
Proof. Fix f : A → B and g : B → A injective.
Define subsets An and Bn recursively by
A0 = A, B0 = B, Bn+1 = f (An ), An+1 = g(Bn )
EXERCISE 6. Show that
(1) An+1 ⊆ T
An for all n.T
(2) f maps n An onto n Bn .
S
Now
S it is easy to check that f maps S{An \ An+1 : n even} 1-1 and
onto S {Bn \ Bn+1 : n odd} and g maps {Bn \ Bn+1 : n even} 1-1 and
onto {An \ An+1 : n odd}. So gluing these maps together along with the
previous exercise gives a bijection from A to B.
• For A and B nonempty sets, |A| ≤ |B| if and only if there is a surjection
from B onto A.
Proof. Suppose there is a surjection g : B → A. Let f : A → B be a
choice function for {g −1 (a) : a ∈ A}. Then f is an injection. Conversely,
if f : A → B is injective, let a0 ∈ A be a fixed element (since A is not
empty). Define g : B → A by g(b) = f −1 (b) if b ∈ ran(f ) and g(b) = a0
otherwise.
• Every finite subset of ω has a maximal element, hence is bounded.
• For all m, n ∈ ω, if m < n then |m| < |n|. This justifies the notation
|A| = n for finite sets A. (By induction on m).
• A is finite iff |A| ≤ |n| for some n ∈ ω iff |A| < ω.
• If A is infinite then |ω| ≤ |A|. I.e., any infinite sets contains a countably
infinite subset.
Proof. Let g be a function with domain the collection of all nonempty
subsets B ⊆ A, such that g(B) ∈ B for all such B. For each s ∈ A<ω ,
we have that s is not surjective (otherwise s would witness that |A| ≤ n
implying that A is finite). Thus by the principle of recursion, there is a
f : ω → A such that for all n f (n) ∈ A \ ran(f n). Any such f is
injective.
• A is infinite if and only if there is an injection from A to A that is not
surjective.
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Note, we have had to invoke a new axiom!
Axiom
of Choice For any family A of nonempty sets, there is a function f : A →
S
A such that f (a) ∈ a for all a ∈ A.
EXERCISE 7. For
invoking AC?
(1) The family of
(2) The family of
(3) The family of
(4) The family of
(5) The family of
which family of sets can you find a choice function without
all
all
all
all
all
nonempty
nonempty
nonempty
nonempty
nonempty
subsets of ω.
subsets of Q
subsets of R
closed subsets of R.
open subsets of R
Further discussion on AC deferred to later.
Well-ordered sets, ordinals and transfinite recursion
Definition 7. A relation R ⊆ A2 is a partial order if it is transitive, reflexive and
antisymmetric.
It is a strict partial order if it is transitive irreflexive and a-symmetric.
A linear or total order if it is a partial order such that for all x, y ∈ A either
xRy or yRx.
It is a well-order if it is linear and every nonempty subset has a minimal element.
E.g., (ω, ∈) is a strict well-order. And we define ≤ on ω by n ≤ m if n ∈ m or
n = m.
Any others?
If A and B are w.o.’s then A + B denotes the disjoint union of A and B ordered
by all elements in A lie below all elements of B.
A × B with the lexicographic order.
etc...
Aω with the lexicographic order is not a well-order.
Theorem 5. (Transfinite Induction) Given A, ≤ a well-ordering. Suppose that
B ⊆ A is such that
• min(A) ∈ B
• ∀a ∈ A if Ia = {x ∈ A : x < a} ⊆ B, then a ∈ B.
Then B = A.
In the following, X <A is the set of functions with domain an initial segment of
A into X.
Theorem 6. (Transfinite Recursion) Given A, ≤ a well-ordering, a set X and an
x0 ∈ X. And given a function g : X <A → X. Then, there is a F : A → X such
that F (min(A)) = x0 and for all a ∈ A, F (a) = g(F Ia ).
Proof. First suppose that A has no maximum.
Let
F = {f ∈ X <A : f (min(A)) = x0 and f (b) = g(f Ib ) for all b ∈ dom(f )}.
Claim 1. For all f, h ∈ F either S
f dom(h) = h or h dom(f ) = f .
In either case we have that F = F is a function. And we claim that dom(F ) =
A and F satisfies the required conclusion.
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Note, that if A has a maximum a∞ , then the above construction gives dom(F ) =
Ia∞ and the proof is completed by extending F by defining by adding a∞ to the
domain and defining the function there to be equal to g(F Ia∞ ).
Theorem 7. Given well-orderings A and B exactly one of the following is true:
(1) A is order isomorphic to B
(2) A is order isomorphic to Ib for some b ∈ B
(3) B is order isomorphic to Ia for some a ∈ A.
Proof. Assume (2) does not hold. Define h : B → A by transfinite recursion.
Let a0 = minA. Let h(min(B)) = a0 .
Having defined h(x) for all x < b so that h is order preserving on its domain, let
h(b) = min(A \ {h(x) : x < b}). Since (2) does not hold, A 6= {h(x) : x < b}, so
h(b) is defined.
Formally, we are doing the following:
For s ∈ A<B if s is order preserving let g(s) = min(A \ ran(s)) otherwise let
g(s) = a0 .
By the principle of transfinite recursion, there is an h : B\A such that h(min(B)) =
a0 and for all b ∈ B h(b) = g(h Ib ).
EXERCISE 8. Prove that h is order preserving and that the range of h is an
initial segment of A (i.e., either all of A or Ia for some a ∈ A).
So either (1) or (3) holds completing the proof.
EXERCISE 9. Without using the Axiom of Choice, prove that there is an uncountable well-ordered set. Hint: Let
W = {(ω, R) : R is a well-ordering of ω}
Mod out by the equivalence relation of order isomorphism to form W/ ∼. Then let
the equivalence classes be ordered by [W 1] ≺ [W 2] if W 1 is order isomorphic to an
initial segment of W 2 for some (any) representatives. This is a well-ordering and
W/ ∼ is uncountable.
With AC, much more can be said:
Theorem 8. TFAE
(1) AC
(2) WO: Every set can be well-ordered.
(3) KZL. Should be Kuratowski Lemma, more widely called Zorn’s Lemma, and
as a compromise some call it the Kuratowski-Zorn Lemma: Suppose that
Z
S is a collection of sets such that for every C ⊆ Z linearly ordered by ⊆,
C ∈ Z. Then there is an ⊆-maximal member of Z.
Outline of proof (look it up!): (3) ⇒ (2) ⇒ (1) ⇒ (3). All of these are fairly
easy and standard. The only one that is hard is that AC implies the KZL. And
this was proven by Kuratowski in 1922 (while Zorn only formulated the principle
and gave applications in a paper in 1935).
vonNeumann Ordinals.
Definition 8. α is an ordinal if α is a transitive set (i.e., x ⊆ α whenever x ∈ α)
and strictly well-ordered by ∈.
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Examples? ω, n for any n ∈ ω. Any others?
Note: if α is an ordinal and x0 is the minimum of α, and y ∈ x0 what about y?
What can we conclude then about x0 ?
So, ∅ must be the minimal element of any ordinal α. Is {∅} ∈ α?
Observations
• If α is an ordinal and β ∈ α, then {x : x ∈ α and x is below β} = {x : x ∈
α and x ∈ β} = α ∩ β = β. So not only is every element a subset, but every
element is the set of its predecessors....
• If α is an ordinal so is α ∪ {α} and this ordinal is denoted α + 1.
• If an ordinal α has a maximal element β then α = β + 1. Any such ordinal
is called a successor. Other ordinals are called limit ordinals.
EXERCISE 10. Prove that the
S union of any set of ordinals is an ordinal.
α = β. And if α is a limit ordinal then
And
that
if
α
=
β
+
1,
then
S
α = α.
• Uniqueness: If two ordinals are order isomorphic, then they are equal.
Indeed, suppose that f : α → β is an order isomorphism between ordinals
α and β. Let γ = min{x ∈ α : f (x) 6= x}. Since each element is the
set of it’s predecessors and, since f is an order isomorphism, the set of
predecessors of f (x) is {f (y) : y ∈ x}. By minimality of x we have f (y) = y
for all y ∈ x, so f (x) = {f (y) : y ∈ x} = x.
• Given any two ordinals α and β, either α ∈ β or β ∈ α.
So, the “ next” ordinal after ω is ω + 1 = ω ∪ {ω}.
Consider the sequence ω, ω + 1, ω + 2, .... If we put them all together into a single
set, we could call this set ω + ω, but why are we justified in asserting the existence
of this set?
Axiom of Replacement Given a formula φ, and a set A suppose for all x ∈ A
there is a unique y such that φ(x, y). Then there is a set B such that for all x ∈ A
there is a unique y ∈ B. such that φ(x, y).
Main application of Replacement here is
Theorem 9. For any well-ordering W there is a unique ordinal α order isomorphic
to W .
Proof. First observation. Uniqueness follows by our previous observation that if
two ordinals are order isomorphic, then they are equal.
Fix W . For each w ∈ W we may ask if there is an ordinal αw order isomorphic
to Iw . If there is such an ordinal for each w, then it must be unique. So by the
Axiom of Replacement there is a set X such that αw ∈ X for every w ∈ W . Now,
let α = {β ∈ X : ∃w ∈ X(β = αw )}.
Claim: α is an ordinal and α is order isomorphic to W .
To prove this note that since α is a set of ordinals, it is strictly well-ordered by
∈. Also, ∅ ∈ α since the initial segment below the minimum of W is indeed empty.
So it suffices to prove that α is transitive. So fix β ∈ α. Pick w such that Iw
is order isomorphic to β. Pick γ ∈ β. Then there is x < w such that x is the
image of γ under the order isomorphism. Thus Ix is order isomorphic to the set of
predecessors of γ, but this is means that γ ∈ α. As required.
12
Given a well-ordered set W , let the order type of W be the unique ordinal order
isomorphic to W .
Ordinal arithmetic
Definition 9. For α, β, γ ordinals, let
• α + β be the order type of the well order formed by putting an isomorphic
copy of β above a copy of α. I.e., {0}×α∪{1}×β ordered lexicographically.
• α · β is the order type of α × β in the reverse lexicographic ordering. This
corresponds to the order type of the well-ordering formed by stacking up β
copies of α.
• αβ is defined recursively: α0 = 1, αβ+1 = αβ · α and for γ a limit ordinal,
αγ = sup{αξ : ξ < γ}.
EXERCISE 11. Ordinal arithmetic satisfies the following:
(1) + is in general not commutative, but it is associative and for β a limit
ordinal α + β = sup{α + ξ : ξ < β}.
(2) Ordinal multiplication is in general neither commutative nor distributive on
the right, but it is associative and distributive on the left: I.e.,
α · (β + γ) = α · β + α · γ
Moreover, for γ a limit we have α · γ = sup{α · ξ : ξ < γ}
(3) αβ+γ = αβ · αγ .
(4) If γ > γ1 ≥ ... ≥ γk , then
ωγ >
k
X
ω γi
i=1
Hint: Prove by induction on γ treating the cases where γ is a successor or
a limit separately.
A curious application to number theory
A Goodstein sequence for a natural number n is constructed recursively as follows. (Illustrated for n = 8)
STEP 1. Write out n in base 2 (and its exponents too)
G(8)1 = 8 = 23 = 22+1
STEP 2. Replace all 2’s by 3’s and subtract 1 and rewrite in base 3.
G(8)2 = 33+1 − 1 = 80 = 2 · 33 + 2 · 32 + 2 · 3 + 2
Step 3. Replace all 3s by 4s, subtract 1, then rewrite in base 4:
G(8)3 = 2 · 44 + 2 · 42 + 2 · 4 − 1 = 2 · 44 + 2 · 42 + 4 + 3 = 295
G(8)4 = 2 · 54 + 2 · 52 + 5 + 2 = 1307
G(4)5 = 2 · 64 + 2 · 62 + 6 + 1 = 2671
2
3
For n = 65 G(65)1 = 25 + 1 = 22 +1 + 1 G(65)2 = 33
P256
4
G(65)3 = 44 +1 − 1 = 4257 − 1 = 1 3 · 4i
Hmmmm.
Goodstein’s Theorem
+1
3
+ 1 − 1 = 33
+1
Theorem 10. (ZF) For every n, there is a k such that G(n)k = 0. I.e., every
Goodstein sequence eventually begins to decrease and terminates.
13
Idea of the proof: In the kth step of the Goodstein sequence, replace each instance
of k + 1 by ω, denote this ordinal O(n)k .
E.g. for the case n = 8:
G(8)1 = 22+1
O(8)1 = ω ω+1 = ω ω · ω
G(8)2 = 2 · 33 + 2 · 32 + 2 · 3 + 2
O(8)2 = ω ω · 2 + ω 2 · 2 + ω · 2 + 2
And O(8)2 < O(8)1
EXERCISE 12. for all n and all k O(n)k+1 < O(n)k .
And O(n)k = 0 if and only if G(n)k = 0
To prove the Claim, Suppose first that n is given and that n is expressed hereditarily in base k + 1. Let B(n)k be the ordinal obtained by replacing each instance
of k + 1 by ω in that expression writing multiplication by integers less than k + 1
in reverse order. E.g., if n = mr (k + 1)r + mr−1 (k + 1)r−1 + ... + m1 (k + 1) + m0 is
written in base k + 1, then B(n)k = ω B(r)k · mr + ω B(r−1)k · mr−1 ... + ω · m1 + m0 .
Moreover, if G(n)k = N = mr (k + 1)r + mr−1 (k + 1)r−1 + ... + m1 (k + 1) + m0
written in base k, then O(n)k = B(N )k .
So to solve the exercise, one would write out G(n)k , and consider the minimal i
such that mi 6= 0. If that happens to be m0 , then it is easy to verify the required
inequality, since in this case O(n)k+1 + 1 = O(n)k . However, if m0 = 0, then one
must calculate G(n)k+1 and the required inequality follows from a previous exercise
on inequalities involving ordinal arithmetic.
Remarkably, the reliance on ordinal arithmetic (and exponentiation of ordinals)
is in some sense necessary. While the Axiom of Infinity cannot be proven from PA,
it is need to prove Goodstein’s theorem, and indeed:
Theorem 11. Goodstein’s Theorem is not a theorem of Peano Arithmetic.
A bit of mathematical logic and some observations on consistency and
provability.
One would need to give a more formal introduction to mathematical logic to
give these ideas a rigorous study, but let us consider two examples of axiomatic
frameworks and what it means for a set of axioms to be consistent and what it
means for a sentence within the framework to be provable.
Example 1. The language of group theory and the axioms of a group. The language
consists of a constant symbol e and a binary function symbol P . The axioms of
Group Theory are
(1) ∀xP (x, e) = P (e, x) = x
(2) ∀x, y, zP (x, P (y, z)) = P (P (x, y), z)
(3) ∀x∃yP (x, y) = e
Here are some sentences in the language of group theory:
• f orallx, y, z[(P (x, y) = P (y, x) = e ∧ P (x, z) = P (z, x) = e) → y = z]
• f orallx, y(P (x, y) = P (y, x)).
• f orallx, y∃z(P (x, z) = y)
14
Which of these are provable from the Axioms of Group Theory, and which are
not provable?
So the second is not provable, because there is a model of group theory (i.e., a
group) where that statement fails. Namely a non-abelian group.
Definition 10. A model for a language is a set along with an interpretation of
all the function, relation and constant symbols in the language. What it means for
a sentence to be satisfied by a model is defined recursively and is the natural one.
E.g., A set G with an associated function · : G × G → G and an element 1G ∈ G
satisfies (GA) if and only if (G, 1G , ·) is a group.
Definition 11. Given a set of sentences Σ in some formal language. Σ is consistent
if there is no sentence φ such that φ ∧ ¬φ is provable from Σ.
Theorem 12. A set of sentences Σ is consistent if and only if there is a model
satisfying all the sentences in Σ.
Theorem 13. Given a set of axioms Σ in some formal language. And given a
sentence φ in that language, there is a proof of φ from Σ if and only if for every
model M for the language, if M satisfies Σ then M also satisfies φ.
So, the Axioms of Group theory are consistent but the sentence ∀x, yP (x, y) =
P (y, x) is not provable from those axioms.
Example 2. The language of Arithmetic and the Peano Axioms. The language has
one function symbol and one constant symbol. The axioms of Peano Arithmetic has
a model, namely the standard model (ω, 0, +1), hence the set of axioms is consistent.
Goodstein’s theorem holds in the standard model, however the proof requires
some strength of ZFC (namely the axiom of infinty) to carry out the proof. However,
there is a model of PA (a “nonstandard” model) where Goodstein’s theorem fails.
Cardinals and the ℵ’s
Since every set can be well-ordered, given a set X there is an ordinal α such that
|α| = |X|. So
Definition 12. For any set X, let |X| denote the minimal ordinal equipotent with
X. E.g., if A is countable then |A| = n for some n or |A| = ω. c denotes the
cardinality of R.
The ordinals that correspond to the cardinality of a set are called cardinals.
EXERCISE 13. Some basic facts about cardinals:
(1) Prove that α is a cardinal if and only if |α| = α.
(2) |α| ≤ α for every ordinal α.
(3) For any ordinal α there is a cardinal γ such that γ > α.
(4) We now have two meanings to each of the notions |A| = |B|, |A| ≤ |B| and
|A| < |B|. Prove that they coincide. E.g., |A| ≤ |B| means, by our original
definition, that there is an injection from A to B. But it also means the
cardinal associated to A is less than or equal to the cardinal associated to
B. These are equivalent notions. Similarly for = and <.
So, for any cardinal κ we denote by κ+ the least cardinal larger than κ.
By recursion on the ordinals, define ωα for all ordinals α as follows:
15
ω0 = ω
Having defined ωα , let ωα+1 = (ωα )+ .
If α is a limit and ωβ has been defined for all β < α, then let
[
ωα =
ωβ
β<α
EXERCISE 14. Prove that for any set X of cardinals,
S
X is a cardinal.
ω < ω1 < ω2 < .... < ωω < ωω+1 < ...
Successor cardinal, limit cardinal etc... all cardinals are limit ordinals.
EXERCISE 15. Find a cardinal κ such that ωκ = κ. What can you say about the
cofinality of the such a cardinal (or the particular κ you found) (see the definition
of cofinality given below).
ω1 is characterized as the unique uncountable ordinal such that all its predecessors are countable.
Q: Where is the continuum in this hierarchy?
A: It is undecidable.
Indeed, Cantor asked whether c = ω1 , and this came to be known as the Continuum Hypothesis (CH, for short). Gödel proved that it is consistent with the
axioms ZFC that CH is true (I.e., if we add CH as an axiom then as long as ZFC is
consistent, so is ZFC + CH). Equivalently, CH can be formulated as the statement
that every subset of R is either countable or equipotent with R.
And Cohen proved that the negation of CH is also consistent. I.e., if ZFC is
consistent, then so is ZFC + ¬CH.
Finding such extra set-theoretic axioms are useful in many applications outside
of set theory. Now there are a plethora of axioms (most established either by the
methods introduced by Gödel or those introduced by Cohen) such as MA, PFA,
OCA, ♦, etc... All of them are of a combinatorial nature, and one can apply
them to establish consistency and independence proofs without understanding the
metamathematical bases for why they are relatively consistent with ZFC.
E.g., it was an open question for many years whether countably compact perfectly normal spaces are actually all compact.
Theorem 14. (Ostaszewski) ♦ implies that there is a countably compact perfectly
normal non-compact (topology on ω1 ).
Theorem 15. (Weiss) M A + ¬CH implies that all countably compact perfectly
normal spaces are compact.
Or from Group Theory Shelah proved: There is a Whitehead group that is not
free assuming M A + ¬CH, but V = L, Gödel’s Axiom of constructibility, implies
all Whitehead groups are free.
Or from C ∗ -algebras: Weaver proved that ♦ implies that there is a C ∗ algebra
with only one irreducible representation but it is not isomorphic to the algebra of
compact operators on some Hilbert space. The companion result showing this is
independent is an open question.
There are a plethora of such results in many areas of mathematics.
Cardinal Arithmetic
16
Since the definitions that follow for cardinal arithmetic uses the same notation
as ordinal arithmetic, we introduce the notation ℵξ = ωξ to emphasize cardinality
and avoid confusion.
Our next set of results allow us to establish some important facts about cardinalities related to the continuum and subsets of the reals but have much more far
reaching applications. E.g, we will show (and generalize) such facts as
Cardinality of sets related to the continuum: The following families all have
cardinality the continuum:
• The set of subsets of ω
• The set of all functions from ω to ω, i.e., ω ω.
• The family of all countable subsets of R, [R]ℵ0
• The family of open subsets of R
• The family of closed subsets of R
• Indeed the family of all Gδ and Fσ and indeed the family of all Borel subsets
of R.
• The family of all continuous functions from R to R
One can prove the above from some basic facts about cardinality that we have
already established, but suggest a systematic approach that generalize to results
on cardinals in general.
CARDINAL ARITHMETIC DEFINED:
Definition 13. For cardinal κ and λ,
(1) κ · λ = |κ × λ|
˙
(2) κ + λ = |κ∪λ|
λ
λ
(3) κ = | κ| = |{f : f : λ → κ}|
Definition 14. For a sequence of cardinals {λi : i ∈ I}
X
[
˙
λi = |
λi |
i∈I
i∈I
and
Y
i∈I
λi = |
Y
λi |
i∈I
I.e., the product of an infinite family of cardinals is the cardinality of its cartesian
product.
So, for example ℵ0 · ℵ0 = ℵ0 (cardinal multiplication) while ω · ω > ω (ordinal
multiplication). Indeed, in general ℵα · ℵα = ℵα for all cardinals:
Proposition 6. For κ, λ, γ and {λi : i ∈ I} cardinals:
(1) P
κ · λ = κ + λ = max{κ, λ}
(2)
i∈I λi = |I| · sup{λi : i ∈ I}
λ γ
(3) (κ ) = κλ·γ
(4) κ ≤ γ implies κλ ≤ γ λ
(5) λ ≤ γ implies κλ ≤ κγ
(6) 2κ ≥ κ+
(7) The second inequality in (4) may be = even if the first inequality is strict.
I.e., For all cardinals 2 ≤ λ ≤ 2ℵ0 , λℵ0 = 2ℵ0 .
Corollary 2. All the statements listed above on cardinality of sets related to the
continuum follow from the previous proposition.
17
Proof. For example, the cardinality of the set O of all open subsets of R has cardinality c: Since Q × Q is countable, the set B of all open S
intervals with rational
endpoints is countable. Define H : P(B) → O by H(A) = A. Since every open
subset of the reals is a union of open intervals with rational coefficients, H is onto
so |O| ≤ |P(B)| = |P(N)| = 2ℵ0 .
And indeed, 2ℵ0 = c since we also have 2ℵ0 ≤ |R| ≤ |O| so we are done.
EXERCISE 16. Prove the other items listed above on cardinality of sets related
to the continuum. For the set of continuous functions, recall that if two continuous
functions agree on the rationals then they are equal.
To prove (1) we observe first the following:
Proposition 7. There is a canonical well-ordering of α × α for each α such β × β
is a proper initial segment of α × α for all β < α.
Proof. Draw a picture.
(x, y) < (a, b) if
(1) max{x, y} < max{a, b} or,
(2) max{x, y} = max{a, b} = m and either
(a) x = a = m and y < b, or
(b) y = b = m and x < a
(c) x = b = m and y < m
It is straightforward to check that for β < α we have that β × β is the set of
predecessor of (β, 0) and so is an initial segment of α × α
Now we use this ordering to prove that for every cardinal κ we have otp(κ×κ) = κ
(in the canonical ordering). This is proven by induction. First note that ω × ω has
order type ω since the initial segments are all finite.
Now suppose that κ is a cardinal and for all cardinals λ < κ we have that the
order type of λ × λ is λ. Note that this implies that λ · λ = λ for all cardinals less
than κ. Now suppose that the type of κ × κ is greater than κ.
Then some initial segment has order type κ so fix (γ, β) ∈ κ × κ so that the
initial segment up to (γ, β) has order type κ. Letting α ∈ κ lie above both γ and
β we have that the initial segment α × α below (α, 0) has cardinality ≥ κ. BUT
|α × α| = |α| · |α| = |α| < κ by our inductive hypothesis.
Corollary 3. κ · κ = κ for all cardinals κ.
There is not much that can be said, in ZFC, about κλ even for concrete instances.
For example, as noted above, if κ ≤ 2λ then we can say that κλ = 2λ . But what is
2ℵ0 ? Or in general 2ℵα ?.
The Continuum Hypothesis (CH) states that 2ℵ0 = ℵ1 , and
The Generalized Continuum Hypothesis (GCH) states that 2κ = κ+ for all
cardinals κ.
Of course, from GCH all values of κλ can be calculated. E.g.,
EXERCISE 17. Assume (CH). Prove that ℵℵn0 = ℵn for all n ∈ ω. Under the
assumption 2ℵ0 = ℵ12 calculate the values of ℵℵn0 for each n ∈ ω.
Theorem 16. (König’s Theorem) Given indexed families of cardinals {λi : i ∈ I}
and {κi : i ∈ I}, such that λi < κi for all i, then
X
Y
λi <
κi
i∈I
i∈I
18
Proof. .....
ωω has the curious property that it contains a countable cofinal subset:
Definition 15. Given an ordinal α, the cofinality of α is the minimal order type
of a cofinal subset, denote cof(α). A cardinal κ is regular if cof(κ) = κ
EXERCISE 18. Cofinality satisfies the following
(1) cof(α) ≤ |α|
(2) cof(α) is a regular cardinal
(3) For all α there is a strictly increasing cofinal embedding of cof(α) into α.
(4) Successor cardinals are regular.
(5) κ is regular if and only if for any family of sets A of size less than κ
consisting of sets of size less than κ, its union has size less than κ. I.e., if
|I| < κ and λi < κ for all i ∈ I, then
X
λi < κ.
i∈I
König’s Theorem implies the following results, which were, for about a century
until Shelah’s pcf-theory, the only significant ZFC constraints on the value of the
exponential function in cardinal arithmetic.
Theorem 17. κ < cof(2κ ) and κ < κcof(κ) hold for all infinite cardinals κ.
Not much else can be said in ZFC about the value of 2ω or 2κ for other cardinals
κ. E.g., almost any imaginable possibilities not ruled out by König’s theorem and
its consequences turns out to be consistent. E.g.,
Theorem 18. It is consistent with ZFC that 2ω = ω1 7 and 2ωn = ωω+1 for all
n > 0 and 2ωα = ωα+114 for all α ≥ ω.
Of course, König’s theorem rules out 2ω = ωω as a possibility since the cofinality
of 2ω must be uncountable.
However, a consequence of Shelah’s relatively recent pcf-theory gives some previously unknown bounds. E.g.,
Theorem 19. (Shelah) ωω ω < ωω4
Set theory of the reals
Some basic facts and notation:
R has an algebraic structure, order-theoretic structure, topology, and Lebesgue
measure.
Order properties:
R is the unique separable dense linear order that is order complete and has no
endpoints.
Indeed, Q is the unique countable dense linear order without endpoints. And R
can be viewed as the Dedekind completion of Q (using Dedekind cuts).
And an order isomorphism between Q and another such countable order extends
to their Dedekind completions.
EXERCISE 19. Every countable linear order is isomorphic to a suborder of Q.
E.g., every countable ordinal is order isomorphic to a subset of Q. However, there
is no order preserving embedding of ω1 into R.
19
Bounded closed subsets are compact.
Example 3. The Cantor set (constructed as a subset of [0, 1] via deleting the open
middle thirds of each closed interval...)
X
s(i)
ω
C=
: s ∈ {0, 2}
⊆ [0, 1]
3i+1
is homeomorphic to 2ω in the product topology
Example 4. P = R \ Q is homeomorphic to ω ω in the product topology (where each
factor ω has the discrete topology).
Proof. Since Zω is homeomorphic to ω ω , we will prove that Pω is homemorphic to
Zω .
Our plan is to construct by recursion on n, for each s ∈ Zn an open interval Is
with rational endpoints so that the following inductive hypotheses are satisfied
(1) For every k ∈ Z, Ihki = (k, k + 1).
(2) for every s and t if s ⊆ t (so t extends s as a function) then It ⊂ Is .
(3) for every n ∈ ω and every s ∈ Zn , {Is ˆ k : k ∈ Z} is a disjoint family of
open intervals
covering P ∩ Is of length < 21n . Hence for each n we have
S
that P ⊆ {Is : s ∈ Zn }.
S
(4) For every rational number r, there is an n such that r 6∈ {Is : s ∈ Zn }.
Having defined these intervals, one defines h : Zω → R by
\
h(f ) = the unique real number x such that x ∈
If n
n
ω
EXERCISE 20.
(1) h(f ) ∈ P for all f ∈ Z
(2) For every x ∈ P there is f ∈ Zω such that h(f ) = x.
(3) The family {Is ∩ P : s ∈ Z<ω } forms a base for the usual topology on P.
(4) h is a homeomorphism.
So we are left with defining this family of intervals. To this end we first note
that for each open interval I = (a, b) we may fix two sequences of rationals (qnI )n≥0
increasing and converging to b and (qnI )n≤0 decreasing and converging to a.
So, to carry out the recursion, first fix an enumeration of all the rationals as
{rn : n ∈ ω} and suppose that we have defined {Is : s ∈ Z<n } so that the above
IH(1)(2) and (3) hold. Let
Pn = {q ∈ Q : q is an endpoint of one of the intervals Is for some s ∈ Z<n }.
Is n Is n
If rn ∈ Pn then for each s ∈ Zn , let Is = (qq(n)
, qq(n)+1 ). One can check that items
≤n
(1)-(3) hold for the family {Is : s ∈ Z }.
In case rn 6∈ Pn . Determine first the unique s ∈ Zn−1 } containing rn and choose
the two sequences of rationals inside Is so that q0Is = rn . Then proceed as above
and again (1)-(3) hold for the same reasons.
S
Finally, (4) holds since at stage n we have guaranteed that rn 6∈ {Is : s ∈ Zn }.
Theorem 20. Every uncountable closed subset of R contains a copy of the Cantor
set.
So every closed set is either countable or equipotent with R. Similarly any Borel
set has this property.
20
Example 5. A Bernstein set is a subset X of the reals with the property that
neither X nor its complement contains an uncountable closed set (equivalently,
neither contains a copy of the Cantor set). There are Bernstein subsets of R.
Theorem 21. Baire Category Theorem. The union of countably many nowhere
dense subsets of R (or C or P) does not cover any open set. Equivalently, the
intersection of countably many dense open sets is dense.
A ⊆ R is meagre if it is the countable union of nowhere dense subsets.
Proposition 8. Some facts about meagre sets of R also true for meagre subsets of
C or the irrationals.
•
•
•
•
Countable sets are meagre
Every meagre set is contained in a meagre Fσ .
a countable union of meagre sets is meagre, hence
R cannot be covered by countably many meagre sets.
This is entirely analogous with measure zero sets (except the version of (2) for
measure zero is that every measure zero set is contained in a measure zero Gδ ).
Indeed, both families are σ-ideals: closed under countable unions, arbitrary subsets and are proper in that the whole real line is not in the ideal.
If CH fails, what can be said collections of size less than c?
Indeed
Definition 16. If we let I denote the ideal of meagre sets or measure zero sets
then we can define
• N ON (I) = minimum cardinality of a set that is not in I.
• ADD(I) = minimum cardinality of a family from I whose union is not in
I.
• COV (I) = minimum cardinality of a family from I that covers R, i.e.,
whose union is R.
Proposition 9. For I the ideal of meagre sets or measure zero sets we have all these
cardinals are less than or equal c and moreover ω1 ≤ ADD(I) ≤ min{N ON (I), COV (I)}
Assuming CH all of them are equal to ω1 = 2ℵ0 , however the actual values
of these invariants are not decidable in ZFC. Some things can be said about the
relationships between these invariants for measure and category. E.g., letting M
denote the meagre ideal and N denote the ideal of measure zero sets, we have
Theorem 22. COV (N ) ≤ N ON (M) and COV (M) ≤ N ON (N ).
Proof. Both inequalities are based on the fact that there is a subset E ⊆ R of
measure zero whose complement is meagre. To see this, enumerate Q = {qn : n ∈ ω}
and for each > 0 we show that there is an open set U ⊇ Q such that µ(U ) < .
containing qn and
This is done by choosing for each n an interval of length 2n+1
letting U to be the union of these intervals. SoTit’s a dense open set of measure
less than . Now choose k → 0 and let E = k Uk . E has measure 0 and its
complement is meagre.
Now, suppose that A is a nonmeagre set of cardinality N ON (M). For each real
x, consider the translate of E, x − E. x − E is non meagre so intersects A. So there
is an a ∈ A and an e ∈ E such that a = x − e. I.e., x = a + e. Therefore x ∈ a + E.
21
So the translates {a + E : a ∈ A} cover the real line and each of the translates has
measure 0. So COV (N ) ≤ N ON (M).
The other inequality is similar.
EXERCISE 21. Prove the other inequality
And most that can be said in ZFC is summarized in Cichoń’s diagram.
COV (N )
ω1
ADD(N )
N ON (M)
COF (M)
b
d
ADD(M)
COV (M)
COF (N )
c
N ON (N )
Definition 17. Consider the relation ≤∗ defined on ω ω given by f ≤∗ g if f (n) ≤
g(n) for all but finitely many n.
This is almost a partial order:
EXERCISE 22. Define f =∗ g if f (n) = g(n) for all but finitely many n. Show
that =∗ is an equivalence relation and that ≤∗ is a well-defined partial order on the
set of equivalence classes ω ω / =∗ .
Definition 18. Given a partial order (P, ≺), B ⊆ P is unbounded if for all p ∈ P
there is a b ∈ B such that b 6≺ p. B ⊆ P is dominating if for all p ∈ P there is a
b ∈ B such that p ≺ b. The unbounding number for P is
b(P ) = min{|B| : B is unbounded in P }
and the dominating number for P is
d(P ) = min{|B| : B is dominating in P }
b is the unbounding number for ω ω under ≤∗ and d is its dominating number.
EXERCISE 23. Let P = ω × ω1 and order P coordinatewise by (n, α) ≤ (m, β) if
n ≤ m and α < β. Show that the unbounding number for P is ω and the dominating
number is ω1 .
Some observations:
(1) For linear orders, the unbounding and dominating number are the same.
(2) b(P ) ≤ d(P ) for any partial order P .
(3) b is uncountable.
(4) For any f ∈ ω ω , the set {g : g(n) ≤ f (n) for all n} is nowhere dense.
EXERCISE 24. Show that b ≤ N ON (M) and COV (M) ≤ d (hint: use (4) and
consider {g : g ≤∗ f })
Theorem 23. ADD(M) ≤ b
Theorem 24. d ≤ COF (M)
Cardinal invariants of the continuum b and d are but two of many important
so called “cardinal invariants of the continuum.”
Another is the “pseudo-intersection number” p
22
Definition 19. Given subsets A, B ⊆ ω, we say A ⊆∗ B if A \ B is finite.
Remarks
(1) There are decreasing sequences ω ⊇ A0 ⊇ A1 ⊇ .... of infinite sets such that
\
An = ∅
n
(2)
(3)
(4)
(5)
however, for any such sequence there is an infinite B such that B ⊆∗ An
for all n.
Indeed, all countable families F ⊆ [ω]ℵ0 with the strong finite intersection
property (intersection of any finite subfamily is infinite) has an infinite
pseudo-intersection.
So, if we let p = min{|F| : F ⊆ [ω]ℵ0 has the sfip but no infinite pseudo-intersection},
then p > ℵ0
p≤b
Moreover p ≤ ADD(M)
Proof. Of last item. Suffices to prove that if κ < p then the intersection of κ many
dense open sets is comeagre.
So fix {Uα : α < κ} a family of dense open subsets of R. For each α let
Dα = Q ∩ Uα .
Lemma 1. There is a dense D ⊆ Q such that D ⊆∗ Dα for all α
Proof. Let B be the countable family of basic open intervals in R. For each basic
open I ∈ B fix DI a pseudo-intersection of {Dα ∩ I : α < κ}.
Enumerate Q as {qn : n ∈ ω}.
Now define fα : B → ω by fα (I) = max{n : qn ∈ DI \ Dα }. Since κ < p ≤ b,
we can find one function f : B → ω dominating all the fα . Then for all but finitely
many I ∈ B, we have that
DI0 = DI \ {qn : n < f (I)}
is fully a subset of Dα and for all the other I we do still have DI0 ⊆∗ Dα . Therefore
for each α < κ we have
[
D = {DI0 : I ∈ B} ⊆∗ Dα
as required. Proving the lemma.
Now fix this D as in the lemma. For each α and for each q ∈ D, if q ∈ Uα let
hα (q) be the minimal n such that (q − 1/n, q + 1/n) ⊆ Uα . Once again since κ < b
there is an h dominating all the hα . Fixing α, note that if we let F be the finite
set including D \ Dα and including {q ∈ D : h(q) < hα (q)}, then
[
UF = {(q − 1/h(q), q + 1/h(q)) : q ∈ D \ F },
then UF ⊆ Uα and UF is dense open. Since there are only countably many finite
subsets of D we have that
\
{UF : F ∈ [D]<ω }
T
is comeagre and contained in Uα . So this set is comeagre as required.
Finally we mention two other important “special subsets of R:”
23
Definition 20. A Lusin set is an uncountable set of reals such that its intersection
with every meagre set is countable. A Sierpinski set is an uncountable set of reals
whose intersection with every measure zero set is countable.
Theorem 25. CH implies the existence of Lusin and Sierpinski sets.
Separability of R and the CCC
A topological space is separable if it contains a countable dense subset.
E.g., R, and its subspaces.
ℵ0
Theorem 26. (Hewitt-Marczewski-Pondiczery) R2
is separable.
ω
Proof. Let 2 be the index set in the product. For each s ∈ 2n let [s] be the set of
all f ∈ 2ω that extends s. And let
ω
D = {f ∈ Q2 : ∃n ∈ ω ∀s ∈ 2n (f is constant on [s])}.
ω
EXERCISE 25. Prove that D is countable and that D is dense in R2 .
Example 6. If κ > 2ℵ0 then Rκ is not separable.
Proof. Suppose that D ⊆ Rκ is countable. For each α < κ, let Dα = {s ∈ D :
s(α) ∈ (0, 1)}. There are more than continuum such α but only continuum many
subsets of D, so there are α 6= β such that Dα = Dβ . Then the basic open set
πα−1 ((0, 1)) ∩ πβ−1 ((1, 2)) ∩ D = ∅. So D is not dense.
However, Rκ has an important related property:
Definition 21. A topological space is said to have the countable chain condition
(CCC), if every disjoint family of nonempty open sets is countable.
This notion is also an important property of partial orders.
Definition 22. In a partial order, (P, ≤), two elements p, q ∈ P are said to be
incompatible if there is no r ∈ P such that both r ≤ p and r ≤ q. A partial order
has the CCC if every family of pairwise incompatible elements (aka an antichain)
is countable.
So, if we order the nonempty open sets of a topological space with the subset
relation, then the two notions coincide (since u and v are incompatible if and only
if u ∩ v = ∅).
It is an easy application of the pigeon hole principle that a separable topological
space has the CCC.
A little harder is that the product of any family of separable spaces has the CCC.
But for this we need the notion of a ∆-system. A family of sets D forms a δ-system
with root r if
(1) r ⊆ d for all d ∈ D
(2) {d \ r : d ∈ D} is a pairwise disjoint family of nonempty sets.
Equivalently we can write that c ∩ d = r for all c 6= d in D.
Theorem 27. Any uncountable family of finite nonempty sets includes an uncountable ∆-system.
24
Proof. If all the sets have the same fixed cardinality n, then it can be proved by
induction on n. By pigeon hole, for some n the family includes uncountably many
sets of the same cardinality n.
Theorem 28. Rκ is CCC for all κ.
Proof. Let {Uα : α ∈ ω1 } be a family of open sets in Rκ . It suffices to find two
open sets with nonempty intersection.
For each α there is a basic open Bα ⊆ Uα . So for each α, there is a finite set
Fα ⊆ κ and for each i ∈ Q
Fα an open Oα (i) such that Bα = {f ∈ Rκ : f (i) ∈
Oα (i)∀i ∈ Fα }. I.e., Bα = {Oα (i) : i ∈ κ} where Oα (i) = R for all i 6∈ Fα .
By the ∆-system lemma, we may assume wlog that {Fα : α ∈ ω1 } is a ∆-system
with root F .
Since QF is dense in RF , so it is separable. So, there are α 6= β such that
!
!
Y
Y
Oα (i) ∩
Oβ (i) 6= ∅
i∈F
i∈F
It follows that Bα ∩ Bβ 6= ∅.
One of the problems that played a central role in the development of many areas
of combinatorial set theory through the last century was Suslin’s Hypothesis related
to a natural conjecture about the real line. We know that separable spaces are CCC
and that there are CCC spaces that are not separable, but Suslin wondered about
linearly ordered spaces. Recall:
Theorem 29. (Folklore?) The real line is the unique, separable, dense-in-itself
complete linear order without endpoints.
Proof. Sketch of the proof:
The rationals are the unique countable dense-in-itself linear order without endpoints (Cantor).
Any such countable linear order has a unique completion (Dedekind).
Suslin’s Hypothesis: The real line is the unique, CCC, dense-in-itself complete
linear order without endpoints.
As it turns out, SH, is consistent with and independent of ZFC. So it is neither
provable nor refutable.
A counter-example to Suslin’s Hypothesis is called a Suslin Line.
So the existence of a Suslin Line is consistent with the axioms of ZFC.
Before we talk more about Suslin Lines, we need to talk about trees.
Definition 23. A tree is a partial order (T, ≤) such that for every t ∈ T the initial
segment It = {s ∈ T : s ≤ t} is well-ordered.
All trees considered will be rooted, meaning they have a unique lower bound below
all elements of the tree.
The height of t in T is the order type of It .
The α’th level of T is Tα = {t : ht(t) = α}.
The height of the tree is the minimal α such that Tα = ∅.
A branch in a tree is a maximal well-ordered subset that is closed downward.
25
Observations
(1) any ordinal is a tree
(2) If s and t are incomparable, then they have no common upper bound.
(3) A tree in its reverse order: a pairwise incomparable family is an antichain.
(4) every level of a tree forms an antichain.
(5) a node is splitting if it has two immediate incomparable successors.
(6) If a tree has a branch with κ many splitting nodes, then it has an antichain
of size κ.
Examples.
(1) 2<ω
(2) X <κ .
Theorem 30. (König) Any infinite tree with all levels finite has an infinite branch.
However, if you try to generalize this up, you will hit a wall, or rather an Aronszajn tree.
Example 7. (Aronszajn) There is a tree T such that the height of the tree is ω1 ,
levels are countable, but there is no uncountable branch.
Our interest in trees now focus on trees of height ω1 so it might be worth thinking
about some easier examples before considering Aronszajn trees (and later Suslin
trees):
Some trees of height ω1
(1) ω1 with the usual order. Levels have size 1, no node is splitting, and there
is a branch of length ω1
(2) The full binary tree 2<ω1 . Levels have size 2ℵ0 , and there are 2ℵ1 many
branches of length ω1 . Similarly the full ω-splitting tree ω <ω1 .
(3) {f ∈ ω <ω1 : f is 1-1} ⊆ ω <ω1 . This tree has no ω1 branches but it is of
height ω1 . Indeed, the α’th level is exactly {f ∈ ω α : f is 1-1}. Levels are
still large, of size 2ℵ0 .
Two constructions:
The classical construction (Aronszajn). Where T ⊆ {s ∈ Q<ω1 : s is increasing and continuous}
ordered by end extension.
One constructs Tα by recursion letting T<ω = all finite increasing sequences of
rational numbers.
We need to preserve the following inductive hypothesis for all α:
(1) For all n < ω, Tn = {s ∈ Qn : s is increasing}.
(2) For all α ≥ ω Tα ⊆ {s ∈ Qα+1 : s is increasing and continuous}.
(3) If α is a limit ordinal, then given β < α, s ∈ Tβ and a rational q > s(β).
Then for any β < γ < α there is t ∈ Tγ such that s ≤ t and t(γ) = q.
(4) For all α |Tα | = ℵ0 .
Given α such that Tα is defined, let Tα+1 = {s ∈ Qα+2 : s α + 1 ∈ Tα }. The IHs
are easily checked to hold.
Given α limit such that Tβ is defined for all β < α so that the IH hold, proceed
as follows:
For each βS< α, fix βn an increasing sequence cofinal in α such that β = β0 .
For each s ∈ β<α Tβ and each rational q > s(β), use IH(3) to build an increasing
sequence ts,n such that
26
(1) s = tq,s,0 ≤ tq,s,1 ≤ ....
(2) tq,s,n ∈ Tβn
(3) (tq,s,n (βn )) → q.
S
Then, letting ts,q = ( n ts,q,n ) ∪ {(α, q)}, we get that ts,q : α + 1 → Q is order
preserving and continuous.
Now let Tα = {Ts,q : ∃β < α(s ∈ Tβ and q ∈ Q and q > s(β))}.
Then Tα S
is countable and the inductive hypotheses are satisfied.
Let T = α<ω1 Tα
Claim 1. T is special. I.e., can be written as the countable union of antichains.
Proof. Easy peasy: for each q ∈ Q, {s ∈ T : s(ht(s)) = q} is an antichain.
Claim 2. Any special tree has no uncountable branches.
Thus, the tree constructed is Aronszajn.
Modern construction (Todorcevic) Using ρ-functions. ???????????????
Lines and trees and trees and lines
There is an important duality between trees and linear orders.
Indeed, from any linear order, one can construct a binary (or countably branching) tree whose nodes are open convex subsets of the linear order. And conversely,
given a tree, there is a natural ordering on the maximal branches, or the nodes,
that linearly orders the tree. And indeed these operations are kind of like inverses.
E.g., there are properties of a linear order that define a so-called “Aronszajn Line”
and we have that the tree constructed from an Aronszajn line is an Aronszajn tree,
and conversely, the line one constructs from an Aronszajn tree is an Aronszajn line.
Moreover, if one starts with a Suslin Line (a ccc, nonseparable linear order
without endpoints) the tree one obtains is a “Suslin Tree,” i.e., it is a ccc tree (all
antichains countable) of height ω1 with no uncountable branches, hence a very nice
kind of Aronszajn tree. E.g., it is very different than the special Aronszajn tree
just constructed.
Constructing a Suslin tree from a Suslin line. Suppose that (S, ≤) is a Suslin
line: so CCC and nonseparable.
We want to assume that S is dense in itself and complete. To do this we can
actually do a bit more: Assuming the existence of a CCC nonseparable linearly
ordered set, there is another one that is complete and has the property that for
every a < b the interval (a, b) is nonseparable.
Indeed, to make sure every subinterval is nonseparable: define an equivalence
relation on S by declaring a ∼ b if the interval (a, b) is separable (or even empty).
Then there is a natural ordering on the equivalence classes, namely [a] ≤ [b] if a ≤ b.
Note that any equivalence class is convex subset of S, hence CCC and it follows
that any equivalence class [a] is separable.
And this linear order on S/ ∼ is CCC and every interval is nonseparable. Indeed,
if ([a], [b]) were separable, then the interval (a, b) in S would be separable. So
[a] = [b].
CCC follows from the fact that if the intervals ([a], [b]) ∩ ([c], [d]) = ∅, then so
also (a, b) ∩ (c, d) = ∅.
Next, taking the Dedekind completion of S we obtain a complete (still CCC and
still nonseparable) Suslin line.
27
Now we readjust notation and assume our original line S satisfied the conclusion
of the exercise and we can begin the construction of the tree T . We will construct
its levels by recursion so that the following holds, for each α:
(1) Tα is
S a maximal countable disjoint family of nonempty intervals of S.
(2) S \ Tα is separable hence has empty interior (and by maximality of Tα it
must have empty interior).
(3) For all β < α and for all s ∈ Tα , there is a t ∈ Tβ such that s ⊂ t. (So the
ordering on the tree is reverse inclusion), and
(4) For all β < α and for all t ∈ Tβ there is s ∈ Tα such that s ⊂ t.
Let’s check how to carry out the recursion:
Assume that Tα has been defined. Then each element s of Tα is an interval (a, b),
so nonempty and not separable. So find a c ∈ s = (a, b) and let s0 = (a, c) and
s1 = (c, b). Do this for each element of Tα and let Tα+1 = {si : s ∈ Tα , i ∈ 2}.
Check the IH(1)-(4), they all hold for α + 1.
Now assume S
that α is a limit and Tβ has been defined for all β < α. By inductive
hypotheses S \ Tβ is separable for all β < α, so
Aα =
[
S \ Tβ
β<α
is closed and separable so nowhere dense.
S
Let Tα be the family of all maximal open convex subsets of S \ Aα . So, Tα =
S \ Aα . Since S is complete, every element of Tα is an interval and so IH(1) and
(2) hold.
EXERCISE 26. Check that IH (3) and (4) hold.
S
So we need to check that T = {Tα : α < ω1 } ordered by reverse inclusion
(s ≤ t if t ⊆ s) is a Suslin tree. This means we must check that it is CCC and has
no ω1 branches.
Note: since the tree is binary (i.e., every node has two immediate successors),
if there were an ω1 branch then it would not be CCC. So it suffices to show it is
CCC.
To see this suppose that s 6 t in T and suppose s and t are incomparable. If
there is an α such that both s, t ∈ Tα , then, by IH(1) s ∩ t = ∅ (as intervals in S).
Moreover, if s ∈ Tα , t ∈ Tβ and α < β, then there is a t0 ∈ Tα such that t0 ≤ t.
And if s and t are incomparable, then s 6= t0 and so s ∩ t0 = ∅ (as intervals in S).
And since t ⊆ t0 we have s ∩ t = ∅. So any antichain in the tree T corresponds to
a pairwise disjoint family of intervals in S. And since S was CCC, so is T .
Some remarks
• The tree we constructed above is binary (i.e., every node has two immediate
successors). It should be clear that the construction can give rise to Suslin
trees with any number of finite or countable splitting at every node.
• The tree constructed is also normal. Every node can be extended to every
level above its level. and no splitting occurs at limit levels.
Constructing a Suslin line from a Suslin tree.
28
There are a couple of ways to do this, but there are two essential constructions:
Linearly order the nodes of the tree, or linearly order the maximal branches. Let’s
do the latter:
First we assume that we start with a very nice Suslin tree (this is for convenience
and is not technically necessary). However
EXERCISE 27. If there is a Suslin tree, then there is a very nice one: I.e., every
node is ℵ0 -splitting, and every node is extended to every level above its level. And
for every node s at limit height, s is uniquely determined by its set of predecessors.
I.e., if {t ∈ T : t ≤ s0 } = {t ∈ T : t ≤ s1 } and ht(s0 ) = ht(s1 ) is a limit ordinal
then s0 = s1 .
By a maximal branch we mean a well-ordered subset that has no proper extension. So each maximal branch is closed downward and consists of nodes at every
level below some limit ordinal α and has no extension into level α. We will say that
such a branch has height α
Note that since levels are countable and our tree is very nice, we have that for
each limit α there are continuum many maximal branches of height α.
To order the branches, we first fix for each s ∈ T an ordering ≤s of s’s immediate
successors in order type the same as Q.
Given a branch b in the tree of height α, for each β < α, denote by b(β) the
unique element of the branch b in the β’th level of T .
Notice that given two branches b and c, if β is the minimal ordinal such that
b(β) 6= c(β), then β is a successor ordinal and if β = γ + 1 then b(γ) = c(γ).
We now define the ordering on the branches by b ≺ c if b(β) ≤s c(β) where
β = γ + 1 is the minimal ordinal such that b(β) 6= c(β), s = b(γ) = c(γ) and ≤s
is the ordering of s’s successors. Now we need to check that this ordering of the
branches is a Suslin line.
It is a linear order: given two branches, they clearly compare under this ordering.
For transitivity, the only nontrivial thing to check, best to draw a picture and
consider cases, but it is not hard to check this. (EXERCISE?)
Non-separable. Suppose that B is a countable family of maximal branches. Then
there is an α such that all branches in B lie below α. Chose a node s above the
α’th level and pick two maximal branches b0 and b1 such that s ∈ b0 ∩ b1 . Check
that no element of B lies between b0 and b1 . Indeed, if c ∈ B then c ≤ b0 iff c ≤ b1 .
CCC: Suppose that ui = (bi , ci ) for i ∈ I is an uncountable family of open intervals.
We show that at least two of them must intersect. For each i fix si to be the
maximal node common to both bi and ci . Since our Tree is very nice there this
such a maximal node.
Case 1. If there are only countably many such si ’s then there is a fixed s at some
level β such that for uncountably many i s = si . Moreover there are two nodes
s0 ≤s s1 immediate successors of s such that for uncountably many i for which
si = s we have bi (β + 1) = s0 and ci (β + 1) = s1 . Since the ordering is in the type
of the rationals, we can find an immediate successor t of s such that s0 < t < s1 .
Now, any maximal branch d extending t must lie in the uncountably many intervals
(bi , ci ) with s0 ∈ bi and s1 ∈ ci .
Case 2. There are uncountably many si ’s. Then since the tree is Suslin, there
are i and j such that si < sj . Now we can check that uj ⊆ ui .
29
In either case, we have two open sets that intersect and so the line is CCC as
required.
EXERCISE 28. Let (S, ≤) be a binary splitting normal Suslin tree. For each
s ∈ S let s0 and s1 denote its two immediate successors and lexicographically order
the nodes of S as follows: Let r ≺ t if either
(1) r ≤ t in the tree ordering, or
(2) r and t don’t compare, s = ∆(r, t) is the maximal node below both r and t
and s0 ≤ r and s1 ≤ t.
Prove that (S, ≺) is a Suslin line.
EXERCISE 29. Suppose that (T, ≤) is an Aronszajn tree. Lexicographically order
it as in the previous exercise. Prove that the resulting linear order has the following
properties:
(1) Every separable subset of (T, ≺) is countable
(2) (T, ≺) does not contain any (increasing) copy of ω1 or a decreasing copy.
Hint: If {yα : α ∈ ω1 } is such a copy in the line, for each α show that
{x ∈ Tα : x ≤ yβ for uncountably many β ∈ ω1 } has exactly one element
xα and these elements form a branch in the tree T .
Closed unbounded and stationary subsets of ω1 To make sense of the combinatorial axiom that allows the construction of a Suslin tree, need to understand
better the properties of subsets of ω1 .
Definition 24. A ⊆ ω1 is closed if for all limit α ∈ ω1 , if A ∩ α is unbounded in
α then α ∈ A. Any such limit α is called a limit point of A. For any set A ⊆ ω1 ,
A ∪ LIM (A) is closed. If A is closed and also unbounded in ω1 (i.e., uncountable)
then we say that A is CLUB.
The next theorem states that the CLUB subsets of ω1 form a σ-filter:
T
Theorem 31. For any countable family C of CLUB subsets of ω1 , C is club.
T
Proof. Let D = C. To see that D is closed is easy: If D is unbounded i α, then
for all C ∈ C, we have that C is unbounded in α and so α ∈ C. Thus α ∈ C for all
C ∈ C. So α ∈ D.
To see that D is unbounded, first note that the intersection of any 2 CLUBs is
CLUB (hence the intersection of any finite family of CLUBs).
Next, fix β and we need Tto find α above β with α ∈ D. Enumerate C as
{Cn : n ∈ ω}. And let Dn = {Ck : k ≤ n}. Then
D0 ⊇ D1 ⊇ D2 ⊇ ...
T
T
and D = C = {Dn : n ∈ ω}.
Now using that each Dn is unbounded, choose, recursively, αn so that β < α0 <
α1 < .... and so that αn ∈ Dn for all n.
Then letting α = sup{αn : n ∈ ω}. We can see that for all n {αk : k > n} ⊆
Dn ⊆ Cn and so Cn is unbounded in α and so α ∈ Cn for all n. And so α ∈ D as
required.
EXERCISE 30. Write out the simpler proof that the intersection of two clubs is
club
30
In the next exercise, let us understand that a f : ω1 → ω1 is strictly monotone
if f (α) < f (β) for all α < β in ω1 . And f : ω1 → ω1 is continuous if for all A ⊆ ω1
countable, if α = sup A then f (α) = sup f 00A.
EXERCISE 31. The set of fixed points of strictly monotone continuous function
f : ω1 → ω1 is a CLUB. Conclude that the set of α such that ωα = α is CLUB.
Also the set of α such that α = ω α is club.
Definition 25. A subset S ⊆ ω1 is stationary, if S ∩ C 6= ∅ for every CLUB
C ⊆ ω1 . S is non-stationary if its complement contains a CLUB.
Note that in this terminology, the family of non-stationary subsets of ω1 forms
a σ-ideal.
Here is the general situation: For an ideal I (measure zero, meagre, nonstationary). I ∗ denotes the family of sets whose complements are in the ideal.
For N ∗ we have the sets of full measure For M∗ we have the co-meagre sets, those
that contain a dense Gδ . And for N S ∗ we have the CLUB subsets of ω1 .
And I + is the family of sets of not in the ideal. These are, in the sense of the
ideal at hand, a notion of largeness. E.g., mathcalN + consists of the sets of positive
measure and the nonmeasurable sets, M+ is the family of non-meagre sets, and
N S + is the family of stationary subsets of ω1 . These families are characterized by
saying: A ∈ I + if and only if A ∩ C 6= ∅ for all C ∈ I ∗ . Even though this is a
measure of largeness, it is not unusual to be able to find uncountable collections of
pairwise disjoint sets from I + .
And if I is a σ-ideal, if S
we take any element A ∈ I + and partition it into
countably many pieces, A = n An , then at least one of the An ∈ I + .
Theorem 32. There is a pairwise disjoint, uncountable family of stationary subsets
of ω1 .
Proof. This is result is obtained from the following often exploited fact (analogies
hold for any successor cardinal): every infinite α ∈ ω1 is countable, so there is a
bijection fα : α → ω. Let Xnβ = {α : fα (β) = n} and note the following properties:
S
(1) For every β {Xnβ : n ∈ ω} = ω1 \ β + 1.
(2) For every n {Xnβ : β ∈ ω1 } is pairwise disjoint.
So, for each β there is an nβ such that Xnββ is stationary. Now, there must be a
single n such that for uncountably many β, nβ = n.
Then {Xnβ : nβ = n} is a pairwise disjoint family of stationary sets.
Next we prove Fodor’s Lemma, aka the “Pressing Down Lemma”:
Theorem 33. Suppose that S is stationary and that f : S → ω1 has the property
that f (α) < α for all α (i.e., f is “pressing down”). Then f is constant on a
stationary set.
Proof. First let me convince you that f must be constant on some uncountable set,
i.e., f −1 (β) must be uncountable for some β.
OK, Now suppose that for all β, there is a CLUB Cβ such that Cβ ∩ f −1 (β) = ∅.
We will get a contradiction by finding a γ ∈ S with the property that γ ∈ Cβ for
all β < γ. Indeed, this would be a contradiction since if γ ∈ Cβ then f (γ) 6= β, so
then f (γ) 6= β for all β < γ contradicting that f is “pressing down.”
So consider D = {γ : γ ∈ Cβ for all β < γ}. We will be done if we can show
that D ∩ S 6= ∅.
31
We first note that D is not empty. Indeed, fix γ0 ∈ ω1 . Then
\
Cβ
D0 =
β<γ0
is club, so there is γ1 ∈ D0 above γ0 . Having defined γ0 < ... < γn , let
\
Cβ
Dn =
β<γn
is club, so there is γn+1 ∈ Dn above γn .
Let γ = sup{γk : k ∈ ω}. And fix β < γ. Then there is n such that β < γn and
so {γk : k ≥ n} ⊆ Cβ . And so γ ∈ Cβ . And so γ ∈ D.
Note that since γ0 was chosen freely, this shows that D is unbounded.
Also, D is closed!! So CLUB. And so D ∩ S 6= ∅ as required to complete the
proof.
Now we are ready to state the combinatorial principle ♦:
The Diamond Principle (♦). There is a sequence {Dα : α ∈ LIM (ω1 )} such
that
(1) Dα ⊆ α for all α ∈ LIM (ω1 )
(2) For all A ⊆ ω1 ,
{α : A ∩ α = Dα } is stationary.
Observations
(1) ♦ implies CH.
(2) The version of ♦ where you ask that a set is captured club often is false. For
example, observe that if A 6= B then {α : A ∩ α = Dα } ∩ {α : B ∩ α = Dα }
is at most countable.
(3) ♦ follows from Gödel’s axiom of constructibility V = L, though ♦ was
formulated and proven to follow from V = L by Jensen in the 1970’s.
Most important application:
Theorem 34. ♦ implies the existence of a Suslin tree.
Proof. The tree order will be constructed on ω1 . To start, let ≤0 be an ordering on
ω · ω that is isomorphic to the complete ω-splitting tree ω <ω . As we construct the
tree, let Tα denote the α’th level of the tree and T<α denote the union of all levels
Tβ for β < α. So we have fixed T<ω in the first step.
Notationally, for α ≥ 0 we define the α’th level so that for all α ≥ ω, T<α = ω · α
and the α’th level Tα = [ωα, ω(α + 1)). Notice that for club many α we have
α = α · ω, and so T<α = α. Call that club C.
So, of course, this just arranges the levels but we still need to define the tree
ordering so that the levels do indeed line up like this. And we need some inductive
hypotheses to make sure we get a Suslin tree in the end. As mentioned above, if
we can make the tree CCC then we know it has no ω1 branches.
Suppose that α is given and that for all β < α we have constructed T<α so that
(1) For all β < γ < α for all s ∈ Tβ there are infinitely many t ∈ Tγ such that
s < t.
32
We need to explain how to extend the ordering on T<α to the α’th level Tα for each
α > ω.
If α = γ + 1 then partition [α · ω, α · ω + ω) into countably many disjoint sets Sn .
Define Sn to be the set of immediate successors of ω · γ + n ∈ Tγ = [ωγ, ωγ + ω).
This defines the tree order up to and including level Tα
If α is a limit, and for each s ∈ T<α we will fix a branch bs in T<α such that
s ∈ bs and bs ∩ Tγ 6= ∅ for all γ < α for all s. Also, we can choose the branches bs
so that for distinct s 6= t we have bs 6= bt . So if we enumerate T<α as {sn :∈ ω},
and we recall that Tα = [ωα, ωα + ω) we can put ωα + n as the upper limit of bsn
thus defining the tree up to level Tα .
Of course we haven’t done anything with the ♦ sequence so there is no guarantee
that the ω1 tree constructed has no ω1 branches or is not CCC.
So, fix a ♦-sequence {Aα : α < ω1 }.
At stage α, we proceed with the construction as above, unless we have the
following two hypotheses:
(1) α = α · ω, and
(2) Aα is a maximal antichain in T<α
In which case we proceed exactly as above making sure that
(∗) for each s ∈ T<α the branch bs is chosen so that bs ∩ Aα 6= ∅.
Let T be the resulting tree.
Claim 3. T is CCC.
Proof. Suppose that A is an antichain in T . By extending A we may assume it is
maximal. By the statement ♦ there is α such that T<α = α and A ∩ α = A ∩ T<α =
Aα is a maximal antichain in T<α . We now claim that Aα is maximal in all of T .
Indeed, for any t ∈ T , if t ∈ T<α then since Aα is maximal in T<α there is a ∈ Aα
that compares to t. And if t ∈ T lies in a level ≥ α, fix t0 ∈ Tα such that t0 ≤ t.
By construction, t0 is the supremum of a branch bs for some s ∈ T<α and by (∗) we
have that bs ∩ Aα 6= ∅. Fixing a ∈ Aα ∩ bs we have that a ≤ t0 ≤ t as required. Martin’s Axiom Martin’s Axiom is an axiom that arose as a unifying statement
of many consistency proofs
Definition 26. Let P, ≤ be a partial order. For p, q ∈ P we say p and q and
compatible, if there is r ∈ P such that r ≤ p and r ≤ q. If p and q are not
compatible, we say they are incompatible and write p ⊥ q. A ⊆ P is an antichain if
it is pairwise incompatible.
Definition 27. A partial order is CCC if all antichains are countable.
EXAMPLES:
(1) If T is a tree with ordering ≺ then the partial order obtained by reversing
the order of T satisfies the property any two incomparable elements are
incompatible. So a Suslin tree under its reverse ordering is a ccc partial
order.
33
(2) Set containment ⊆ on various families of sets give important exmaples of
partial orders. E.g.,
(a) The set of all nonempty subsets of ω under containment. Two sets
are incompatible if and only if they are disjoint. This is a ccc partial
order.
(b) The set of all infinite subsets of ω under containment. Two sets are
incompatible if and only if their intersection is finite. In this case we
say that they are almost disjoint. This partial order is not ccc follows
from
Theorem 35. There is an uncountable pairwise almost disjoint family
of infinite subsets of ω.
Namely: Consider Q instead of ω and fix, for each real number r Ar to
be the range of a nontrivial sequence converging to r. Then Ar ∩ Ax
is finite for all r 6= x.
(c) The set of open subsets of [0, 1]. This is ccc since the rationals are
dense.
(3) F n(I, J) = {s : dom(s) ∈ [I]< ℵ0 , s : dom(s) → J}. The set of finite partial
functions from I to J.
Theorem 36. F n(I, J) is CCC if and only if J is countable.
Proof. If J is uncountable, fix i ∈ I and consider for each j ∈ J, sj : {i} →
J defined by sj (i) = j. Then the collection of all these s0j s is an uncountable
antichain.
If J is countable, the proof is based on the ∆-system lemma (see above).
Let {pα : α ∈ ω1 } ⊆ F n(I, J). By the ∆-system lemma, we may assume
{dom(pα ) : α ∈ ω1 } forms a ∆-system with root d. Since J is countable,
there are only countably many functions into J with domain d. So there is
a single function q : d → J such that {α : pα d = q} is uncountable. Any
pα and pβ that agree on d are compatible.
(4) Consider the set of measurable subsets of R. Define ≤∗ by A ≤∗ B if A \ B
is measure zero. This is not a partial order since there are A 6= B such
that A ≤∗ B and B ≤∗ A (e.g. if both have measure zero). But under
the natural equivalence relation where A ∼ B if A ≤∗ B and B ≤∗ A, ≤∗
induces a partial order on the equivalence classes. The resulting partial
order is CCC. (exercise).
(5) Similarly we may define ⊆∗ on [ω]ℵ0 . This is not a partial order but induces
an equivalence relation in the same way, and the resulting order on the
equivalence classes is not CCC (because of a large almost disjoint family).
Definition 28. A subset D ⊆ P of a partial order is said to be dense if ∀p ∈ P
there is a q ∈ D such that q ≤ p.
Examples:
(1) Let T be an ω1 tree and assume T has the property that for all β < γ < ω1
and for all t ∈ Levβ (T ) there is s ∈ Levγ (T ) such that t ≤ s (we built our
trees to satisfy this). Then consider the partial ordering of T under the
reverse ordering. For each α the following sets are dense: Dα = {t ∈ T :
ht(t) ≥ α}.
34
(2) Consider the partial order O of open subsets of R under containment. For
each > 0, the set D = {U : U ⊆ I for some interval I of length < } is
dense.
Also, for any nowhere dense set K, DK = {U : U ∩ K = ∅} is dense.
(3) In the partial order of measurable sets under inclusion (well, the partial
order of equivalence classes of measurable sets as described above), the set
{[K] : K ⊆ R is closed and µ(K) > 0} is dense.
(4) For i ∈ I, Di = {s ∈ F n(I, J) : i ∈ dom(s)} is dense and for j ∈ J,
Ej = {s ∈ F n(I, J) : i ∈ ran(s)} is dense.
(5) For any f : ω → ω, Df = {s ∈ F n(ω, ω) : ∃n ∈ dom(s) : s(n) > f (n)} is
dense.
The final notion we need to formulate Martin’s Axiom is that of a filter in a
partial order. Recall that a filter on a set X is a family of subsets of X that is
closed under intersections and supersets and does not contain the emptyset. If
one considers the partial order on (P (X), ⊆) the following definition of filter is
equivalent:
Definition 29. For a partial order P a subset G is called a filter if
(1) For p, q ∈ G there is a r ∈ G such that r ≤ p and r ≤ q.
(2) For all p ∈ G and all q ∈ P if p ≤ q then q ∈ G.
Returning to examples,
(1) If G is a filter in an ω1 tree in the reverse order, then G is a chain that is
closed downward in the tree. If G ∩ Dα 6 for all α then G is an ω1 branch.
T
(2) If G is a filter in the partial order of all open subsets of [0, 1] then {u :
u ∈ G} =
6 ∅. And if G ∩ D1/n 6= ∅ then this intersection consists of exactly
one point.
S
(3) If G is a filter in F n(I, J) then g = G is a function. If G ∩ Di 6= ∅, then
i ∈ dom(g) and if G ∩ Ej not empty then j ∈ ran(g).
Now we can state Martin’s Axiom (MA) and indeed much of the work in showing
some applications has been done already:
MA(κ) For any CCC partial order P and for any family D of κ many dense subsets
of P, there is a filter G ⊆ P that meets every dense set D ∈ D.
And Martin’s Axiom (MA), is the statement that M A(κ) holds for all κ < 2ℵ0 .
Proposition 10. We have the following (ZFC) results:
(1) M A(ω) is true and so CH implies MA. However, slightly more generally
this does not require the assumption of CCC in the statement of MA:
Lemma 2. Rasiowa-Sikorski Lemma: For any partial order P (ccc or not)
and for any countable family D of dense subsets, there is a filter G ⊆ P
meeting all the dense sets in D.
(2) M A(2ℵ0 ) is false.
(3) If λ < κ then M A(κ) implies M A(λ).
Now we are ready to prove the first application:
Theorem 37. M A(ω1 ) implies there are no Suslin trees
35
Proof. Suppose that T is a Suslin tree. By thinning it out as we did when constructing a Suslin line, we may assume that
(∗) ∀s ∈ T ∀α > ht(s) ∃t ∈ Levα (T )(s ≤ t)
Now we use the tree under its reverse ordering and apply MA to it. By the
assumption (∗) we know that the sets
Dα = {t ∈ T : ht(t) > α}
is dense for every α ∈ ω1 . So, by M A(ω1 ) we have a filter F ⊆ T meeting each of the
Dα . We already checked that any such filter is actually an ω1 -branch. Contradicting
that T is Suslin.
More can be said here related to a natural question: Given two partial orders
P0 and P1 , The product P0 × P1 have a natural coordinatewise partial ordering. Is
the property of being CCC preserved under such products?
Theorem 38. If T is a Suslin tree then T × T is not CCC
Proof. For each t ∈ T , let t0 and t1 be two distinct immediate successors. Then
{(t0 , t1 ) : t ∈ T } is an antichain.
Theorem 39. M A(ω1 ) implies that the product of any two CCC partial orders is
CCC.
Proof. We do this in two steps. First lemma:
Lemma 3. If P is CCC and if A ⊆ P is of size ω1 then there is a P 0 ⊆ P of size
ω1 , containing A which is also CCC.
Proof. Define Pn for n ∈ ω1 as follows: P0 = A. Having Pn , for all pairs p, p0 ∈ Pn
if p, p0 are compatible, fix rp,p0 a lower bound witnessing this and let
Pn+1 = {rp,p0 : p, p0 ∈ Pn compatible}
Then P 0 =
S
n
Pn satisfies the conclusion of the lemma.
Suppose that P and Q are CCC and A ⊆ P × Q is uncountable. By going to a
subset, assume |A| = ω1 . Then by the lemma there are P 0 ⊆ P and Q0 ⊆ Q such
that
(1) |P 0 | = |Q0 | = ω1 , and
(2) A ⊆ P 0 × Q0 , and
(3) P 0 and Q0 are CCC.
So if we can show the product of two CCC posets of size ω1 are CCC then we would
have the product of any two CCC posets is CCC.
We now define a poset P to have property-K if every uncountable subset has an
uncountable centred subset. where centred = every finite subset has a lower bound.
Lemma 4. The product of a CCC partial order with a Property K partial order is
CCC
Proof. Suppose P is CCC and Q is Property K. Suppose that A ⊆ P × Q is
uncountable. If the projection of A onto the Q coordinate were countable, then we
can easily find two compatible elements of A, Otherwise the project is uncountable,
and by property K we can go to an uncountable centred subset of that. So, there is
A0 ⊆ A which is uncountable and with the property that for any (a, b), (c, d) ∈ A0
36
we have that b and d are compatible. Now, if there is a single p ∈ P such that for
some q 6= q 0 both (p, q) and (p, q 0 ) ∈ A0 then we have two compatible elements of
A0 . Otherwise {p : ∃q(p, q) ∈ A} is uncountable. So by CCC of P there are two
compatible elements p, p0 in this set, so if q, q 0 are such that (p, q), (p0 , q 0 ) ∈ A we
have two compatible elements of A, thus P × Q is CCC.
And the proof of the theorem will be completed when we prove that
Proposition 11. M A(ω1 ) implies that every CCC partial order has Property K
Proof. Let P have size ω1 and suppose it is CCC. Let {pα : α < ω1 } be an uncountable subset. We first note that there is an α, such that for all β > α
{γ > β : ∀q ≤ pβ q is compatible with pγ }
is uncountable. Otherwise we can recursively define an uncountable antichain in
A as follows. If there is no such α, let β0 be given and fix qβ0 < pβ0 such that
{γ > β : qβ is compatible with pγ } is countable and fix α0 to be a bound for this
set. So qβ0 ⊥ pβ for all β > α0 . Having defined βξ , αξ and qβξ for all ξ < η such
that
(1) For all ξ < ν < η βξ < αξ < βν
(2) qβξ ≤ pβξ
(3) qβξ ⊥ pβ for all β > αξ
Then, by our assumption, we may choose βη above the supremum of the βξ for
ξ < η, and we may fix qβη < pβη such that {γ > βη : qβη is compatible with pγ } is
countable and fix αη to be a bound for this set. The resulting sequence {qβη : η ∈
ω1 } is an antichain. (EXERCISE: Check this)
So we may fix α as above and we let
Q = {p ∈ P : p ≤ pβ for some β > α}
Then Q is CCC and
Dγ = {p ∈ P : p ≤ pβ for some β > γ}
is dense in Q for all γ > α. (EXERCISE: Check this)
And so a filter meeting all the Dγ gives a centred family G with the property
that G ∩ {pα : α < ω1 } is uncountable.
Once we know that all CCC posets have property K assuming M A(ω1 ) we can
prove in fact that
Theorem 40. M A(κ) implies that all CCC posets of size κ are σ-centred.
Proof. Let Q be CCC of size κ. Let
P = {s ∈ F n(Q, ω) : ∀n ∈ ran(s), s−1 (n) is centred}
By what we saw in the first proof, we have two things to do: prove that P is
CCC and delineate a family of dense subsets that assure that we can define from the
filter the family of subsets of Q showing it is a countable union of centred subsets.
P is CCC: Fix {sα : α ∈ ω1 } ⊆ P . By going to a subset we may assume that the
domains form a ∆-system with root d and that sα d = sβ d for all α, β ∈ ω1 .
And since each sα has finite range we may assume that for all α the range of sα is
a fixed finite set F . So clearly sα ∪ sβ is a function for all α and β. so we need to
37
−1
make sure that for some α 6= β and for all n ∈ F we have that s−1
α (n) ∪ sβ (n) is
centred. Then sα ∪ sβ would be an element of P and a lower bound for both sα
and sβ .
To this end, for each α and each n ∈ F let qα,n be a lower bound for s−1
α (n) and
let
q α = (qα,n : n ∈ F ) ∈ QF
By our assumption of M A, we know that Q has property K and so QF is CCC
and so there are α 6= β such that q α and q β are compatible. Hence for all n ∈ F
qα,n and qβ,n are compatible. And any lower bound of qα,n and qβ,n is a lower
−1
−1
−1
bound for s−1
α (n) ∪ sβ (n) and so sα (n) ∪ sβ (n) is centred. I.e., sα and sβ are
compatible.
One group of applications of MA can be described as generalizing statements
about ”countable” to ”cardinality ¡c”
For example: MA implies that any set of size < c is meagre, and any set of size
< c is null. And R is not the union of < c many null or meagre sets. etc...
By our analysis of the Cichon diagram, all these results follow from:
Theorem 41. MAσ-centred (κ) implies that p > κ. Hence MA implies p = c.
Before we prove that result, let’s look at some slightly easier MA results to apply.
Theorem 42. MAcountable implies that R is not the union of fewer than c many
nowhere dense sets. Hence COV (M) = c.
Proof. Let’s use our partial order of basic open subsets of R ordered by inclusion.
This is a countable partial order so is clearly CCC. Fix {Kα : α < κ} a family of
nowhere dense subsets of R where κ < c. Now Dα = {U : U is basic T
open and U ∩
Kα = ∅} is dense. As we saw earlier, if G is a filter in B then X = {U : U ∈ G}
is not empty, and if G ∩ Dα for all α, then X ∩ Kα = ∅ for all α.
Now let’s prove that p = c assuming MA.
Proof. Let F be a family of infinite subsets of ω with the sfip with |F | = κ < c. We
need to find a pseudo intersection. So let
P = {(s, A) : s ∈ [ω]<ω , A ∈ [F ]<ω }
where P is ordered by (s, A) ≤ (t, B) if s ⊇ t A ⊇ B and
s\t⊆B
So the first coordinate should be thought of as an approximation to the infinite
pseudo-intersection, and the second coordinate shoudl be thought of as a “promise”
in the sense that any extension of (t, B) promises that any new elements added to
the approximation must lie in all of the sets listed in B.
Our last application of M A:
Theorem 43. MAω1 implies that all Aronszajn trees are special, i.e., can be written
as a countable union of antichains.
38
Proof. Suppose that T is an ω1 -tree with no uncountable branches (i.e., an Aronszajn tree). We employ the natural poset of finite approximations to a specialization, namely a function f : T → ω such that f −1 (n) is an antichain for all n,
namely we let
P = {p ∈ F n(T, ω) : for all n p−1 (n) is an antichain}
So a finite partial function p ∈ F n(T, ω) is in P if and only for all t, s ∈ dom(p) if
s 6⊥ t then p(s) 6= p(t).
We need to show that P is CCC. To this end let {pα : α ∈ ω1 } be an uncountable
subset. By going to an uncountable subset A ⊆ ω1 we may assume that {dom(pα ) :
α ∈ A} is a ∆-system with root d, there is a fixed p : d → ω such that pα d = p
for all α ∈ A.
Finally, letting qα = pα dom(pα ) \ d, it suffices to check that there are α 6= β
such that qα 6⊥ qβ . By thinning out some more we may assume that for a fixed
finite subset r ⊆ ω, we have that ran(qα ) = r for all α ∈ A
Now, to show that qα 6⊥ qβ we need to show that qα−1 (k) ∪ qβ−1 (k) is an antichain
for all k ∈ r. To prove this we first prove:
Lemma 5. The partial order of all finite antichains of T ordered by a ≤ b if a ⊇ b
has the CCC.
To see that this suffices, note that under M A(ω1 ) we have that this partial order
would also have property K. Therefore, for any uncountable sequence {aα : α ∈ ω1 }
of finite antichains, there would be an uncountable B such that for all α, β ∈ B we
would have aα ∪ aβ is an antichain. Therefore if r = {ki : i < n}, we define Ai by
recursion on i < n by letting Ai ⊆ Ai−1 uncountable so that
qα−1 (ki ) ∪ qβ−1 (ki ) is an antichain ∀α, β ∈ Ai
Then for any α, β ∈ An we have that qα 6⊥ qβ as required.
so it suffices to prove the lemma.
Proof. By thinning out {aα : α < ω1 } we may assume that |aα | = n for all α. For
n = 1 the result follows since T has no uncountable branches.
In general, let aα = {tα (i) : i < n} and assume, by thinning out if necessary
that all elements of aα lie at a height below all elements of aβ whenever α < β.
Fix an ultrafilter F on ω1 such that |X| = ω1 for all X ∈ F . For α < β, since
aα ∪ aβ is not an antichain, we may fix iα,β and kα,β such that tα (i) ≤ tβ (k).
For each α, there are iα and kα such that Xα = {β : iα,β = iα and kα,β = kα } ∈
F.
We now have an i and k such that A = {α : iα = i and kα = k} is uncountable.
We now claim that {tα (i) : α ∈ A} is a chain, contradicting that T is Aronszajn.
To see this, fix α < α0 both in A. And fix β ∈ Xα ∩ Xα0 . So, both tα (i) and tα0 (i)
lie below tβ (k) and so they must be comparable.