Set Theory-an Introduction
1. Intro
● A set theoretist is a mathematician who admits not to
know what the real numbers are.
● Some commonly used axioms/theorems/statements/logical reasoning is dangerous. One has to know the danger and should
not ignore it.
● Naive set theory (Cantor: everything with a property is a set:
any definable collection is a set) and Russel’s paradox: The
set of all sets ⊃ the set of all sets not containing itself as an
element. Let us call a set ”abnormal” if it is a member of
itself, and ”normal” otherwise. For example, take the set of all
squares in the plane. That set is not itself a square in the plane,
and therefore is not a member of the set of all squares in the
plane. So it is ”normal”. On the other hand, if we take the
complementary set that contains all non-(squares in the plane),
that set is itself not a square in the plane and so should be
one of its own members as it is a non-(square in the plane). It
is ”abnormal”. Now we consider the set of all normal sets, R.
Determining whether R is normal or abnormal is impossible: if
R were a normal set, it would be contained in the set of normal
sets (itself), and therefore be abnormal; and if R were abnormal,
it would not be contained in the set of all normal sets (itself),
and therefore be normal. This leads to the conclusion that R is
neither normal nor abnormal: Russell’s paradox.
● There is a model of R such that there is an -δ discontinuous
function that is sequentially continuous. From ZF it cannot
be proven that a sequentially contiuous function on a compact
interval attains its maximum.
2. What is a set?
● Main obstacle: A set cannot contain itself as an element.
● solutions: certain axiomatics: Sierpinski: a set is of higher
hierarchy than its elements
● More common: Zermelo-Fraenkel axioms
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A set contains elements (objects): a ∈ A
Two sets are equal iff they have the same elements.
No set is its own element.
Given a condition we do not know beforehand that there is an
object fulfilling it. So it is convenient to define ∅ (empty set)
as the set that does not contain any element.
Sets might be elements of other sets.
Subsets are sets (collection of some elements in the set). The
empty set is a subset of all sets.
Unions of sets are sets: A a set, for each α ∈ A there is a set
Bα then the union ⋃α Bα is defined as the set containing all the
elements of at least one of Bα .
Complements are sets B ∖ A = C is a set.
Intersection of sets: A a set, for each α ∈ A there is a set Bα
then the intersection ⋂α Bα is defined as the set containing all
the elements that are elements of all Bα .
Complements are sets B ∖ A = C is a set.
Cartesian product of sets: A × B is the set of ordered pairs
(x, y), x ∈ A, y ∈ B.
Exponents of sets: AB is the set of all functions from B into
A. Example: AN is the set of all sequences in A.
3. Consistency of ZF
By Gödel’s theorem ZF’s consistency cannot be proven within ZF.
For this one needs the existence of ”large” cardinals (another axiom in
set theory).
However, redundancy is known.
4. Classes
● A class is a collection of sets (or sometimes other mathematical
objects) that can be unambiguously defined by a property that
all its members share.
● A class that is not a set (informally in Zermelo–Fraenkel) is
called a proper class.
● Examples:
– The class of all sets
– The class of all one-element sets
– The class of all groups, rings, fields, vector spaces, etc.
● One way to prove that a class is proper is to place it in bijection
with the class of all ordinal numbers.
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5. Equivalence relations
Let M be a set.
Definition 5.1. A relation ”∼” on M × M is an equivalence relation if
(1) Reflexivity: a ∼ a for all a ∈ M .
(2) Symmetry: a ∼ b,
Ô⇒
b ∼ a.
(3) Transitivity: a ∼ b and b ∼ c
Ô⇒
a ∼ c.
Definition 5.2. Partition into classes:
M = ⋃ Mα ,
α≠β
Ô⇒ Mα ∩ Mβ = ∅.
α
Mα – classes.
Lemma 5.1. Any equivalence relations gives a partition into classes
and vice versa.
Proof.
( Ô⇒ ) Ma = {x ∈ M ∶ x ∼ a}, Ma = Mb = Mα
(reverse) a ∼ b
⇐⇒
⇐⇒
a ∼ b.
a ∈ Mb .
Two classes either coincide or are disjoint!
6. Equivalence of sets
Definition 6.1. A ∼ B iff there is a bijection f ∶ A → B.
Remark 6.1. Two finite sets are equivalent iff they have the same
number of elements (generalization!)
Example 6.1. NN ∼ [0, 1] ∖ Q via continued fraction expansion. Note
that NN is the set of all integer sequences.
Definition 6.2. A set is of infinite power (infinite) if it is not equivalent to any finite set.
Definition 6.3. A set is countable iff it is equivalent to N. If a set
is neither finite nor countable it is called uncountable.
Example 6.2.
● Z is countable.
● The even numbers are countable.
● Q is countable.
● Any subset B of a countable set A is countable or finite. (enumerate A ∶ a1 , a2 , a3 , ⋯ and let B ∶ an1 , an−2 , an3 , ⋯ either the
enumeration of B is finite or a correspondence to ω)
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● Any countable union of countable sets with specified bijections to N is countable (write a table for each An ∶ ⋯, an,i , ⋯
and mimic the enumeration of Q.
Definition 6.4. A set is called Dedekind finite if it does not contain
a countable set as a subset. Otherwise it is called Dedekind infinite.
Lemma 6.1 (ωAC equivalent). Any infinite set is Dedekind infinite.
Proof.
● Choose arbitrary a1 ∈ M (OK!).
● Choose a2 ∈ M ∖ a1 (OK!).
● Continue⋯ since there are always elements left because M is
infinite (OK???????).
We run into the same problems as with sequential continuity. There
is a model of R where R contains infinite sets that are Dedekind finite! This gives also an example of a discontinuous function that is
sequentially continuous!
Theorem 6.1. I ∶= [0, 1] ∩ R is uncountable (in ZF without AC!).
Proof.
● Any x has a unique binary expansion with infinitely many 0’s.
So I ∼ B ⊂ {0, 1}N . (The latter is the set of all 0–1 sequences).
● {0, 1}N ∼ C1/3 ⊂ I (use base 3 expansion).
● This shows by the Cantor-Bernstein Theorem that I ∼
{0, 1}N .
● Use Cantor’s diagonal argument to show that N ≁ {0, 1}N : Consider a table [ak,l ]∞
k,l=1 with countably many lines indexed by k
and in each line countably many entries indexed by l and each
entry either 0 or 1. This table reflects a numeration (bijection
with N) of 0 − 1-sequences. Let 0∗ = 1 and 1∗ = 0. Then the sequence (a∗n,n )∞
n=1 is not in the table and hence there is no desired
bijection of N ⇄ {0, 1}N .
7. The Cantor–Bernstein Theorem
Theorem 7.1 (Cantor–Bernstein). Let A ∼ B1 , B ∼ A1 , A1 ⊂ A and
B1 ⊂ B. Then A ∼ B.
Proof. Let f ∶ A → B1 and g∶ B → A1 be the (any) corresponding bijective maps. Consider chains a ↦ b iff a ∈ A, b ∈ B1 ⊂ B and f (a) = b.
Similarly, b ↦ a iff b ∈ B, a ∈ A1 ⊂ A and g(b) = a.
● Each chain is ”infinite to the right”
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● Each element of A, respectively B is contained in exactly one
chain.
● There are (disjoint) 3 possibilities: a chain C is ”infinite in both
directions” (type 1), the ”least element” belongs to A (type 2)
or the ”least element” belongs to B (type 3).
● f maps the elements of A that belong to chains of type 1 or 2 in
a 1-to-1 way into B 1,2 ⊂ B. The remaining elements of A are in
chains of type 3 and g −1 maps A ∖ { chains of type 1 or type 2}
in a 1-to-1 fashion onto B ∖ B 1,2 .
8. The power of a set, Cardinals
Definition 8.1. A cardinal number is an equivalence class of (nonempty) sets.
Definition 8.2. The power or cardinality of a set A is the corresponding cardinal number denoted by m(A).
Remark 8.1. Some facts.
● The cardinality of a finite set is the natural number equal to the
number of elements contained in the set.
● The set N of all natural numbers has the cardinality denoted by
ℵ0 .
● The set R of all real numbers has the cardinality denoted by c.
● There are 4 possibilities:
– A ∼ B1 ⊂ B and B ∼ A1 ⊂ A
– A ∼ B1 ⊂ B but ∀A1 ⊂ B
A1 ≁ B
– B ∼ A1 ⊂ A but ∀B1 ⊂ A
B1 ≁ A
– ∀A1 ⊂ A
A1 ≁ B and ∀B1 ⊂ B
B1 ≁ A.
In the first case (Cantor-Bernstein) m(A) = m(B), in the
second we write m(A) < m(B), in the third m(A) > m(B).
The most interesting case is the fourth. Without any further axioms it can happen or not. The assumption that the
fourth case does not happen, i.e. we can compare any two
sets (Trichotomy) is equivalent to the Axiom of Choice
that we will study later.
For a set M we write P(M ) for its power set, i.e. the set of all
(including the empty set!) subsets of M . We note that for a finite set
M with n elements its power set has exactly 2n elements.
The next theorem shows that there is no ”largest set”.
Theorem 8.1 (Cantor). m(M ) < m(P(M )).
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Proof.
● M ∋ x ⇄ {x} ∈ P(M ) is a bijection onto its image. So M and
P(M ) are compareble.
● Assume there is a bijection x ⇄ f (x) = Mx ∈ P(M ).
● Consider the set X ∶= {x ∈ M ∶ x ∉ Mx } ⊂ M , i.e. X ∈ P(M ).
● Like in Russel’s paradox X ≠ f (y) for all y ∈ M . If it were than
y cannot be in X nor in its complement!
Remark 8.2. We summarize:
● There is no ”largest” cardinal.
● The collection of all cardinal numbers is a proper class!
● Notation: For a finite set A with n elements we write m(A) =
n.
● Notation: m(P(M )) = 2m(M ) in analogy of finite sets.
● 2ℵ0 = c, where c = m(R) - the power of the continuum.
● m({0, 1}N ) = c.
● m(NN ) = c. For both last statements consider the equivalence to
the real numbers.
● The continuum hypothesis states that there is no cardinal
number m such that ℵ0 < m < c.
● One can prove that m(Borel sets on R) = c All subsets of C1/3
have outer Lebesgue measure 0 and hence are Lebesgue measurable. So
m(Lebesgue measurable sets in R) > c = m(Borel sets in R).
9. Cardinal arithmetics
Let n, k, l be cardinals. By the very definition we can choose sets
B, C, D, pairwise dijoint (no two of them have common elements), such
that
n = m(B), k = m(C), l = m(D).
We define
● l + n = m(D ∪ B),
● l ⋅ n = m(D × B),
● ln = m(DB ).
This definitions are well-stated since for equivalent sets B ′ , C ′ , D′ one
has m(D ∪ B) = m(D′ ∪ B ′ ), m(D × B) = m(D′ × B ′ ) and m(DB ) =
′
m(D′B ), since the ′ -sets are coming from ”renameing” the elements via
the corresponding bijections. Only for addition we need the assumption
that the involved sets are disjoint!
We have the following rules:
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● n + l = l + m since B ∪ D = D ∪ B.
● n ⋅ l = l ⋅ n since B × D ∼ D × B.
● n ⋅ (k + l) = n ⋅ k + n ⋅ l since B × (C ∪ D) ∼ (B × C) ∪ (B × D).
Lemma 9.1. The notation m(P(A)) = 2m(A) is justified by the computation rules. Moreover,
m(A)
m(P(P(A))) = 22
m(P(P(P(A)))) = 22
2m(A)
etc.
Proof. Let B = {0, 1}. Then B has cardinality equal to 2. Then the
set B A has cardinality 2m(A) . By definition the set B A consists of all
functions f ∶ A → {0, 1}, i.e. for a ∈ A, f (a) = 0 or f (a) = 1. Setting
Af = {a ∈ A ∶ f (a) = 1} we have that f is the charcteristic (indicator)
function of the subset Af ∈ P(A) and vice versa. So we obtained a
1-to-1 correspondence between the subsets of A and the elements of
BA.
For the proofs of the following statements we assume that the chosen
representatives of a cardinal are pairwise disjoint.
Lemma 9.2.
mn1 +n2 = mn1 ⋅ mn2 .
Proof. Follows since we can biject the functions from f ∶ N1 ∪ N2 → M
with the ordered pairs of functions given by (f1 = f ∣N1 , f2 = f ∣N2 ). Lemma 9.3.
k
(mn ) = mn⋅k .
k
Proof. (mn ) is the cardinality of the set of function from K to the set
of functions from N to M . I.e. f (k) = gk (n) = mk,n ∈ M , k ∈ K, n ∈
N . On the other hand mn⋅k is the cardinality of the functions h from
N × K ∋ (n, k) → m ∈ M . We bijectively set f ∼ h iff gk (n) = h(n, k) for
all k ∈ K, n ∈ N .
Lemma 9.4. For m ≤ n and k ≤ l we have
m + k ≤ n + l,
m ⋅ k ≤ n ⋅ l.
Proof. Obviously,
m(M ∪ K) ≤ m(N ∪ L)
and
m(M × K) ≤ m(N × L).
Lemma 9.5.
(m1 ⋅ m2 )n = mn1 ⋅ mn2 .
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Proof. Follows since we can biject the functions from f ∶ N → M1 × M2
given by f (n) = (m1 , m2 ) with the elements of M1N × M2N , i.e. ordered
pairs of functions given by (f1 (n) = m1 , f2 (n) = m2 ).
Lemma 9.6. If n ≥ ℵ0 , n -finite then
n = n + n = n + ℵ0 .
Proof. By the assumption we find a sets N , N1 , m(N ) = n, N = N1 ∪ N
and N1 ∩ N = ∅ (note that n ≥ ℵ0 means that there are representatives of n that contain the natural numbers since the two cardinals are
comparable) . We define bijections between N and N ∪ {1, 2, ⋯, n} and
N ∪ N, respectively, by taking the identity on N1 and bijective map N
to N ∪ {1, 2, ⋯, n} or N to N ∪ N, respectively. Those bijections have
been considered before (sometimes called ”Hilbert’s hotel”).
Lemma 9.7.
ℵ20 = ℵ0 .
Proof. The statement can be interpreted as m(N × N) = m(N), i.e. we
have to biject (ordered) pairs of natural numbers to natural numbers.
An effective way is to choose N × N ∋ (m, n) ⇄ 2m (2n + 1) ∈ N.
Lemma 9.8. For n ≥ ℵ0 and k ≤ 2n
2n + k = 2n .
Proof. Since n ≥ ℵ0 we have n + 1 = n and
2n ≤ 2n + k ≤ 2n + 2n = 2 ⋅ 2n+1 = 2n .
10. CH and GCH, the continuum hypothesis and the
generalized continuum hypothesis
CH: There is no cardinal n such that
ℵ0 = m(N) < n < c = 2ℵ0 = m(R).
GCH: For any cardinal number m there is no other cardinal n such
that m < n < 2m .
11. Well-ordering
Definition 11.1. (Partial) Ordering a ≤ b is a relation, i.e. a subset
of M × M with:
● a ≤ a (refelxivity)
● a ≤ b, b ≤ a Ô⇒ a ≤ c (transitivity)
● a ≤ b, b ≤ a Ô⇒ a = b (antisymmetry)
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Example 11.1. N, R, P(M ) with subset relation, ⋯
Definition 11.2. A set M is totally ordered if for any distinct a, b ∈
M either a < b or b < a.
Example 11.2. N, R but not P(M ).
Definition 11.3. A well-ordering of a set M is a total ordering
such that any subset M1 ⊂ M , i.e. M1 ∈ P(M ), has a least element,
i.e. ∃x ∈ M1 such that for all a ∈ M1 we have x ≤ a.
Example 11.3. N but not R.
Definition 11.4. A subset M1 ⊂ M of a (partially) ordered set M is
called a chain if it is totally ordered, i.e.∀a, b ∈ M1 either a ≤ b or b ≤ a.
Definition 11.5. An element a ∈ M of a (partially) ordered set M is
said to be an upper bound for a subset M1 ⊂ M if ∀x ∈ M1
Ô⇒
x ≤ a. A lower bound is defined analoguously. A subset with an
upper/lower bound is said to be bounded from above/below.
An element a ∈ M of a (partially) ordered set M is maximal if
a ≤ x Ô⇒ a = x.
Definition 11.6. An element a ∈ M1 ⊂ M of a subset of an (partially)
ordered set is said to be comparable in M1 if
∀x ∈ M1
Ô⇒ a ≤ x or x ≤ a.
Remark 11.1. A subset M1 ⊂ M of a (partially) ordered set M is a
chain iff any of its elements is comparable in M1 .
12. The Axiom of Choice and related statements
AC: For any collection of non-empty sets {Aα }α∈I , Aα , I are sets,
there is a function f ∶ {Aα }α∈I → ⋃α Aα such that for any α we have
f (Aα ) ∈ Aα .
Remark 12.1. In contrast to ZF this axiom allows to build sets without
specifying the elements. One can choose one shoe from an infinite
collection of pairs of shoes (choose always the left!) but one needs AC
to choose from pairs of socks.
Since any set contains elements AC is not needed for a finite collection of sets!
AC is not harmless at all but it has many convenient applications.
Sometimes one uses AC even when it is not needed (definition of a
differentiable structure).
One should always be aware when one uses AC!
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12.1. Statements equivalent to AC.
The AC is equivalent to the following statements. Each of which has
its advantages in different applications.
Theorem 12.1 (Maximal chain theorem of Hausdorff). In any partially ordered set, every totally ordered subset (chain) is contained in a
maximal totally ordered subset (chain).
AC implies the Maximal chain theorem of Hausdorff.
● First change ≤ to ⊂ by defining M ⊃ Mx ∶= {y ∈ M ∶ y ≤ x}.
● Let X be a non-empty collection (it contains ∅) of subsets of
M with the properties: Every subset of a set in X belongs to X
and the union of each chain in X is contained in X.
● Let f ∶ P(M ) ∖ ∅ → M be a choice function for M (AC!). For
each A ∈ X let A∗ ∶= {x ∈ M ∶ A ∪ {x} ∈ X}. Define g∶ X → X by
g(A) = A if A∗ ∖ A = ∅ or otherwise g(A) = A ∪ {f (A∗ ∖ A)}.
Then g(A) contains at most 1 element more than A.
● We want to prove that g(A) = A for some A ∈ X. That will
impliy the theorem.
● We say set a subcollection J of X is a tower if
– ∅∈J
– If A ∈ J then g(A) ∈ J.
– If C is a chain in J then ⋃A∈C A ∈ J.
● The intersection of towers is a tower. Let J0 be the intersection
of all towers, i.e. the smallest. We are going to prove that it is
a chain.
● Let C be comparable in J0 .
● Assume A ⊊ C then g(A) ⊂ C. Otherwise A ⊊ C ⊊ g(A) contradicting that g(A) has at most one more element than A.
● Let U be the collection of sets in A ∈ J0 such that A ⊂ C or
g(C) ⊂ A. We want to show that U is a tower.
– ∅ ∈ U.
– A ⊊ C then (previously) g(A) ⊂ C, i.e. g(A) ∈ U.
– A = C then g(C) = g(A) (i.e. g(C) ⊂ g(A)) and g(A) ∈ U.
– g(C) ⊂ A then g(C) ⊂ A ⊂ g(A) and g(A) ∈ U.
– The union of the elements over a chain is by the definition
of U contained in U.
● U = J0 .
● If C is comparable then so is g(C) by the previous considerations.: If A ∈ J0 = U so either A ⊂ C ⊂ g(C) or g(C) ⊂ A.
● g maps comparable sets to comparable sets. The union of comparable sets over a chain is comparable. That implies that
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comparable sets constitute a tower and hence J0 consists of
comparable sets only, i.e. J0 is a chain itself.
● Since J0 is a chain and a tower the union A over all elements of
J0 is in J0 . Therefore g(A) ⊂ A since A includes all sets in J0 .
On the other hand A ⊂ g(A). Therefore A = g(A).
Theorem 12.2 (Zorn’s Lemma). Every non-empty partially ordered
set in which every chain (i.e., totally ordered subset) has an upper
bound contains at least one maximal element.
Maximal chain theorem of Hausdorff implies Zorn’s Lemma.
Take a chain C in the set and an upper bound a for it. Then a ∈ C
and is a maximal element. Otherwise there is an element b in the set
such that a < b ∉ C and the chain C is not maximal.
Theorem 12.3 (Zermelo’s Well-Ordering Principle (WOP)). Every
set can be well-ordered.
Zorn’s lemma implies WOP.
● If A, B are two well-ordered sets then A is said to be a continuation of B if B ⊂ A and the subset inclusion preserves the order
in B, A.
● If a collection of well-ordered sets forms a chain C with respect
to continuation then there is a unique well-ordering of U, the
union of the sets in the chain, that is a continuation of the
well-ordering of all sets in C. This well-order is defined in the
following way: take a, b ∈ U. The there are a ∈ A ∈ C, b ∈ B ∈
C. By the continuation property either A = B or one is the
continuation of the other. That defines the order between a
andb. It is clearly a well-ordering.
● Consider U the collection of well-ordered subsets of M , i.e subsets together with a (chosen) well-ordering. Then U is partially
ordered by continuation. This collection contains the empty
set. If C is a chain in U then its union of the sets in C is an
upper bound. Hence, there is a maximal well-ordered set M ′
in U. This set must be equal to M since otherwise we could
add the ”missing” element x ∈ M ∖ M ′ to the chain by x > y,
∀y ∈ M ′ .
WOP implies AC.
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Consider U = ⋃α∈I Aα and choose (no AC needed for this!) a wellordering on the set U . Since Aα ⊂ U we can well-define f (Aα ) as the
minimal element of Aα ⊂ U . This is a choice function.
Theorem 12.4 (Trichotomy of Cardinals). If two sets are given, then
either they have the same cardinality, or one has a smaller cardinality
than the other.
Proof. This proof comes in section 14.
Theorem 12.5 (Tychonov’s theorem). Let {Kα }α∈I , I any index set,
Kα compact spaces. The the product ∏ Kα endowed with the product
α∈I
topology is a compact space.
Proof. The proof is postponed till section 17.
Remark 12.2. If the spaces Kα are assumed to be Hausdorff Tychonov’s theorem follows from the Ultrafilter Lemma without using AC
as can be seen from the forthcoming proof.
There are other statements commonly used in analysis, topology,
combinatorics,⋯ that are equivalent to AC. We will list them without
proofs.
Theorem 12.6. Every vector space over any field F has a basis. In
particular (not equivalent to AC!), R has a basis (Hamel basis) as a
vector space over Q. The latter statement will be proved in section 17.
Remark 12.3. This theorem does not hold if one considers modules
instead of vector spaces: ∏∞
n=1 Z is not a free Z-module, i.e. has no
basis as a module over Z.
Theorem 12.7. Every ring with a unit has a basis.
Theorem 12.8. The closed unit ball of the dual of a normed vector
space over R has an extreme point.
Theorem 12.9. Any connected graph has a spanning tree.
12.2. Statements implied by AC but strictly weaker. The following are some important statements that are not implied by ZF but
can be proven in ZF+AC and do not imply AC.
● Any union of countably many countable sets is countable. (Compare with the statement in section 6! It is equivalent with ωAC.)
● An infinite set is Dedekind-infinite (equivalent to ωAC).
● The existence of (Lebesgue-) non-measurable subsets of R (follows from the Ultrafilter Lemma, see section 15).
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●
●
●
●
The Banach-Tarski paradox.
Every subgroup of a free group is free.
The additive groups of R and C are isomorphic.
The Hahn-Banach theorem (Follows from the Ultrafilter Lemma):
Any (semi-)norm bounded linear functional on a linear subspace
U of a (semi-)normed vector space V over R or C has a linear
extension to all of V .
● The Ultrafilter Lemma (see section 15).
● Every set can be totally ordered (follows from the Ultrafilter
Lemma but is strictly weaker!)
● The Alaoglu theorem (equivalent to the Ultrafilter Lemma):
The closed unit ball of the dual of a normed vector space is
compact with respect to the weak* topology.
● The Baire Category Theorem (equivalent to the Axiom of dependent choices) : A non-empty complete metric space is not
the countable union of nowhere-dense closed sets.
● The open mapping and closed graph theorems (rely on the Baire
Category Theorem).
● Every infinite-dimensional topological vector space has a discontinuous linear functional.
● Every Tychonov space has a Stone-Čech compactification (Relies on the Ultrafilter Lemma).
Another interesting statement is the following
Theorem 12.10 (Ulam). Under CH (or weaker) there is no finite, σadditive, non-atomic measure such that any subset of R is measurable.
There are also stronger axioms that imply AC. The most familiar one
is the generalized continuum hypothesis (GCH). On the other hand, CH
is independent of ZFC (Cohen).
13. Ordinal numbers
Lemma 13.1. The collection On of all ordinals is a proper class.
Definition 13.1. An ordinal α ⪰ 1 is called a limit ordinal if it has
no predecessor, i.e. there is no ordinal β with α = β + 1.
Definition 13.2. Let ξ be a limit ordinal and
α1 ⪯ α2 ⪯ ⋯ ⪯ αβ ⪯ ⋯,
β≺ξ
be an increasing sequence of ordinal numbers. Its limit is defined as:
γ ∶= lim αβ ∶= min{α ⪰ αβ ∶ ∀β ≺ ξ}.
β≺ξ
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Lemma 13.2. (ωAC implies:) Given any increasing sequence of infinite ordinals (αn )n∈N , αn ≺ ω1 , ∀n ∈ N, then its limit γ is a countable
ordinal, i.e. γ ≺ ω1 .
Proof. We have
γ = lim ⪯ α1 + α2 + ⋯ + αn + ⋯
n≺ω0
and hence,
∞
m(γ) ≤ m ( ⋃ N) = m(N) = ℵ0 .
n=1
Hence, γ ≺ ω1 .
13.1. A fixed point theorem.
Definition 13.3. A normal function is a function f ∶ On → On from
the class On of all ordinal numbers to itself with the properties:
1. f (α) ≺ f (β) whenever α ≺ β (strict monotonicity).
2. For all limit ordinals γ we have f (γ) = sup{f (α) ∶ α ≺ γ}
(continuity).
Lemma 13.3. Let f be a normal function. Then
∀α ∶ α ⪯ f (α).
Proof. Assume the contrary and let α0 be the least ordinal such that
f (α0 ) ≺ α0 . Then β ⪯ f (β) for all β ≺ α0 . Let β0 = f (α0 ). Then
f (β0 ) = f (f (α0 )) ≺ f (α0 ) = β0 .
This contradicts the minimality of α0 .
Lemma 13.4. Let f be a normal function and A a non-empty set of
ordinal numbers. Then
f (sup A) = sup{f (α) ∶ α ∈ A}.
Proof. If sup A is a successor ordinal (i.e. it is not a limit ordinal) then
sup A ∈ A. Hence,
f (sup A) ⪯ sup{f (α) ∶ α ∈ A}.
On the other hand by monotonicity
f (sup A) ⪰ f (α)
∀α ∈ A.
If sup A = γ is a limit ordinal then
f (sup A) = sup{f (α) ∶ α ≺ γ} = sup{f (α) ∶ α ∈ A}.
15
Theorem 13.1. For any ordinal α there is an ordinal β ⪰ α such that
f (β) = β.
Proof. For given α we define a sequence (αn )n∈N by α0 = α, αn+1 = f (α).
Let β = supn∈N αn . Then β ⪰ α and
f (β) = f (sup αn ) = sup{f (αn ) ∶ n ∈ N} = sup{αn+1 ∶ n ∈ N} = β
n∈N
since the sequence (αn )n∈N is monotone increasing.
14. Cardinals and alephs
Definition 14.1. The cardinal number of a well-ordered set is called
an Aleph: ℵ.
Definition 14.2. An ordinal number ωa is an initial ordinal for an ℵ
if ℵ = m(ωa ) and ∀ω with m(ω) ≥ ℵ follows ω ≥ ωa . Initial ordinals are
canonical representatives of cardinal numbers ℵ.
Lemma 14.1. For α ∈ I let ωα be initial ordinals. Then γ ∶= ⋃α∈I ωα
is an initial ordinal.
Proof. Since any initial ordinal is an ordinal and the union of ordinals
is an ordinal we only have to prove that γ is initial, i.e. m(γ) > m(β)
for all β ≺ γ. The latter means that β ∈ γ and hence, ∃α ∈ I such that
β ∈ ωα Ô⇒ β ≺ ωα ⊂ γ. Assume that m(γ) = m(β) (we always have
m(γ) ≥ m(β)!) then the restriction of the bijection γ ∼ β to ωα is an
injection from ωα ↪ β. Hence, ωα ≤ β. Since, β ≺ ωα Ô⇒ β ≤ ωα we
should have β ∼ ωα . But then ωα is not an initial ordinal.
Remark 14.1. The ℵ’s inherit a well-ordering from their initial ordinals.
Lemma 14.2. The function f ∶ On → On defined recursively by
f (α) ∶= min{ω ∶ ω is an initial ordinal and ω ≻ f (β), ∀β ≺ α
is a normal function.
Proof. Clearly the monotonicity requirement is fulfilled.
By definition of f we have for γ = supβ≺α β that
f (γ) ⪰ sup f (α).
β≺α
It remains to show that supβ≺α f (α) is an initial ordinal. But this
follows from Lemma 14.1
Lemma 14.3. There are large cardinals ℵα in the sense that their
initial ordinals ωα fulfill
α = ωα .
16
Proof. The function f (α) = ωα is a normal function. So we can apply
Theorem 13.1.
Remark 14.2.
● The least fixed point for f from Lemma 14.3 is ωωω⋱
● There are many more normal functions. F.e. the function
∶ On → On defined by
(α) = ω α
is normal. Their fixpoints, i.e. ordinal numbers fulfilling
α = ωα,
are called - numbers. The least - number is 0 = ω ω
ω⋰
Lemma 14.4 (Hartogs, Sierpinski). The following statement holds
without the use of AC. For any set M there is a set f (M ) ∈ P(P(P(M )))
such that
● f (M ) is well-ordered,
● m(f (M )) ≰ m(M ),
2m(M )
● m(f (M )) ≤ 22
.
Proof.
● Let C ∈ P(P(M )) be the subset of the subsets of M (chains)
that are well-ordered with respect to ⊂. We have C ∈ P(P(P(M ))).
● For C = {⋯ ⊂ Mα ⊂ ⋯} ∈ C we consider its equivalence class [C]
of all chains that are order-equivalend with C.
● Let C ∈ P(C) be the set of all such equivalence classes. C is
well-ordered itself (by the magnitude of the order types of their
equivalence classes and we set f (M ) = C.
● Since C ⊂ P(C) and C ⊂ P(P(M )) we get f (M ) ⊂ P(P(P(M ))).
2m (M )
.
● By construction: f (M ) is well-ordered and m(f (M )) ≤ 22
Assume m(f (M )) ≤ m(M ). Then C ∼ M1 ⊂ M and this would
induce a well-ordering of M1 of the same type ωC as C. Now
the chain of initial segments
M1 ∶= {{(M1 )1 ⊂ (M1 )2 ⊂ ⋯ ⊂ (M1 )α } ∶ α ≺ ωC = ω(M1 )}
is order equivalent to an initial segement of C and hence, M1 ∈ C
and has order tupe
ω(M1 ) = ωC = ω(f (M ))
a contradiction.
17
Corollary 14.1. From this follows (without AC): There is no injective net in any set (a map f ∶ Ordinal numbers → M with f (α) = xα ,
α along all ordinal numbers and xα ≠ xβ if α ≠ β). This gives an alternative proof of Zorn’s lemma (use AC to construct an injective net
if there is no maximal element). Roughly speaking: ”There are more
ordinal numbers than any set can have elements.”
Lemma 14.5 (Zermelo). Without AC: If A is an infinite well-ordered
set then
(m(A))2 = m(A).
Proof. We have seen that ℵ20 = ℵ0 . Assume that the statement is wrong.
Since the ℵ’s are well-ordered there is a smallest one, α sucht that
ℵ2α > ℵα and ℵ2β = ℵβ for all β ≺ α. Let ωα be its initial ordinal. Let
P ∶= {(µ, ν) ∶ µ ≺ ωα , ν ≺ ωα },
and for λ ≺ ωα
Pλ ∶= {(µ, ν) ∶ µ + ν = λ}.
● P = ⋃λ≺ωα Pλ : We need to show that
µ ≺ ωα , ν ≺ ωα Ô⇒ µ + ν ≺ ωα .
But this follows by the definition of α:
m(µ) + m(ν) ≤ m(λ)2 = m(λ) < ℵα .
● We well-order P :
– µ ≺ µ′ Ô⇒ (µ, ν) ≺ (µ′ , ν ′ )
– ν ≺ ν ′ Ô⇒ (µ, ν) ≺ (µ, ν ′ ).
● Obviously m(Pλ ) ≤ m({µ ∶ µ < λ}) since for each µ ≺ λ there is
exactly one order type ν such that µ + ν = λ. Hence,
m(Pλ ) ≤ m(λ + 1) = m(λ) < ℵα .
● For any given initial segment P(µ,ν) of P there are at most λ′ =
µ + ν ”diagonals” Pλ′ not exceeding (µ, ν) ≺ ωα . Therefore
m(A)2 = m(P ) ≤ m(λ′ + 1)2 = m(λ′ )2 = m(λ′ ) ≤ ℵα
a contradiction. Thus P ≺ ωα but then ℵ2α ≤ m(P ) = ℵα contradicting the definition of α.
Lemma 14.6. If n ≥ ℵ0 then n + n < 2n (without AC!).
The statement is obviously true for 3 ≤ n ∈ N.
18
Proof. Let n ≥ ℵ0 . Since N ∋ x → {x} ∈ P(N ) and N ∋ x → N ∖ {x} ∈
P(N ), N ∖ {x} ≠ {y} are bijections we have n + n ≤ 2n . We also have:
n + n = 2 ⋅ n ≤ n2 .
Hence it suffices to show that n2 ≱ m(P(N )). Assume there is an
injection
g∶ P(N ) ↪ N × N.
Let Y ⊂ N be an infinite subset with a well-ordering ≺R .
● By Lemma 14.5 there is a bijection
h = h(Y,≺R ) ∶ Y ⇄ Y × Y.
● Let
A = A(Y,≺R ) ∶= {y ∈ Y ∶ g −1 (h(y)) is defined and y ∉ g −1 (h(y))}.
If g(A) ∈ Y × Y then there is a y0 ∈ Y such that h(y0 ) = g(A).
But then the question y0 ∈ A or y0 ∉ A? leads to a contradiction
(diagonal argument as by Russel or Cantor). Hence,
● g(A) ∉ Y × Y .
● We define a map on infinite well-ordered subsets of N with
F ((Y, ≺R )) ∈ N ∖ Y
(1)
by
F ((Y, ≺R )) = π(g(A(Y,≺R ) ))
where π maps an element (x, z) ∈ (N × N ) ∖ (Y × Y ) to x if
x ∉ Y or to z otherwise.
● We will start with N ⊋ Y0 ∼ N. The latter equivalence is possible
since m(N ) ≥ ℵ0 . So Y0 is of the form
y1 < y2 < ⋯ < yn < ⋯
We put F ((Y0 ≺N )) = x ∈ N ∖ Y0 and for n ∈ N we put
F (y1 < ⋯ < yn ) = yn+1 .
● We call a set Y ⊂ N equipped with a well-ordering ≺R a g-wellordered set if for any y ∈ Y
F ({x ∈ Y ∶ x ≺R y}) = y.
● (Y0 , ≺N ) is a g-well-ordered set.
● If (Y, ≺R ) and (Z, ≺N ) are both g-well-ordered sets then one is
an initial segment of the other. For:
– Since we can compare any two well-ordered sets w.l.o.g.
there is an order-preserving injection i ∶ (Y, ≺R ) → (Z, ≺S )
mapping Y onto an initial segment of Z.
19
– We need to show the i = id∣Y . Assume not let x ∈ Y be the
least element such that i (x) ≠ x. Then
{y ∈ Y ∶ y ≺R x} = {z ∈ Z ∶ z ≺S i (x)}.
But by the g-well-ordering property
i (x) = F ({z ∈ Z ∶ z ≺S i (x)}) = F ({y ∈ Y ∶ y ≺R x}) = x,
a contradiction.
● Let now W be the union of all g-well-ordered sets. By the
preceding statement there is a (unique) g-well-ordering ≺g on
W.
● F ((W, ≺g )) ∈ W for otherwise W ∪ {F (W )} can be equipped
with a g-well-ordering ≺′g by setting ≺′g =≺g on W and F (W )
is larger than any element in W (this is a well-defined g-wellordering since F (W ) ∈ N ∖ W ). This contradicts the definition
of W .
● On the other hand by 1 we should have F (W, ≺g )) ∈ N ∖ W – a
contradiction.
Theorem 14.1. Trichotomy for cardinals is equivalent to the axiom
of choice.
Proof.
● By the WOP we can well-order any two sets A, B. Then one
is order-equivalent to an initial segment of the other and the
order-equivalence gives a bijection of one set onto a subset of
the other.
● If one can compare a set A with any other set so one can do so,
in particular, with any well-ordered set. By Theorem 14.4 we A
is comparable with f (A) and the possibility m(A) ≥ m(f (A))
is excluded. Hence, m(A) < m(f (A)) and we inherit on A a
well-ordering from f (A).
Theorem 14.2. The statement n2 = n for all infinite cardinals n is
equivalent to the axiom of choice.
Proof. The AC is equivalent that any cardinal number is an ℵ. Thus
Lemma 14.5 shows
AC
Ô⇒
n2 = n
∀ cardinals n.
Conversely,
n2 = n,
k2 = k,
(n + k)2 = n + k.
20
Hence,
n + k = n2 + 2n ⋅ k + k2 = n + 2n ⋅ k + k ≥ n ⋅ k.
and
n ⋅ k = (n1 + 1) ⋅ (k1 + 1) = n1 ⋅ k1 + n1 + k1 + 1 ≥ 1 + n1 + k1 + 1 ≥ n + k
Therefore n+k = n⋅k and there are (pairwise disjoint) sets N1 , N2 , K1 , K2 ,
N1 ∼ N2 , K1 ∼ K2 , m(Ni ) = n, m(Ki ) = k such that N1 × K1 ∼ N2 ∪ K2 .
We will choose K1 = f (N1 ) from Lemma 14.4
There are 2 possibilities:
● ∃n1 ∈ N1 such that ∀k1 ∈ K1 we have (n1 , k1 ) ⇄ n2 ∈ N2 . Then,
k = m(K1 ) ≤ m(N2 ) = n.
This possibility is excluded since k = m(f (N1 )) ≰ m(N1 ) = n.
● ∀n1 ∈ N1 ∃k1 ∈ K1 such that (n1 , k1 ) ⇄ k2 ∈ K2 . For fixed
n1 ∈ N1 we choose g(n1 ) to be the smallest k ∈ K1 (note that
K1 is well-ordered) such that (n1 , k) ⇄ k2 ∈ K2 . We set M =
{(n1 , g(n1 ))} and get
n = m(N1 ) = m(M ) ≤ m(K2 ) = k = m(K1 ).
The latter means that N1 is equivalent to an initial segment of
f (N1 ) = K1 and hence, well-ordered.
Lemma 14.7. The generalized continuum hypothesis implies:
If k + n = 2n , n ≥ ℵ0 then k = 2n .
Proof. We have k ≤ 2n and
n ≤ n + n ≤ 2n + 2n = 2n+1 = 2n .
But n + n < 2n and by GCH n + n = n.
Now 2n ⋅ 2n = 2n+n = 2n = n + k. Hence, there are sets N, K and a
bijection h∶ P(N )×P(N ) ⇄ N ∪K, m(N ) = n, m(K) = k and N ∩K = ∅.
Since n < 2n there is a subset N0 ⊂ N such that ∀M ⊂ N we have
(N0 , M ) ∉ h−1 (N ).
Therefore h∣(N0 ,P(N )) ↪ K is an injection and
k = m(K) ≥ m(P(N )) = 2n .
Theorem 14.3. The generalized continuum hypothesis implies the axiom of choice.
21
Proof. Let n be an arbitrary non-finite cardinal. We need to prove that
under GCH n is an Aleph, i.e. if m(N ) = n then N can be well-ordered
(for finite n there is nothing to prove!).
Since n is not finite the cardinal k = n + ℵ0 ≥ ℵ0 is non-finite too. If
k = m(K) and K can be well-ordered so can N itself since it corresponds
to a segment of K (i.e. k ≥ n).
We have
k
2k
ℵ0 ≤ k < 2k < 22 < 22 .
Also
k + 2k = 2k ,
k
k
2k
k
2k + 22 = 22 ,
2k
22 + 22 = 22 .
2k
By Theorem 14.4 m(f (K)) ≤ 22 and therefore
2k
k
k
2k
m(K) + 22 ≤ 22 + 22 = 22 .
2k
If m(f (K)) + 22 = 22 then by Lemma 14.7:
k
2k
m(f (K)) = 22 > k ≥ n.
Since f (K) is well-ordered this induces a well-ordering on n.
k
k
2k
Suppose 22 ≤ m(f (K)) + 22 < 22 then GCH implies
k
k
k
22 ≥ m(f (K)).
22 = m(f (K)) + 22 ,
Hence,
k
k
m(f (K)) + 2k ≤ 22 + 2k = 22 .
Again: If m(f (K)) + 2k = 22 then by Lemma 14.7:
k
k
m(f (K)) = 22 > k ≥ n.
Again this induces a well-ordering on n.
k
Suppose 2k ≤ m(f (K)) + 2k < 22 then GCH implies
2k = m(f (K)) + 2k ,
2k ≥ m(f (K)).
Hence,
m(f (K)) + k ≤ 2k + k = 2k .
Again: If m(f (K)) + k = 2k then by Lemma 14.7:
m(f (K)) = 2k > k ≥ n.
Again this induces a well-ordering on n.
Finally we should have by GCH:
m(f (K)) + k = k thus m(f (K)) ≤ k,
a contradiction.
22
Proof of Ulam’s theorem 12.10. Let µ ba a finite, non-atomic, σ-additive
measure on R.
Since under CH and AC c = m(R) = ℵ1 is the first uncountable
Aleph there is a well-ordering ≺ on R such that for each y ∈ R the set
Ay ∶= {x ∈ R ∶ x ≺ y} is at most countable. Choose 1-1 mappings (AC)
f (⋅, y)∶ Ay ⇄ Nx ⊂ N. So f is an integer-valued function defined for all
pairs (x, y), y ∈ R, x ≺ y). By definition
x ≺ x′ ≺ y
Ô⇒
f (x, y) ≠ f (x′ , y).
For x ∈ R, n ∈ N we define
Fxn ∶= {y ∈ R ∶ x ≺ y and f (x, y) = n}.
We form a table
Fx11 Fx12 ⋯ Fx1 ⋯
Fx21 Fx22 ⋯ Fx2 ⋯
⋯ ⋯ ⋯ ⋯ ⋯
Fxn1 Fxn2 ⋯ Fxn ⋯
⋯ ⋯ ⋯ ⋯ ⋯
with ℵ0 rows and ℵ1 columns. It has the following properties.
● The sets in a row are pairwise disjoint. Otherwise suppose
y ∈ Fxn ∪ Fxn′ . Then x ≺ y, x′ ≺ y and f (x, y) = f (x′ , y) = n. Since
f (⋅, y) is a bijection it follows x = x′ .
● ⋃n∈N Fxn = R ∖ Xx where m(Xx ) ≤ ℵ0 , i.e. the union of each
column fills R up to a countable set. For y ≻ x then y ∈ Fxn for
some n, i.e. the one for which f (x, y) = n. Hence
R ∖ ⋃ Fxn = {y ∈ R ∶ y ⪯ x}.
n∈N
The set on the right-hand-side is countable by the definition of
≺.
Since µ(R) < ∞ in each row
m({x ∈ R ∶ µ(Fxn ) > 0}) ≤ ℵ0 .
Therefore there are at most countably many sets of positive measure
in the entire table. Because there are ℵ1 colums there is a x0 ∈ R such
that µ(Fxn0 ) = 0 for all n ∈ N. Then by σ-additivity
0 = µ (⋃ Fxn0 ) = µ (R ∖ Xx0 ) = µ(R).
n
The last equality is due to the assumption that µ is non-atomic and
m(Xx0 ) ≤ ℵ0 .
23
15. ultrafilter
Definition 15.1 (Filter). A filter F on a set M is an element of
P(P(M )) such that for all N, K ∈ P(M )
● M ∈ F, but ∅ ∉ F.
● N ∈ F, N ⊂ K Ô⇒ K ∈ F.
● K ∈ F, N ∈ F Ô⇒ N ∩ N ∈ F.
We can introduce a partial order on filters of a set M .
Definition 15.2. We say that the filter F ′ is finer than the filter F
if for any N ∈ F it follows N ∈ F ′ . We will write F ⊂ F ′ . This gives a
partial order on the set of all filters on a (fixed) set.
Definition 15.3. A non-empty family (Aα )α∈I of subsets ∅ ≠ Aα ∈
P(M ) is said to have the finite intersection property if for any
finite subset {α1 , α2 , ⋯, αn } ∈ P(I), n ∈ N the intersection ⋂nk=1 ≠ ∅.
Definition 15.4. A non-empty family (Aα )α∈I of subsets ∅ ≠ Aα ∈
P(M ) is said to have the centered finite intersection property
if ⋃α∈J Aα = ∅ for some subset J ⊂ I then there is a finite subset
{α1 , α2 , ⋯, αn } ∈ P(J), n ∈ N such that already the finite intersection
⋂nk=1 = ∅.
Lemma 15.1. Let M ∈ P(P(M )) have the finite intersection property.
Then
n
F ∶= {N ⊂ M ∶ ∃M1 , M2 , ⋯, Mn ∈ M such that ⋂ Mk ⊂ N }
k=1
is a filter on M . This is the smallest filter containing M and M is
called a filter basis.
′
Proof. Since ∅ ≠ ⋂nk=1 Mk ∪ ⋂nk=1 Mk′ ⊂ N ∩ N ′ ⊂ M the axioms of a
filter are fulfilled. Any filter containing M must also contain any finite
intersection of elements of M and hence also all the supsets of these
intersections.
Example 15.1.
● The principal filter px of an element x ∈ M is the family
px ∶= {N ⊂ M x ∈ N }.
● In a topological space X the system Ux of all neighborhoods of
a point is a filter, called the neighborhood filter
Vx ∶= {Y ⊂ X ∶ x ∈ U (x) ⊂ Y, U (x) open}.
24
M
● Let M be an infinite set. The cofinite filter Fcof
in is defined
as
M
Fcof
in ∶= {N ⊂ M ∶ ∅ ≠ N, m(M ∖ N ) < ∞}.
Definition 15.5 (Ultrafilter). A filter F on M is called an ultrafilter
if any set N ∈ P(M ) either N ∈ F or (M ∖ N ) ∈ F.
Remark 15.1.
● Any principal filter px is an ultrafilter since x ∈ N, x ∈ K Ô⇒
x ∈ N ∩ K.
● Reversely if for a filter F we have {x} ∈ F then F = px because
it must contain all supsets of {x} and cannot contain their complements.
● An ultrafilter that is not principal (if it exists!) is called free.
M
● Fcof
is an ultrafilter if and only if M cannot be partitioned into
two disjoint infinite subsets, i.e. if M ∼ M1 ∪ M2 , M1 ∩ M2 = ∅
M
and ∀n ∈ N, m(Mi ) > n then Fcof
is not an ultrafilter. The latter
M
is true since otherwise exactly one of Mi belongs to Fcof
but both
are not cofinite. If there is no such partition of M into two
infinite subsets then any set is either finite or its complement is
finite, i.e. any subset or its complement (if the set is a proper
M
subset) belongs to Fcof
.
– In ZF without choice there can be amorphous sets, that
means infinite sets such that each of its subset is either
finite or cofinite.
– Amorphous sets cannot be totally ordered, for otherwise the
set
A ∶= {x ∈ M ∶ m({y ≤ x}) > n and m({x < z}) > n, ∀n ∈ N} ≠ ∅
and M is the disjoint union of {y ≤ x} and {x < z}.
– The ultrafilter lemma implies the existence of a total order
on any set. Hence there are no amorphous sets assuming
UFL.
● Note that exactly one of the two sets N and M ∖ N is contained
in an ultrafilter since N ∩ (M ∖ N ) = ∅ ∉ F.
● An ultrafilter p can be interpreted as a finitely additive 0-1measure of X, measuring all subsets:
– We set
µp (N ) = {
1 N ∈p
,
0 M ∖N ∈p
∀N ⊂ M.
25
– We have from the ultrafilter properties that µp is defined
for all N ⊂ M
– µp ({∅}) = 0 and µp (M ) = 1, since no filter contains {∅},
but always contains M .
– If Ni ∩ Nj = ∅, i ≠ j, 1 ≤ i, j ≤ n then either ⋃nk=1 Nk ∈ p
and exactly one of the sets Nk ∈ p, or ⋃nk=1 Nk ∉ p and
Nk ∉ p, ∀k. This gives finite additivity
n
n
k=1
k=1
µp ( ⋃ Nk ) = ∑ µp (Nk ).
●
●
●
●
●
So the family of ultrafilters on a given set M can be interpreted
as the family of all finitely additive 0-1- measures on P(M ).
N ∪ K ∈ p, p is an ultrafilter Ô⇒ N ∈ p or K ∈ p.
K ⊂ N, N ∉ F Ô⇒ K ∉ F for any filter F.
Let p be an ultrafilter. If M = ⋃nk=1 Nk , Nk ∩ Nl = ∅, k ≠ l then
there is exactly one 1 ≤ k ≤ n such that Nk ∈ p and Nl ∉ p, l ≠ k.
Any ultrafilter on a finite set is principal, since a finite set is
the finite union of its distinct elements.
If p is a free ultrafilter on an infinite set M (if it exists) and
the subset N = {x1 , ⋯, xn } ∈ P(M ) is a finite set then N ∉ p.
For otherwise
n
N = ⋃ {xk }, {xk } ∩ {xl } = ∅, k ≠ l.
k=1
● An ultrafilter p on M is free if and only if it contains the cofinite
M
filter Fcof
in .
M
For if p is free and N ∈ Fcof
in then m(M ∖ N ) < ∞, hence
(M ∖ N ) ∉ p and N ∈ p. On the other hand (we may assume
that M is an infinite set since otherwise the cofinite filter does
M
not exist) the cofinite set M ∖ {x} ∈ Fcof
in is not an element of
px .
The next lemma establishes an alternative definition of an ultrafilter.
Lemma 15.2. A filter F on M is an ultrafilter if and only if F is
maximal, i.e. there is no filter F ′ finer than F.
Proof.
● ( Ô⇒ ) Assume F ⊊ F ′ and ∅ ≠ N ∈ F ′ ∖ F. Since F is an
ultrafilter we have (M ∖ N ) ∈ F ⊂ F ′ and this implies that
N ∈ F ′ and (M ∖ N ) ∈ F ′ - a condradiction.
● ( ⇐Ô ) Assume F is not an ultrafilter but maximal. Hence,
there is a set N such that neither N nor (M ∖ N ) is contained
in F. Assume further there is a K ∈ F with K ⊂ M ∖ N . But
26
then (M ∖ N ) ∈ F which is a contradiction to our assumption.
So every K ∈ F intersects N , i.e. ∀B ∈ F Ô⇒ B ∩ N ≠ ∅. So
the family F ∪ {N } has the finite intersection property and can
be extended to a filter containing N . So F cannot be maximal.
The existence of free ultrafilters is not a priori. It is impossible to
derive it from ZF only. It is impossible to construct an ultrafilter
even when its existence is established. The next theorem shows that
AC implies the existence of free ultrafilters.
Theorem 15.1 (Ultrafilter Lemma I). AC implies: Any filter F on a
set M is a subset of an ultrafilter p on M . Note that the ultrafilter p
does not have to be unique!
Proof. Let C be a chain (with respect to the ”finer” ordering) of filters
on M containing F. First we observe that FC ∶= ⋃G∈C {N ∈ G} is a filter:
● ∅ ∉ G ∀G ∈ C Ô⇒ ∅ ∉ FC .
● (N ⊂ FC Ô⇒ ∃G ∈ C ∶ N ∈ G) Ô⇒ N ⊂ K ∈ G and therefore
K ∈ FC
● (N, K ∈ FC Ô⇒ ∃G ∈ C ∶ N, K ∈ G) Ô⇒ N ∩ K ∈ G and
therefore N ∩ K ∈ FC .
By Theorem 12.1 (equivalent to AC) C ⊂ Cmax and Cmax is a maximal
chain.
Therefore Fmax ∶= ⋃G∈Cmax {N ∈ G} is itself a filter, finer than any
other in Cmax by maximality. So by Lemma 15.2 it is an ultrafilter
containing F.
An equivalent (seemingly stronger) statement is
Theorem 15.2 (Ultrafilter Lemma II). (AC implies:) Any family M
of subsets of M with the finite intersection property can be extended to
an ultrafilter.
Proof. Any filter has the finite intersection property and any family of
sets having the finite intersection property can be extended to a filter
by Lemma 15.1.
Remark 15.2. The equivalent Theorems 15.1 and 15.2 can be stated
as an axiom UFL if one wants to avoid AC. This axiom is strictly
weaker than AC as will be indicated in the proof of Theorem 12.5.
Theorem 15.3. The Ultrafilter lemma implies (without AC) that any
set can be totally ordered.
27
Proof. We consider a strict partial order ”<” on a set M . It is clear
that this exists.
0) On any finite set M we can extend the partial order ”<”to a
total order ”<M ” : it is clearly true for m(M ) = 1. Assume
we can extend the partial order for a set of n elements. Let
M ′ = M ∪ {xn+1 }, m(M ′ ) = n + 1. Let y be the maximal (with
respect to <M ) element in M such that y < xn+1 and z the
minimal element such that x < z. Then by compatibility of <
and <M we have y <M z. Put for w ∈ M either w <M ′ xn+1 if
w <m y or w = y, or x <m′ w if y < w and for all u, w ∈ M let
u <M ′ w ⇐⇒ u <M w. This gives the desired total order of
M ′ and we conclude by the induction principle (for N).
i) We consider the set of maps S ∶= {f ∶ Gf ⊂ M ×M → {0, 1}} such
that
a) (x, y), (y, z), (x, z) ∈ Gf , f (x, y) = f (y, z) = 1 Ô⇒ f (x, z) =
1.
b) (x, y), (y, x) ∈ Gf Ô⇒ (f (x, y) = 1 ⇐⇒ f (y, x) = 0).
c) (x, y) ∈ Gf , x < y Ô⇒ f (x, y) = 1.
ii) For each finite subset G ⊂ M × M, m(G) < ∞ there is a f ∈ S
with Gf = G. We just take f to be the characteristic function
of the extended totel order <G on G, i.e. f (x, y) = 1 if x <G y
and f (x, y) = 0 otherwise.
iii) f ∈ S ⇐⇒ f ∣G ∈ S for any finite subset G ⊂ Gf .
– ( Ô⇒ ) follows immediately from the definition of restriction.
– ( ⇐Ô ) let (x, y), (y, x), (y, z) ∈ Gf . We consider the
following finite subsets of Gf : Ga = {(x, y), (y, z)}, Gb =
{(x, y), (y, x)}, Gc = {(x, y)}. So a), b), c) follow from
applying the restriction properties to Ga , Gb and Gc , respectively.
iv) There is an element f0 ∈ S such that Gf0 = M × M .
– For finite F, G ⊂ M × M let SF = {f ∈ S ∶ F ⊂ Gf }. Then
SF ∩ FG = SF ∪G ≠ ∅: For each f ∈ SF ∪G ,we have S ∪ Gf ,
hence F ⊂ Gf and G ⊂ Gf and f ∈ SF and f ∈ SG . Reversely, f ∈ SF ∩ SG implies F ∈ Gf and G ∈ Gf . Thus
F ∪ G ⊂ Gf .
– This means {SG ∶ G ⊂ M × M, m(G) < ∞} has the finite
intersection property. Hence there is an ultrafilter p on S
that includes {SG ∶ G ⊂ M × M, m(G) < ∞}.
– For (x, y) ∈ M × M we have
p ∋ {f ∈ S(x,y) } = {f ∈ S(x,y) ∶ f (x, y) = 0} ∪ {f ∈ S(x,y) ∶ f (x, y) = 1}.
28
Hence exactly one of the sets {f ∈ S(x,y) ∶ f (x, y) = i}, i =
1, 2 is contained in p. We define i(x,y) equal to 0 or 1 so
that {f ∈ S(x,y) ∶ f (x, y) = i(x,y) } ∈ p.
– let G ⊂ M × M, m(g) < ∞. Then the finite intersection
⋂ {f ∈ S(x,y) ∶ f (x, y) = i(x,y) } ∈ p.
(x,y)∈G
Thus ⋂(x,y)∈G {f ∈ S(x,y) ∶ f (x, y) = i(x,y) } ≠ ∅. Chooseany
f0 ∈ ⋂(x,y)∈G {f ∈ S(x,y) ∶ f (x, y) = i(x,y) }. So f0 ∈ S and
G ⊂ Gf0 and f0 ∣G ∈ S. Since this holds for any finite subset
of S by iii) Gf0 = S.
v) Define (x <M y) ⇐⇒ f0 (x, y) = 1. By the preceding arguments this will give a compatible total order on M .
Theorem 15.4. The Ultrafilter Lemma implies the existence of nonmeasurable sets.
Proof.
● To each x ∈ R ∩ [0, 1] we associate its binary expansion containing infinitely many zeros, i.e. x ⇄ (xn )n∈N , xn ∈ {0, 1} ∀n ∈ N.
● We consider the sequence (Xn )n∈N of independent (with respect
to Lebesgue measure L ) random variables
Xn (x) ∶= xn .
● We define an equivalence relation ∽ on [0, 1]:
x∽y
⇐⇒
m({n ∈ N ∶ xn ≠ yn }) < ∞.
● A subset A ⊂ [0, 1] is called a tail set if A respects ∽, i.e. if
∀y ∽ x
x ∈ A ⇐⇒ y ∈ A.
● Kolmogorov’s 0-1-law states: If A is a (Lebesgue) measurable
tail event then
L (A)2 = L (A), i.e. L (A) = 0 or L (A) = 1.
● The inversion j ∶ [0, 1] → [0, 1] defined by xn → xn + 1 (mod 2)
leaves the Lebesgue measure invariant, i.e.
L (B) = L (j −1 (B))
∀ measurable B ⊂ [0, 1].
● Let p be an ultrafilter on N. We define
Ap ∶= {x ∈ [0, 1] ∶ {n ∈ N ∶ xn = 1} ∈ p}.
29
● Ap is a tail set: Assume x ∈ Ap and y ∽ x. Then
{n ∈ N ∶ xn = 1} ⊂ {n ∈ N ∶ yn = 1} ∪ {n ∈ N ∶ xn ≠ yn } ∈ p.
The union on the right-hand-side is a disjoint union and the set
{n ∈ N ∶ xn ≠ yn } is finite. Hence {n ∈ N ∶ yn = 1} ∈ p.
● j (Ap ) = [0, 1] ∖ Ap : First we note that since N ∈ p the sequence
(1)n∈N ∈ Ap . Moreover ∀x ∈ [0, 1], j (x) = y we have xn = 1 ⇐⇒
yn = 0. Hence, we have the disjoint union
N = {n ∈ N ∶ xn = 1} ∪ {n ∈ N ∶ yn = 1}.
The latter implies that x ∈ Ap if and only if j (x) ∉ Ap .
● We conclude that 0 < L (Ap ) = L ([0, 1] ∖ Ap ) = 21 < 1 if Ap were
measurable. This contradicts Kolmogorov’s 0-1-law. Hence, Ap
is not measurable.
16. Non-standard analysis
In this section we want to ”construct” a model of the real numbers
that is ”more complete” than the standard reals. From the point of
view of analysis we have to deal with ”infinitesimal small” or ”unboundedly large” quantities. From model-theoretic point of view
we also might consider a countable set of consistent inequlities like
1
0<x< .
n
While any finite subset of these equations has a non-empty set of solutions the countable collection lacks a solution. Non-standard analysis
is to solve these problems but keep the basic structure of R. The new
non-standard model(s) ∗ R called the hyperreals is ”constructed”.
Let us first collect the properties from R that are important for us
and that we want to survive in the non -standard model. The reals R
are a field with addition + and multiplication ⋅ with ”neutral” elements
0 and 1 respectively. Moreover, R is totally ordered by ≤. Another
important property is the Archimedian property: ∀x, y > 0 there is an
n ∈ N such that nx > y. Unfortunately if we require this property we
are not able to extend the real numbers.
Let us first try to find a model where we have infinitesimal small
and unboundedly large (non-standard) ”numbers and still keeep the
algebraic operations. We recall that in standard analysis sequences
are used to ”define” infinitesimal small or unboundedly large quantities. We also can term-wise add and multiply sequences. So let these
30
sequences x ∶= (xn )n∈N , xn ∈ R ∀n ∈ N, be the new non-standard numbers, i.e. we start with RN = ∏n∈N R and define
x +N y = (xn + yn )n∈N ,
x ⋅N y = (xn yn )n∈N .
That suffices to keep the algebraic properties. What about the total
order? Naively we can define
x ≤N y
⇐⇒
xn ≤ yn , ∀ n ∈ N.
However, how do we compare ((−1)n )n∈N with ((−1)n+1 )n∈N ? There are
infinitely many indices where one is larger than the other and vice versa.
Of course we can just declare an order between these two sequences.
But there are many of those pairs of sequences and it might be hard to
define it consistently. In fact it is impossible to do so in a constructive
way.
The (elegant) way out is to use free ultrafilters. By their maximality
property they allow to compare any sequences. So lets fix a free ultrafilter p ∈ βN ∖ i (N). We will define an order between two arbitrary
sequences in the following way: Divide the indices of the two sequences
into two parts:
S = S(x, y) = {n ∈ N ∶ xn ≤ yn }
and
G = G(x, y) = {n ∈ N ∶ xn ≥ yn }.
Then S ∪ G = N and hence at least one of them is contained in p. We
would like to say
x≤y
⇐⇒
S∈p
and
x≥y
⇐⇒
G ∈ p.
To get the axioms of an order we have to have that
x ≤ y and y ≤ x Ô⇒ x = y.
So we need to introduce equivalence classes
x ∼p y ⇐⇒ {n ∈ N ∶ xn = yn } ∈ p.
Finally we let
R ∶= RN / ∼p
be the set of all those equivalence classes. Here we see that we do not
obtain anything new if we use a principal ultrafilter since in the case of
p = pn we get ∗ R ≃ πn (RN ) = R where πn is the projection to the n-th
coordinate.
Heuristically this means that the ultrafilter points on the set of coordinates that are important. It is clear that we preserve all algebraic
operations. We need to check that we arrived at a total ordering ≤p
∗
31
as the quotient of ≤N by the equivalence relation ∼p . Let be given two
non-standard numbers, i.e. equivalence classes of sequences. Let x and
y be their representatives. Then
x = y ⇐⇒ x ∼p y,
i.e. they are equal only if they are in the same equivalence class, i.e.
N ∖ (S ∪ G) ∈ p. Otherwise S ∪ G ∈ p and exactly one of G or S is
contained in p. That means we can compare any two non-equivalent
sequences. The latter fact is due to the maximality of p.
16.1. Uniqueness of the field ∗ R. We constructed the hyper-real
numbers with the help of an ultrafilter p. The question arises whether
the hyper-reals depend on this choice. The answer is in general: Yes!
However if we assume (CH) then all the obtained fields are isomorphic.
In this section we will prove this statement using ideas of Hausdorff and
Erdős, Gillman, Henriksen. Throughout this section we will assume
(AC) and (CH):
ℵ1 = c = 2ℵ0 .
We first introduce a notion of a special class of well-orderings (going
back to Hausdorff.
Definition 16.1. Let for an ordinal α denote ℵα the corresponding
aleph. We say that a total order ≤ on a set M is a ηα - order if
● ∅ ≠ N, K ⊂ M , m(N ) < ℵα , m(K) < ℵα and N < M Ô⇒ ∃x ∈
M such that N < c < M .
● If ∅ ≠ N ⊂ M , m(N ) < ℵα Ô⇒ ∃x, y ∈ M such that x < N < y.
We are mainly interested in η0 - and η1 - orders.
Example 16.1. (R, ≤) is η0 but not η1 ordered. The properties are
clearly fulfilled for any finite sets N, K ⊂ R but not for Z ⊂ R since
there is neither a lower nor an upper bound. The first condition is
violated for sets of the form N = {0}, M = { n1 ∶ n ∈ N}.
We will make use of dense sets of lower cardinality. For this we will
need the following
Lemma 16.1. In a totally ordered topological space (X, Torder ) where
U ∈ Torder ⇐⇒ U = {x ∈ X ∶ ∃a, b ∈ X such that a < x < b} any ηα
ordering on X is a ηα ordering on any dense subset Y ⊂ X.
Proof. Applying the definition of an ηα ordering to x < K or x > K we
see that there is a y ∈ X such that x < y < K or x > y > K and hence an
entire interval x < z < y respectively y < z < x that must contain points
from the dense set Y ⊂ X.
32
Theorem 16.1. (∗ R,∗ ≤)p is a η1 ordering.
Proof. let p be fixed. Assume we have two countable sets N < K ⊂∗ R.
We enumerate them
⎧
⎧
x11 ⋯x1n ⋯
y11 ⋯yn1 ⋯
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⋯
⋯
⎪
⎪
⎪
⎪
⎪
⎪
N =⎨
,
K=⎨
.
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
xk1 ⋯xkn ⋯
y1k ⋯ynk ⋯
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⋯
⎩
⎩⋯
Since N < K there is a set U ∈ p such that
xkn < ynk
∀k ∈ N and ∀n ∈ U.
We choose ∗ z = z1 ⋯zn ⋯ by
max xkn < zn < min ynk .
1≤l≤n
1≤l≤n
Then for every k ∈ N there is an m(k) ∈ N such that xkn < zn < ynk
for all n > m(k). Since the cofinite sets are contained in p (p is free!)
we conclude N ∗< ∗z ∗< K. The other condition is proved with similar
arguments.
Corollary 16.1. (∗R,≤) is not (order) isomorphic to (R, ≤), since ≤ is
not a η1 -ordering.
Theorem 16.2. Let m(P ) = m(Q) = ℵ1 and ≤P , ≤Q be η1 - orderings on
P and Q respectively. Then (P, ≤P ) and (Q, ≤Q ) are order-equivalent.
Proof. According to the assumption (AC) there are a well-orders ≺P
and ≺Q such that for all p ∈ P and q ∈ Q the sets
{x ∈ P ∶ x ≺P p}
and
{y ∈ Q ∶ y ≺Q q}
are countable.
● We will use a transfinite algorithm, i.e. the number of cycles
(each of which consists of a finite number of steps) is infinite
and ordered by ordinal numbers α. F.e. if we run it ω0 +1 cyclesteps that means we have run it for all integers (i.e. infinitely
often) and when we finished with that we run it one more time.
33
Step 0 Let p1 , respectively q1 be the least elements in (P, ≺P ),
respectively (, ≺Q ). We define
f (p1 ) = q1 .
Step 1 Let q (1) ∈ Q be the least element such that there is no p ∈ P
such that f (p) = q (1) . Let
P< (q (1) ) ∶= {x ∈ P ∶ f (x) is defined already and f (x) <Q q (1) },
P> (q (1) ) ∶= {x ∈ P ∶ f (x) is defined already and f (x) >Q q (1) }.
If there is a p(1) ∈ P such that P< (q (1) ) <P p(1) <P P> (q (1) )
we set
f (p(1) ) = q (1) .
Step 2 Let p(2) ∈ P be the least element such that f (p(2) ) is not
defined yet. Let
Q< (p(2) ) ∶= {f (x) ∈ Q ∶ f (x) is defined already and x <P p(2) },
Q> (p(2) ) ∶= {f (x) ∈ Q ∶ f (x) is defined already and x >P p(2) }.
If there is a q (2) ∈ Q such that Q< (p(2) ) <Q q (2) <Q Q> (q (2) )
we set
f (p(2) ) = q (2) .
Step 3 Return to step 1.
● Lets enumerate the number of cycles run by the algorithm by
an ordinal β cyc .
● The algorithm does not terminate at a β0cyc ≺cyc ω1 .
– Assume the algorithm has run for all β cyc ≺cyc β0cyc . Since
β0cyc ≺cyc ω1 it has been running only countably many cycles. Therefore at each cycle enumerated by β cyc we have
(1)
(2)
(1)
(2)
previously defined f at points pβ cyc , pβ cyc , qβ cyc , qβ cyc . All
together these are at most countably many points.
– Then all the four sets P< (q (1) ), P> (q (1) ), Q< (p(2) ), Q> (p(2) )
are at most countable by assumption and P< (q (1) ) <P P> (q (1) )a
and Q< (p(2) ) <Q Q> (p(2) ). By the η1 - ordering properties
(1)
(2)
(1)
(2)
of ≤P and ≤Q the numbers pβ cyc , pβ cyc , qβ cyc , qβ cyc can be de0
0
0
0
fined and step 1 and step 2 can be performed (That we
(1)
(1)
(1)
defined pβ cyc , qβ cyc = f (pβ cyc ) in step 1 does not changed
0
0
0
countability for step 2 in the same cycle).
34
Assume we have performed all β cyc ≺cyc ω1 cycles and there
is a p ∈ P left for which f is not yet defined. Let p0 be the
least of it (w.r.t. ≺P ). Since by the choice rule (least element)
∀β cyc ≺cyc γ cyc ≺cyc ω1 we have
pβ cyc ≺P pβ cyc ≺P pγ cyc ≺P pγ cyc ≺P p0
(1)
(2)
(1)
(2)
(Note that the latter inequalities are w.r.t. ≺P !). But this
implies that p0 has its initial segment w.r.t. ≺P of order type
⪰P ω1 contradicting the assumption. A similar argument shows
that we have at step ω1 of cycle runs exhausted the set Q.
● By the choices in step 1 and step 2 we ensured that f is transforming ≤P into ≤Q and hence is order-preserving.
● f ∶ (P, ≤P ) ⇆≤ (Q, ≤Q ) is an order-preserving bijection.
Theorem 16.3. Let p, q ∈ βN ∖ i (N) be two free ultrafilters. Then
assuming (CH) (∗ R,∗≤)p is order-equivalent to (∗ R,∗≤)q .
Proof. Since R ⊂j ∗ Rr ⊂ ∏n∈N R and ℵ0 = m(N) < 2ℵ0 = c we have
c = m(R) ≤ m(∗ Rr ) ≤ m (⋃n∈N R) = c for any ultrafilter r. Now (CH)
implies that m((∗ R,∗≤)p ) = m((∗ R,∗≤)q ) = ℵ1 . Since both models of ∗ R
(with respect to p and q) are η1 - ordered Theorem 16.2 implies the
assertion.
Remark 16.1. One can generalize the main argument to the operations
∗+ and ∗⋅ on ∗ R with respect to p and q. For this one fixes a dense Hamel
basis for each of ∗ Rp and ∗ Rq . Then one has to control a countable
number of (polynomial) equations over Q on countable arguments from
the basis. This is by transfinite induction by a (CH) induced wellordering on R and ∗ R with similar (but more involved) arguments as in
the proof of Theorem 16.2. Finally we will use a version of Lemma 16.1
to step from Q to R and from a dense set of equations to its closure.
We then arrive at the following statement.
Theorem 16.4 (Erdős, Gillman, Henriksen). Assume (CH). Let p, q ∈
βN ∖ i (N) be two free ultrafilters. Then the two fields
(∗ R,∗≤,∗+,∗⋅,∗ 0,∗ 1)p ≃ (∗ R,∗≤,∗+,∗⋅,∗ 0,∗ 1)q
are isomorphic.
17. Ultrafilter in topology
Definition 17.1. A set is called a topological space if there is a
collection of sets T ∈ P(P(X)) with the properties
35
● ∅ ∈ T and X ∈ T .
● Uα ∈ T Ô⇒ ⋃α Uα ∈ T .
● Ui ∈ T , i = 1, ⋯, n Ô⇒ ⋂ni=1 Ui ∈ T .
The sets U ∈ T are called open sets. Their complements are called
closed. A set V ⊂ X with x ∈ U ⊂ V, U ∈ T is called a neighborhood
of x.
Definition 17.2. A topological space (X, T ) is called a Hausdorff or
T2 -space if for all distinct x, y ∈ X there are open sets x ∈ U , y ∈ V such
that U ∩ V = ∅.
Definition 17.3. A subset Y ⊂ X of a topological space is called compact if for every open cover ⋃α Uα ⊃ Y there are indices α1 , ⋯, αn such
that ⋃ni=1 Uαi ⊃ Y , i.e. any cover has a finite subcover.
Lemma 17.1. Any closed subset of a compact space is compact.
Proof. Let {Cα ∶ α ∈ I} be an arbitrary cover by open sets of the closed
set F ⊂ X and X is compact. Then {Cα ∶ α ∈ I} ∪ {X ∖ F } is an open
cover of the compact set X. So we can extract a finite subcover that
covers X and hence also F .
Definition 17.4. A subset Y ⊂ X of a topological space is called sequentially compact if any sequence (xn )n∈N , xn ∈ Y contains a converging (to a point of Y ) subsequence limk→∞ xnk = y ∈ Y .
As we will see later these two definitions of compactness are independent.
Sometimes it is convenient not to consider the collection of all open
sets but rather some specific subsets that nevertheless carry the needed
information.
Definition 17.5. A collection B ⊂ T is called a subbasis for the topology if every open set U ∈ T is the arbitrary union of finite intersections
of elements in B, i.e. the topology T is generated by the subbasis under
finite intersections and arbitrary unions.
Definition 17.6. Let (Xα , Ti ), α ∈ I be a family of topological spaces.
The product topology (Tychonov product topology) T of the product
X = ∏ Xα is generated by the subbasis
α∈I
Uβ ∶= ∏ Vα ,
α∈I
where
V α = Xα
{
Vβ ∈ Tβ
α≠β
.
Definition 17.7. A topological Hausdorff space (X; T ) is said to be
totally disconnected if there is a subbasis B such that every element
U ∈ B is at the same time open and closed.
36
Remark 17.1. The above definition is stronger than the standard definition of a totally disconnected space. It is usually called a totally
separable space. However in the case of (locally) compact spaces this
definition is the same.
Example 17.1.
● Let X be any set. Then T = {∅, X} defines the trivial topology. If X contains at least two points this topology is not Hausdorff. Also any subset is compact (there are only two open sets
altogether) but only ∅ and X are closed!
● Let X be any set and T = P(X) (This topology is called the
discrete topology.). This topology is Hausdorff. Any subset
is at the same time open and closed. The only compact sets
are the finite subsets (Cover an infinite set by its points. This
open cover has no finite subcover.). Such spaces are totally
disconnected (even so not compact).
X
● Let X be any infinite set and T = Fcof
∪ {∅} (This topology is
called the cofinite topology.). This topology is not Hausdorff,
any non-empty U, V ∈ T have a non-empty (infinite) intersection. Any subset is compact: Let Y ⊂ X and ⋃ Uα ⊃ Y be
an open cover, i.e. Uα are cofinite. Choose Uα1 ≠ ∅. Then
Y ∖ Uα1 = {x1 , ⋯, xn } - a finite set. Now we choose Uαi ∋ xi ,
1 ≤ i ≤ n a finite subcover.
● Any finite space with the discrete topology is a compact totally
disconnected Hausdorff space.
● The Cantor set is the ”standard” example of a compact totally
disconnected Hausdorff space.
Definition 17.8. Let (Xi , Ti ), i = 1, 2 be topological spaces and f ∶ X1 →
X2 a mapping. F is said to be continuous if
∀V ∈ T2 Ô⇒ f −1 (V ) ∈ T1 .
Remark 17.2. Since f −1 (A ∖ B) = f −1 (A) ∖ f −1 (B), ∀A, B ⊂ X2 , we
have the following alternative definition of continuity:
f is continuous if and only if the inverse f −1 (F ) of any closed set
F ⊂ X2 is closed in X2 .
Definition 17.9. A (semi-) group (G, ⋅) is called a left topological
(semi-) group if it carries a Hausdorff topology such that the (left)
group operation g → h ⋅ g and g, h ∈ G, is continuous for any fixed h.
The following proposition can be found in any book on topology.
37
(1)
Proposition 17.1. X is a separable metric space Ô⇒ X is a Haus(2)
(3)
dorff space Ô⇒ All compact subsets of X are closed Ô⇒ The family
of compact subsets has the centered finite intersection property.
Proof.
(1)
Ô⇒ For x ≠ y in X set U (x) = {z ∈ X ∶ d(x, z) < 21 d(x, y)} and
U (y) = {z ∈ X ∶ d(y, z) < 12 d(x, y)}. Then x ∈ U (x), y ∈ U (Y )
and U (x) ∩ U (y) = ∅.
(2)
Ô⇒ let K ⊂ X be compact. If K = X the statement is true since X
is closed by definition. So assume K ⊊ X and x ∈ X ∖ K. For
y ∈ K let Uy (x), Ux (y) be open sets such that Ux (y)∩Uy (x) = ∅.
Then ⋃y∈K Ux (y) is an open cover of K and we can extract a
n
finite subcover ⋃N
k=1 Ux (yk ) ⊃ K. The set U (x) = ⋂k=1 Uyk (x) is
open and disjoint from K. Hence, X ∖ K is open.
(3)
Ô⇒ Let Kα , α ∈ I be a family of compact sets with ⋂α∈I Kα = ∅.
Choose a non-empty compact set K ∈ {Kα ∶ α ∈ I}. Since Kα
is closed the sets K ∖ Kα are open and
⋃ (K ∖ Kα ) = K ∖ ( ⋂ Kα ) = K.
α∈I
We extract a finite subcover
α∈I
n
⋃i=1 (K
∖ Kαi ) ⊃ K. Since
n
n
K ⊂ ⋃(K ∖ Kαi ) = K ∖ ⋂ Kαi
we conclude K
i=1
∩ ⋂ni=1 Kαi
i=1
= ∅.
Definition 17.10. Let F be a filter on X. we say that F converges
to x ∈ X if any neighborhood U (x) ∋ x is in the filter, i.e. U (x) ⊂ F.
We say that x is a limit point of F and denote it by x ∈ F − lim.
A filter basis M coverges to a point x ∈ X if a containing filter F ⊃ M
converges to x.
Remark 17.3. M − lim ⊂ F − lim for any filter F ⊃ M.
Example 17.2.
● If px is the principle filter on any topological space X ∋ x it
converges to the point x only.
● If X is a non-finite set equipped with the cofinite topology T =
X
X
Fcof
then any point x ∈ X is a limit point of Fcof
.
Proposition 17.2. If (X, T ) is a Hausdorff space then any flilter has
at most one limit point.
38
Proof. Assume that F − lim ∋ x ≠ y ∈ F − lim. Since X is Hausdorff we
find two neighborhoods x ∈ U (x) ∈ F, y ∈ U (y) ∈ F, U (x) ∩ U (y) = ∅ ∉
F.
Theorem 17.1. A topological space X is compact if and only if any
ultrafilter p on P(X) converges.
Proof. ( Ô⇒ ) Let F be an ultrafilter that does not converge: ∀x ∈
X ∃x ∈ U (x) ∉ F. Assuming (any neighborhood contains an open
neighborhood) that allU (x) are openwe get a cover ⋃x∈X U (x) ⊃ X. by
compactness we can extract a finite subcover ⋃nk=1 U (xk ) ⊃ X. since X ∈
F there is at least one 1 ≤ k ≤ n such that U (xn ) ∈ F - a contradiction.
( ⇐Ô ) assume that X is not compact: ∃{Uα ∶ α ∈ I}, Uα open, such
that for all finite subsets {α1 , ⋯, αn } ∈ P(I) we have X ∖ ⋃ k = 1n Uαk ≠
∅, or ⋂nk=1 (X ∖ Uαk ) ≠ ∅. So {X ∖ Uα ∶ α ∈ I} has the finite intersection property and can be extended to an ultrafilter p by UFL. By
the covering property for each x ∈ X there is (at least) one x ∈ Uα .
But if p ∈ p − lim then Uα ∈ p but on the other hand (X ∖ Uα ) ∈ p - a
contradiction.
Lemma 17.2. Let f ∶ X → Y and p an ultrafilter on X. Then
f (p) = {Z ⊃ f (A) ∶ A ∈ p}
is an ultrafilter on Y .
Proof. Obviously , ∅ is not in the image of any set A ⊂ X, hence
∅ ∉ f (p).
If C, D ∈ f (p), i.e. C ⊃ f (A), D ⊃ f (B), A, B ∈ p then
C ∪ D ⊃ C ∩ D ⊃ f (A) ∩ f (B) ⊃ f (A ∩ B) ∈ f (p)
and
(Y ∖ C) ⊂ (Y ∖ f (A)) ⊂ f (X ∖ A) ∉ f (p).
This shows that f (p) is a filter. We are going to show that it is actually
an ultrafilter.
Let Y ⊃ N ∉ f (p), i.e. f −1 (N ) ∉ p, i.e. L = (X ∖ f −1 (N )) ∈ p. Since
L ⊂ f −1 (f (L)) we have f −1 (f (L)) ∈ p, hence f (L) ∈ f (p). Since
f (L) ∩ N = f (X ∖ f −1 (N )) ∩ N = f ({x ∈ X ∶ f (x) ∉ N }) ∩ N = ∅
we have f (L) ⊂ (Y ∖ N ) and so X ∖ N ∈ f (p).
Remark 17.4. Let X be a topological space. A sequence (xn )n∈N is a
map f ∶ N → X, f (n) = xn . Let p be an ultrafilter on N. Then f (p) is
an ultrafilter on X. We can define
p − lim xn = x ∶= f (p) − lim .
n
39
By the very definition of the (ultra-) filter limit of f (p) this means
∀x ∈ U (x) ∈ T ,
∃N ∈ p such that f (N ) ⊂ U (x).
But f (N ) = {xn ∶ n ∈ N }. On the other hand
f (N ) ⊂ U (x)
⇐⇒
N = {n ∈ N ∶ f (n) ∈ U (x)} ∈ p.
This leads to the equivalent definition
p−lim xn = x
n
⇐⇒
(∀x ∈ U (x) ∈ T
Ô⇒ {n ∈ N ∶ xn ∈ U (x)} ∈ p).
We will note that if limn→∞ x−n = x exists in the classical meaning and
p is any free ultrafilter then p limn xn = x, since for any free ultrafilter
N
⊂ p.
Fcof
17.1. Stone–Čech compactification of N. In this section we will
introduce a construction that is very helpful in analysis, algebra or
combinatorics. It is essentially the same as the construction of the
Stone space before, but is more concrete and connected to topology.
While the general theory holds for ”most” topological spaces we will
restrict ourselves toN. The Stone-Čech compactification βN of N is a
nice way to collect topological and algebraic properties of ultrafilters
on N.
In this section we equip N with the discrete topology Tdiscrete .
Definition 17.11. We say that a subset Y ⊂ X of a topological space
(X, T ) is dense iff ∀U ∈ T we have Y ∩ U ≠ ∅.
Definition 17.12. Let (X, T ) be an arbitrary topological space. We
say the pair (i , K) is a compactification of X if K is a compact
Hausdorff space and i∶ X ↪ K is a continuous injection such that i(X)
is dense in K.
Definition 17.13. The Stone-Čech compactification βN of the
natural numbers N is defined by
● βN consists of all ulltrafilters p on N.
● The topolgy TβN is given by
– A basis is defined as: U ⊂ βN is declared to be open if it is
of the form
U = ⟨N ⟩βN ∶= {p ∈ βN ∶ N ∈ p},
N ⊂ N.
– V ∈ T iff V is a union of elements in the basis.
Lemma 17.3. TβN is a topology.
Proof.
● ∅ = {p ∈ βN ∶ N = ∅ ∈ p} ∈ TβN .
40
● If U1 , ⋯, Un are open sets (elements of the basis are sufficient)
then
n
∩nk=1 Uk = ⋂{p ∶ Nk ∈ p} = {p ∶ Nk ∈ p ∀k} = {p ∶ ⋂ Nk ∈ p} ∈ TβN .
k=1
This is a consequence of
n
⋂ Nk ∈ p
⇐⇒
Nk ∈ p ∀1 ≤ k ≤ n
k=1
and ⋂nk=1 Nk ⊂ Nk ∀k.
● Any union of open sets is open by definition.
● βN = {p ∶ p is an ultrafilter on N} = {p ∈ βN ∶ N = N ∈ p}.
Lemma 17.4. The open sets ⟨N ⟩βN are also closed, hence clopen.
Proof. βN ∖ ⟨N ⟩βN = ⟨N ∖ N ⟩βN since an ultrafilter either contains N or
its complement but not both.
Lemma 17.5. The injection i∶ N ↪ βN given by i(n) = pn is continuous.
Proof.
● i is a bijection onto its image: i (n) = i (m) ⇐⇒ pn = pm since
pn = {L ⊂ N ∶ n ∈ L} = {L ⊂ N ∶ m ∈ L} = pm Ô⇒ n = m.
● Let U = ⟨N ⟩βN , n ∈ N be a basic open set containing i (n). Then
i −1 (U ∩ i (N)) = N.
Since N carries the discrete topology Tdiscrete any subset N ⊂ N
is open.
Lemma 17.6. The continuous injection i∶ N → βN has a dense image,
i.e. ∀U ∈ TβN we have i (N) ∩ U ≠ ∅.
Proof. Let U = ⟨N ⟩βN , N ∈ P(N) be a basic open set. If N ≠ ∅ it
containes some n ∈ N . But then pn ∋ N and U ∋ pn = i (n).
We will now introduce a semi-group structure ”⊕”, respectively
”⊗” on βN that is inherited from the semi-group structure on N. In
thios case both addition ”+” and multiplication ”⋅” work. To motivate the definition we recall that an ultrfilter p can be interpreted as a
finitely additive 0-1-measure µp ∶ P(N) → {0, 1}. for measureson spaces
with a (semi-) group operation we have the canonocally induced operation for measures: the convolution of measures µp⊕,⊗q ∶= µp ∗ µq where
41
”∗” is induced by ”+” or ”⋅”. We recall that for two measure µ, ν on a
semi-group (G, ⋅) the convolution is defined as
µ ∗ ν(A) ∶= ∫ (∫ 1A⋅x−1 (y) dµ(y)) dν(x).
G
G
More precisely,
M ⊃N ∈p⊕q
⇐⇒
{m ∈ N ∶ N − m ∈ p} ∈ q
and
M ⊃N ∈p⊗q
⇐⇒
{m ∈ N ∶ N
1
∈ p} ∈ q
m
where N − m ∶= {k ∈ N ∶ k + m ∈ N } and N m1 ∶= {k ∈ N ∶ km ∈ N}.
Lemma 17.7. Each of the operations ⊕ and ⊗ turn βN into a (additive, respectively multiplicative) semi-group.
Proof. We proceed for ⊕ only. The proof for ⊗ is analogously.
● p ⊕ q is an ultrafilter:
– ∅ ∉ p ⊕ q: otherwise ∅ ∈ q- acontradiction.
– If K ∈ p ⊕ q then L ∈ p ⊕ q, ∀L ⊃ K: Since for each m ∈ N
we have K − m ⊂ L − m we get
q ∋ {m ∈ N ∶ K − m ∈ p} ⊂ {m ∈ N ∶ L − m ∈ p} ∈ q.
– If K, L ∈ p ⊕ q then K ∩ L ∈ p ⊕ q: Since for each m ∈ N we
have (K ∩ L) − m = (L − m) ∩ (K − m) we get
q ∋ {k ∈ N ∶ K − k ∈ p} ∩ {l ∈ N ∶ L − l ∈ p}
= {m ∈ N ∶ (K ∩ L) − m ∈ p} ∈ q.
– ∀N ⊂ N Ô⇒ N ∈ p ⊕ q or N ∖ N ∈ p ⊕ q.
Assume N ∉ p ⊕ q then
{m ∈ N ∶ N − m ∈ p} ∉q ⇐⇒ N ∖ {m ∈ N ∶ N − m ∈ p} ∈ q
⇐⇒ {m ∈ N ∶ N − m ∉ p} ∈ q
⇐⇒ {m ∈ N ∶ N ∖ (N − m) ∉ p} ∈ q
⇐⇒ {m ∈ N ∶ (N ∖ N ) − m ∉ p} ∈ q
⇐⇒ N ∖ N ∈ p ⊕ q.
42
● (p ⊕ q) ⊕ r = p ⊕ (q ⊕ r).
N ∈ (p ⊕ q) ⊕ r ⇐⇒ {m ∈ N ∶ N − m ∈ p ⊕ q} ∈ r
⇐⇒ {m ∈ N ∶
∶ {k ∈ N ∶ N − m − k ∈ p} ∈ q} ∈ r
⇐⇒ {m ∈ N ∶
∶ {l − m ∈ N ∶ N − l ∈ p} ∈ q} ∈ r
⇐⇒ {m ∈ N ∶
∶ {l ∈ N ∶ N − l ∈ p} − m ∈ q} ∈ r
⇐⇒ {l ∈ N ∶ N − l ∈ p} ∈ q ⊕ r
⇐⇒ N ∈ p ⊕ (q ⊕ r).
Lemma 17.8. βN is a compact left topological semi-group.
Proof.
● βN is compact: Assume that this is not the case, i.e. there is
an open cover
⋃ ⟨Nα ⟩βN ⊃ βN,
Nα ⊂ N
α∈I
such that no finite subcover can be extracted, i.e. ∀α2 , ⋯, αn ∈ I
we have
n
⋂ (βN ∖ ⟨Nαk ⟩βN ) ≠ ∅.
k=1
We proceed
βN ∖ ⟨Nαk ⟩βN = βN ∖ {p ∶ N ∈ p} = {p ∶ N ∉ p} = {p ∶ N ∖ Nαk ∈ p}
hence,
n
n
k=1
k=1
∅ ≠ ⋂ (βN ∖ ⟨Nαk ⟩βN ) = ⋂ {p ∶ N ∖ Nαk ∈ p} = {p ∶ ∀k N ∖ Nαk ∈ p}
n
= {p ∶ ⋂ (N ∖ Nαk ) ∈ p}
k=1
By UFL the familty {N ∖ Nα ∶ α ∈ I} extends to an ultrafilter
p ∈ βN. Now p ∈ ⟨Nα ⟩βN for some α ∈ I and also N ∖ Nα ∈ p - a
contradiction.
● βN is Hausdorff: Let βN ∋ p ≠ q ∈ βN. Then there is a set N ⊂ N
such that N ∈ p but N ∉ q (we cannot have p ⊂ q since p is a
43
maximal filter). But then N ∖ N ∈ q (q is an ultrafilter too).
Hence,
p ∈ ⟨N ⟩βN ,
q ∈ ⟨N ∖ N ⟩βN
and
⟨N ⟩βN ∩ ⟨N ∖ N ⟩βN = ∅.
● ⊕ (respectively ⊗) is continuous from the left. We will prove
the case of ⊕. The other is similar.
For fixed p0 ∈ βN consider the map ∶ βN → βN given by (p) =
p0 ⊕p. Let N ⊂ N and consider the basic clopen set ⟨N ⟩βN ∈ TβN .
If ⟨N ⟩βN ∩ (βN) = ∅ then −1 (⟨N ⟩βN ) = ∅ which is open. So we
assume that ⟨N ⟩βN ∋ p0 ⊕ p for some ultrafilter p ∈ βN:
N ∈ p0 ⊕ p ⇐⇒ {n ∈ N ∶ N − n ∈ p0 } ∈ p.
But then
N0 ∶= {n ∈ N ∶ N − n ∈ p0 }
is an element of p. Now
−1
(⟨N ⟩βN ) = {q ∈ βN ∶ N0 ∈ q} = ⟨N0 ⟩βN
is open.
Lemma 17.9. βN is totally disconnected. The points i (n) = pn are
isolated points.
Proof.
● The open sets ⟨n⟩βN ∋ pn actually equals pn since it is the only
ultrafilter containing the point n (n ∈ p ⇐⇒ p = pn ). Hence,
pn is isolated.
● Let ⟨N ⟩βN be a non-empty set of the basis of TβN . Then by the
ultrafilter properties
βN ∖ ⟨N ⟩βN = ⟨N ∖ N ⟩βN .
Hence, any set of the basis of TβN is at the same time open and
closed.
Lemma 17.10. pn ⊕ pm = pn+m and pn ⊗ pm = pnm .
44
Proof.
N ∈ pn ⊕ pm
⇐⇒
{k ∈ N ∶ N − k ∈ pn } ∈ pm
⇐⇒
{k ∈ N ∶ n ∈ N − k} ∈ pm
⇐⇒
⇐⇒
m ∈ {k ∈ N ∶ n + k ∈ N }
n+m∈N
⇐⇒
⇐⇒
{n + m} ∈ pn ⊕ pm
pn ⊕ pm = pn+m .
Lemma 17.11. βN ∖ i (N) is a compact left topological semi-group.
Proof.
● For each n ∈ N the point i (n) = pn is open (and closed) in βN.
Hence,
i (N) = ⋃ pn
n∈N
is open as a union of open sets. Hence, βN ∖ i (N) is a closed
subset of a compact Hausdorff space and therefore compact.
● It is left to prove that p.q ∈ βN ∖ i (N) Ô⇒ p ⊕ q ∈ βN ∖ i (N).
Assume at contrary that p ⊕ q = pn for some n ∈ N. Then
n ∈ p ⊕ q, i.e.
{m ∈ N ∶ {n} − m ∈ p} ∈ q.
But {{n} − m ∶ m ∈ N} is a finite set (as a subset of N) and
hence, the ultrafilter p must be principal.
Theorem 17.2 (Ellis). Any compact left topological semi-group (Γ, ⋅)
contains an idempotent element.
Proof.
● By Hausdorff’s maximal Chain Theorem any chain of compact
left topological semi-groups ⋯ ⊃ Gα ⊃ ⋯, Gα ⊂ Γ is contained in
a maximal chain Cmax .
● By the centered finite intersection property of compact sets (Γ
is compact and Hausdorff!) the intersection
G0 ∶=
⋂ G ≠ ∅.
G∈Cmax
It is also closed as the intersection of closed (since compact in
a Hausdorff space) sets and hence, compact.
45
● G0 is minimal since G0 ⊂ G, ∀G ∈ Cmax and Cmax is a maximal
chain.
● It is straightforward to check (by definition) that G0 is a left
topological semi-group.
● Fix p0 ∈ G0 . Then p0 ⋅G0 ⊂ G0 is closed under multiplication and
hence, compact (by continuity of g, g(p) = p0 ⋅ p) left topological
semi-group contained in G0 . By minimality of G0 we have p0 ⋅
G0 = G0 and p0 ∈ p0 ⋅ G0 . Consequently,
p0 ∈ S ∶= {q ∈ G0 ∶ p0 ⋅ q = p0 } = g −1 (p0 ) ∩ G0 .
● S is a compact left topological semi-group:
p0 ⋅ (q − 1 ⋅ q2 ) = (p0 ⋅ q1 ) ⋅ q2 = p0 ⋅ q2 = p0 ,
q1 , q2 ∈ S.
● Hence, S = G0 and p0 ∈ S. The latter implies p0 ⋅ p0 = p0 and
moreover, S = {p0 } since {p0 } is then a compact left topological
semi-group itself.
Theorem 17.3. βN contains an idempotent element that is a free
ultrafilter, i.e.βN ∖ i (N) contains an idempotent element.
Proof. Apply Theorem 17.2 and Lemma 17.11 to βN ∖ i (N).
Remark 17.5. The Stone-Čech compactification of N (even for all
discrete spaces) is universal in the sense that given a compact hausdorff
space K and a continuous map f ∶ N → K there is a unique continuous
continuation βf ∶ βN → K such that f = βf ○ i. This continuation is
defined by: Let p ∈ βN then f (p) is an ultrafilter on K. Since K is
compact Hausdorff f (p) converges to a single point w ∈ K. We set
βf (p) = w.
This map has the desired properties.
This property actually defines the Stone-Čech compactification uniquely
up to homeomorphism. It is the ”largest” compactification in this sense.
17.2. Proof of Tychonov’s Theorem.
Tychonov’s Theorem is equivalent to AC. (AC Ô⇒ Tychonov’s Theorem.)
First we note that by AC K ≠ ∅ if Kα ≠ ∅, α ∈ I: Let f ∶ {Kα } →
⋃α∈I Kα , f (Kα ) ∈ Kα be a choice function. Then (fα ) ∶= (f (Kα ))α∈I ∈
K. Hence, by UFL there is an ultrafilter on K extending (fα )α .
Let p be an ultrafilter on K = ∏α∈I Kα . The projections πα ∶ K → Kα
are continuous and πα (p) is an ultrafilter on Kα . Since Kα is compact
this ultrafilter converges. Using AC we can choose simultaneously for
46
all α ∈ I a point xα ∈ πα (p − lim) (In case that all the spaces Kα are
Hausdorff we do not need AC!!!). We will prove that (xα )α ∈ p − lim
(the ultrafilter converges), i.e. any neighborhood U ((xα )α ) is contained
in p.
● Let Uα ∶= {(yα )α ∈ K ∶ yβ ∈ Kβ , Yα ∈ U (α) ⊂ Kα }.
● By the definition of the product topology
n
U ((xα )α ) ⊃ Uα1 ,⋯,αn ((xα )α ) = ⋂ Uαk ((xα )α ).
k=1
● π −1 () = Uαk ((xα )α ).
● By compactnes of Kα and Theorem 17.1
παk (p) ∋ U (αk ) (xαk ) ⊃ παk (Ak )
with Ak ∈ p. Therefore
n
n
k=1
k=1
U ((xα )) ⊃ ⋂ πα−1k (U (αk ) (xαk )) ⊃ ⋂ Ak ∈ p.
We conclude that any ultrafilter on K converges.
(Tychonov’s Theorem Ô⇒ AC.)
The main point is to prove that the product of non-empty compact
sets is non-empty knowing that it is compact. So let {Xα ≠ ∅}α∈I be a
family of non-empty sets. We consider the product ∏α∈I Xα and want
to prove that it is non-empty. In that case it contains a point (xα )α∈I
with xα ∈ Xα . That gives us the desired choice function.
We want to introduce a topology on Xα that makes it a compact
space. However it shows to be more convenient not to use the trivial
topology on Xα but rather to enlarge our sets.
● We define Kα ∶= Xα ∪ {∞α }.
Xα
● We equip Kα with the topology Tα = Tcof
∪{∞α }, i.e. we add to
the open sets from th cofinite topology the singleton {∞α } as
an open (and closed) set. This turns Kα inot a compact space.
● The sets Xα are closed and πα are continuous. Hence, π −1 (Xα )
is closed.
● We have by the definition of πα :
∏ Xα = ⋂ π −1 (Xα ).
α∈I
α∈I
We will prove that each of the inverse images is non-empty and
has the finite intersection property.
47
● Let αk ∈ I, k = 1, ⋯, n. Then Xα1 × ⋯ × Xαn ≠ ∅ since it is the
finite product of non-empty sets. Consider the point
xα = xα k
(xα )α∈I = {
xα = ∞ α
α = αk , k = 1, ⋯, n
else
where (x+alpha1 , ⋯, xαn ) ∈ Xα1 × ⋯ × Xαn . Here we used the
additional points ∞α ∈ Kα to avoid the choice for the remaining
infinite number of coordinates. Obviously
(xα )α∈I ∈ πα−1k (Xαk ),
k = 1, ⋯, n.
● For any choice of (α1 , ⋯, αn ) the set
n
⋂ πα−1k (Xαk ) ≠ ∅.
k=1
● By the centered finite intersection property of closed subsets of
a compact set we have
∏ Xα = ⋂ πα−1 (Xα ) ≠ ∅.
α∈I
α∈I
Theorem 17.4. Tychonov’s Theorem for compact Hausdorff spaces is
equivalent to the Ultrafilter Lemma.
Proof.
( ⇐Ô ) follows from the proof of Tychonov’s Theorem.
( Ô⇒ ) Given an arbitrary non-empty set X with a filter F we
want to show the existence of an ultrafilter p ⊃ F. For this we will
construct (with the help of UFT) a convenient representation of families of subsets as elements of a totally disconnected compact Hausdorff
space (Stone space). Then we will use topological properties to find
the desired ultrafiltet. The Stone space itself is a very important construction in general topology. It is the initial step for constructing
the Stone-Čech-compactification of a completely regular Hausdorff
space (Tychonov space). This construction is a rather abstract generalization of the one from section 17.1 for the case of N.
● We identify families of subsets of X with
2P(X) ∼ Y = {0, 1}P(X) = {f ∶ P(X) → {0, 1}}.
We equip 2P(X) with the following topology: we identify
{0, 1}P(X) ∼
∏ {0, 1}
A∈P(X)
48
where {0, 1} carries the discrete topology and 2P(X) the inherited product topology. Then by Tychonov’s Theorem for compact Hausdorff spaces (Kα = KA = {0, 1}) the space 2P(X) is a
non-empty compact Hausdorff space. For each A ∈ P(X) the
sets {f ∈ 2P(X) ∶ f (A) = 1} and {f ∈ 2P(X) ∶ f (A) = 0} are
complementary open subsets of 2P(X) and form a subbasis of
the topology on 2P(X) . Hence , ∀A ∈ P(X) the two sets are
at the same time open and closed. That implies that 2P(X) is
totally disconnected.
● We denote by
St(2P(X) ) = {f ∶P(X) → {0, 1} ∶ ∀A, B ⊂ X holds
f (A ∪ B) = 1 ⇐⇒ f (A) = 1 or f (B) = 1,
f (A ∩ B) = 1 ⇐⇒ f (A) = 1 and f (B) = 1,
f (X ∖ A) = 1 ⇐⇒ f (A) = 0}.
So the points of St(2P(X) ) are those maps that map a subset A to a family of sets that contains A and is closed under
finite unions and intersections and also does not contain complements.
X
X
● St(22 ) is a non-empty closed subspace of 22 and hence, compact and totally disconnected.
– St(2P(X) ) ≠ ∅: Given x ∈ X consider the function
fx (B) = {
fx (B) = 1 B ∋ x
fx (B) = 0 otherwise
Then fx ∈ St(2P(X) ).
–
St∪ ∶= {f ∶P(X) → {0, 1} ∶ ∀A, B ⊂ X ∶ f (A ∪ B) = 1 ⇐⇒
⇐⇒ f (A) = 1 or f (B) = 1} ⊂ St(2P (X) )
is closed:
St∪ = ⋂ ({f (A ∪ B) = 1, f (A) = 1} ∪ {f (A ∪ B) = 1, f (B) = 1})∩
A,B⊂X
∩ ⋂ ({f (A ∪ B) = 0, f (A) = 1} ∪ {f (A ∪ B) = 0, f (B) = 1})
A,B⊂X
is the intersection of closed (and open) sets and hence
closed itself.
– Similarely St∩ and St∖ are closed subsets of the compact
space 2P(X) .
– St(2P(X) ) = St∪ ∩ St∩ ∩ St∖ is closed.
49
– St(2P(X) ) is a compact (since it is a closed subset of a
compact space) totally disconnected Hausdorff space.
X
● We define a map ∶ P(X) → Cl(22 ) ⊂ P(St(2P(X) )), where
Cl(Y ) denotes all clopen (open and closed) subsets of Y , by
(A) = {f ∈ St(2P(X) ) ∶ f (A) = 1}. Then for any A ⊂ X the
image (A) is clopen.
● has the properties:
– ∀A ⊂ X the image (A) is clopen by the definition of the
product topology.
– (A) = ∅ then A = ∅: If x ∈ A then fx (A) = 1 and fx ∈ (A).
– ∀A, B ⊂ X we have (A ∪ B) = (A) ∪ (B):
(A ∪ B) = {f ∶ f (A ∪ B) = 1} = {f ∶ f (A) = 1 or f (B) = 1}
= {f ∶ f (A) = 1} ∪ {f ∶ f (B) = 1} = (A) ∪ (B).
– ∀A, B ⊂ X we have (A ∩ B) = (A) ∩ (B):
(A ∩ B) = {f ∶ f (A ∩ B) = 1} = {f ∶ f (A) = 1 and f (B) = 1}
= {f ∶ f (A) = 1} ∩ {f ∶ f (B) = 1} = (A) ∩ (B).
– ∀A ⊂ X we have (A) = 1 and (X ∖ A) = 0 or (A) = 0 and
(X ∖ A) = 1:
(X ∖ A) = {f ∶ f (X ∖ A) = 1} = {f ∶ f (A) = 0}
= St(2P(X) ) ∖ (A).
– is injective: Assume that for A, B ⊂ X we have (A) = (B).
Then
∅ = (A) ∖ (B) = (A) ∩ (St(2P(X) ) ∖ (B))
= (A ∩ X ∖ B).
Hence, ∅ = A ∖ B and similarly ∅ = B ∖ A.
● Let F be a filter on X. Since 2P(X) is Hausdorff and (A) is
compact (since it is a closed (and open) subset of a compact
set) the family {(A) ∶ A ∈ F} has the centered finite intersection
property and the finite intersection property (since F is a filter
and preserves finite intersections). Hence, ⋂A∈F (A) ≠ ∅ and
there is a point p ∈ ⋂A∈F (A). Let p ∶= {A ⊂ X ∶ p ∈ (A)}.
● For A, B ∈ P(X) and A ∪ B ∈ p we have (A ∪ B) = (A) ∪ (B):
(A ∪ B) = {f ∶ f (A ∪ B) = 1} = {f ∶ f (A) = 1} ∪ {f ∶ f (B) = 1}
= (A) ∪ (B).
Hence, p ∈ A or p ∈ B and consequently A ∈ p or B ∈ p.
● Therefore p is an ultrafilter.
50
17.3. Some more statements from topology and analysis.
Example 17.3.
● X = [1, ⋯, ω1 ) equipped with the order topology is not compact
but sequentially compact:
– The open cover X = ⋃α<ω1 [1, ⋯, α) has no countable (and
therefore no finite) subcover.
– Any sequence (αn )n∈N , α ≺ ω1 is by Lemma 13.2 contained
in some [1, ⋯, β ′ ≺ ω1 ) and hence has a least upper bound
γ ≺ ω1 . Hence, γ ∈ X. That γ is the limit point for some
subsequence (αnk )k∈N .
● βN is compact but not sequentially compact.
– βN is compact by Lemma 17.11
– Consider the sequence (pn )n∈N of principal ultrafilters corresponding to n ∈ N. We need to show that it does not
contain a convergent subsequence (pnk )k∈N . Let p be an arbitrary point of βN. Then by the properties of an ultrafilter
at least one of the two disjoint infinite sets
Se = {n2 , ⋯, n2k , ⋯},
So = {n1 , ⋯, n2k+1 , ⋯}
does not belong to p, say Se . Since βN is a Hausdorff space
there is a neighborhood U (p) not containing the infinite
subsequence (pnl )l∈Se and the subsequence (pnk )k∈N does not
converge to p, i.e. does not converge at all since p was
arbitrary.
17.4. The Banach-Alaoglu Theorem.
Definition 17.14. Let M be a set. We can define a Banach space
l∞ (M ) by
l∞ (M ) = {(xα )α∈M ∈ ∏ (−∞, ∞) ∶ ∥(xα )α∈M ∥ ∶= sup ∣xα ∣ < ∞}.
α∈M
α∈M
l∞ (M )∗
The dual space
consists of all continuous linear and bounded
∞
functionals to R on l (M ), i.e.
l∞ (M )∗ ∶= {f ∶ l∞ (M ) → R ∶ f is linear, continuous and bounded }
(here continuity is with respect to the product topology). We introduce
the weak∗ topology on l∞ (M )∗ by a subbasis given by
−1
B ∶= {T(x
((a, b)) ∶ (a, b) is an interval in R , (xα ) ∈ l∞ (M )}
α)
and T(xα ) (f ) = f ((xα )) ∈ R. The unit ball B ⊂ l∞ (M )∗ is defined as
B ∶= {f ∈ l∞ (M )∗ ∶ ∥f ∥ = 1}
51
where ∥f ∥ ∶= sup{∣f ((xα ))∣ ∶ supα ∣xα ∣ ≤ 1}.
Remark 17.6. The space l∞ (M ) can be considered as the space of
all bounded functions φ∶ M → R. It contains all the functions taking
only 0 or 1 as values. Those functions code the subsets of M , i.e.
P(M ) ⊂ l∞ (M ) in a natural way. Hence any element f ∈ l∞ (M )∗
gives by linearity rise to a finitely additive bounded (signed) measure
on M where all subsets are measurable. In particular the set
U ∶= {f ∈ B ⊂ l∞ (M )∗ ∶
∶ f ((xα )) ∈ {0, 1} ∀{(xα ) ∈ l∞ (M ) ∶ xα ∈ {0, 1} ∀α}, f ((1)α ) = 1}
represents finitely additive 0-1-measures on P(M ), hence ultrafilters.
Theorem 17.5 (Banach-Alaoglu). If X is a normed space then the
unit ball B ∗ in its dual X ∗ is weak∗ -compact.
The statement of this theorem follows from the following.
Theorem 17.6. The Banach-Alaoglu Thjeorem 17.5 is equivalent to
the Ultrafilter Lemma 15.1.
Proof.
( Ô⇒ )
– let B be the unit ball in l∞ (M )∗ equipped with the weak∗ topology. By Theorem 17.5 it is compact. The subset
U ∶= {f ∈ B ∶ f ((xα )) ∈ {0, 1}
is clearly weak∗ -closed.
– There is a canonical embedding of δ∶ M ↪ U given by
1 xβ = 1
δ(β) = fβ ((xα )) = {
,
0 xβ = 0
( ⇐Ô )
i.e. assigning the corresponding principal filter pβ .
– Given any fliter F on M the collection F ∶= {δ(F )}
of
F ∈F
closed sets has the finite intersection property. Since U is
compact the intersection ⋂F ∈F δ(F ) ≠ ∅.
– If f ∈ ⋂F ∈F δ(F ) then f ((xα )) ∈ {0, 1} if xα ∈ {0, 1} ∀α and
f ((1)α ) = 1 and hence represents an ultrafliter p containing
F (since for β ∈ F, F ∈ F we have xβ = 1).
– For x ∈ X consider the interval Ix = {a ∈ R ∶ ∣a∣ ≤ ∥x∥}
and let I ∗ = ∏x∈X Ix . By Tychonov’s Theorem for compact Hausdorff spaces I ∗ is compact (we remark that by
Theorem 17.4 this Theorem is equivalent to (UFL)).
52
– On the other hand I ∗ are the functions {φ ∶ X → R ∶
∣φ(x)∣ ≤ ∥x∥ ∀ x ∈ X}. Hence, B ∗ is exactly the subset of
linear functions in I ∗ .
– It is left to show that I ∗ ∖B ∗ is open in the weak∗ topology.
For let φ ∈ I ∗ ∖ B ∗ be a non-linear function. Hence, there
are x1 , x2 ∈ X and 0 ≠ λ1 , λ2 ∈ R such that
φ(λ1 x1 + λ2 x2 ) − λ1 φ(x1 ) − λ2 φ(x2 ) ∉ (−, )
for some > 0. But then for any ψ in the open set (with
z = λ1 x1 + λ2 x2 )
O ∶= {ψ ∈ I ∗ ∶ ∣ψ(z) − φ(z)∣ < , ∣ψ(xi ) − φ(xi )∣ <
, i = 1, 2}
3
3λi
we have ψ ∉ B ∗ , i.e. φ ∈ O ⊂ I ∗ ∖ B ∗ .
17.5. Every vector space has a basis.
Definition 17.15. A family B = {bα ∈ R ∶ α ∈ I} is called a basis iff
∀x ∈ R
∃r1 , ⋯, rn ∈ Q and bα1 , ⋯, bαn ∈ B
such that
n
x = ∑ rk bαk
k=1
and this representation (called a representation over Q with basis
B) is unique.
More generally, let K be a field and V be a linear space over K.
Definition 17.16. A family B = {bα ∈ V ∶ α ∈ I} is called a basis iff
∀x ∈ V
∃r1 , ⋯, rn ∈ K and bα1 , ⋯, bαn ∈ B
such that
n
x = ∑ rk bαk
k=1
and this representation is unique.
Remark 17.7. Assume such a basis for R over Q exists. Then ωAC
implies that m(B) > ℵ0 . Clearly, m(B) ≤ m(R) = c and also m(B) > n,
∀n ∈ N (f.e., there are infinitely many rationally independent algebraic
√
numbers like p, p - a prime number). (By ωAC) B contains a subset
comparable to N and hence, ℵ0 ≤ m(B) ≤ c. But (again using ωAC)
since B is a basis for R:
⎛∞ ⎛
⎞⎞
m(R) ≤ m ⋃
Qn = ℵ 0
⋃
⎝n=1 ⎝bα1 ,⋯,bαn ∈B ⎠⎠
53
if m(B) = ℵ0 .
Theorem 17.7. Every linear space V over a field K has a basis.
Proof.
● For x1 , ⋯, xn ∈ V we define
n
K(x1 , ⋯, xn ) = { ∑ ri xi ∶ ri ∈ K, 1 ≤ k ≤ n} .
k=1
● For an ordinal number β we define
∞
Kβ = ⋃ {K(bα1 , ⋯, bαn ) ∶ αk ≺ β, 1 ≤ k ≤ n} .
n=1
● By the well-ordering principle we can well-order V:
x1 ≺ x2 ≺ ⋯ ≺ xω ≺ xω+1 ≺ ⋯xα ≺ ⋯
● We define
B ∶= {xα ∶ xα ∉ Kα } .
● Any x = xγ ∈ V can be represented in at most one way (not
paying attention to the order of summation) with the basis B
and coefficients in K. For if otherwise
n
m
k=1
k=1
∑ rk bαk = xγ = ∑ sk bβk ,
bαk , bβk ∈ B, rk , sk ∈ K
we have (w.l.o.g. m ≥ n, sm ≠ 0 and βm ≻ αk , βk , k ≠ m)
n
m−1
rk
sk
bα k − ∑
bβk ∈ Kβm
k=1 sm
k=1 sm
bβ m = ∑
which is impossible by the definition of B.
● We are left to prove that any x ∈ V can be represented this way:
If xγ is the least number ( w.r.t. ≺) that cannot be represented
this way then xγ ∉ B since every element 1 ⋅ b = b ∈ B. But if
xγ is least and not chosen to be in B then
n
n
ml
m
xγ = ∑ rk xαk = ∑ rk (∑ sl bβ (k) ) = ∑ tk bγk ,
(k)
k=1
k=1
l=1
l
γk ≺ γ, tk ∈ K
k=1
a contradiction.
As a special case of Theorem 17.7 we derive:
Theorem 17.8 (Hamel). There exist a Hamel basis in R, i.e. a basis
of R as a vector space over Q.
54
Corollary 17.1. There are (discontinuous) non-linear solutions to
Abel’s functional equation f (x + y) = f (x) + f (y), ∀x, y ∈ R.
Proof.
● Any continuous function satisfying the functional equation is
linear: Let f (1) = a. Then
n
f (n) = ∑ f (1) = na
k=1
and
p
p
qf ( ) = f (q ) = f (p) = pa
q
q
Ô⇒
p
p
f( )=a .
q
q
By continuity f (x) = ax.
● It is possible to prove that any measurable function f satisfying
the equation is linear.
● We define for b ∈ B
r x = rb + ∑nk=1 rk bαk , bαk ≠ b
f (x) = {
.
0 else
● Since B is a basis f is defined for all x ∈ R and also fulfills the
equation.
● f is not linear since it is not identically 0 and has more than
one (in fact uncountably many) x ∈ R with f (x) = 0.
● Hence, f is not continuous (in fact not measurable).
18. Ultrafilter in infinite combinatorics and ergodic
theory
Definition 18.1. Let f ∶ X → X be a continuous map of a compact metric space X. A subset Y ⊂ X is called compact minimal invariant
if it is invariant, i.e. f (Y ) = Y , and compact and there is no invariant
compact subset Y ′ ⊊ Y , f (Y ′ ) = Y ′ .
Remark 18.1.
● If Y is minimal compact invariant then any orbit in it is dense:
∀ > 0, ∀x ∈ Y and ∀y ∈ Y ∃ n ∈ N such that d(f n (x), y) < .
This is since the orbit closure of any orbit (i.e. the intersection
of all closed sets containing {f n (x) ∶ n ∈ N}) is invariant by
continuity of f .
55
● Any periodic orbit {f k (x) ∶ k ∈ N, f n (x) = x} is minimal compact invariant.
● Let α ∈ R ∖ Q then for the irrational rotation f ∶ x → x + α
(mod 1) the entire circle T = R/Z is minimal compact invariant.
● If α = pq ∈ Q then for the rational rotation f ∶ x → x + α (mod 1)
and any x ∈ T the orbit {x+ kq ∶ 0 ≤ k ≤ q−1} is minimal compact
invariant. Hence there are uncountably many different minimal
compact invariant subsets.
Theorem 18.1 (Furstenberg). Let f ∶ X → X be a continuous map of
a compact metric space X. Then there is a minimal compact invariant
set K ⊂ X.
Proof. We consider a chain C
⋯ ⊃ Cα ⊃ ⋯ ⊃ Cβ ⊃ ⋯
of compact invariant sets C. Since the intersection of non-empty compact sets is non-empty and the intersection of invariant sets is invariant
it follows by Hausdorff’s Maximal Chain Theorem that there is a maximal element in C with Y ⊂ C, ∀C ∈ C and Y has no proper compact
invariant subset. Hence, Y is minimal compact invariant.
Proposition 18.1 (Corollary to Theorem 17.8). There is a proper
non-empty subset A ⊂ T1 = R/Z which is periodic under every rotation,
i.e.
∀y ∈ R ∃n ∈ N
Ô⇒
A + ny (mod 1) = A.
Proof. Let B be a Hamel basis of R and let
n
A = {x ∈ R ∶ x = ∑ ak bαk ak ∈ Z}.
k=1
Let y ∈ R. Then
m
pk
bβ k .
k=1 qk
y=∑
If n = ∏m
k=1 qk then A = A + ny. Projecting A to the circle gives the
desired set.
Theorem 18.2 (Ramsey). Any complete finitely colored graph has a
complete infinite monochromatic subgraph.
Proof. Let the complete infinite graph
(V, E = (V × V )/(x, y) ∼ (y, x) ∖ {(x, x) ∶ x ∈ V })
56
be finitely colored, i.e. there is a function f ∶ E → {1, ⋯, n} for some
fixed n ∈ N. Since V is an infinite set, there is a free ultrafilter p on V .
We set Vi (x) ∶= {y ∈ V ∶ f (x, y) = i}. We have V ∖ {x} ∈ p and
n
V ∖ {x} = ⋃ Vi (x)
i=1
and this union is disjoint. Next we define g(x) = j, where j is the
unique index such that Vj (x) ∈ p. Let Gi ∶= {x ∈ V ∶ g(x) = i} then
again
n
V = ⋃ Gi
i=1
and this union is disjoint. Let Gi0 be the unique set such that Gi0 ∈ p.
Choose x1 ∈ Gi0 . Continue (weaker than AC) by choosing
n−1
xn+1 ∈ (Gi0 ∩ ⋂ Vi0 (xk )) ≠ ∅
k=1
since it is the finite intersection of elements of a (free) ultrafilter. Note
also that xk ∉ Vi0 (xk ) implies that xk ∉ {x1 , ⋯, xk−1 }
The complete graph with vertices xk , k ∈ N is monochromatic: ∀ m >
n, xm ∈ Vi0 (xn ).
Theorem 18.3 (A finite Ramsey Theorem). Let n, r ∈ N be given.
Then there is a number R = R(n, r) ∈ N such that any coloring by r
colors of the complete graph KR with R vertices contains a monochromatic complete subgraph Kn with n vertices.
Proof. Assume the contrary and let for arbitrary k ∈ N the set ck consist
of those r-colorings of the complete graph Kk of k vertices that do not
have a monochromatic complete subgraph Kn (with n vertices). Let
cm
k ∶= { colorings of Kk+m ∶ there is no monochromatic Kn }.
We enumerate the vertices of Kl by {1, 2, ⋯, l}. Then Kk ⊂ Kk+m
induced by {1, ⋯, k} ⊂ {1, ⋯, k, ⋯, k+m}. Let cm ∣k denote the restriction
m
of cm
k to Kk . By the assumption ck ≠ ∅ and hence by the centered
finite intersection property of finite sets (We equip a finite set with the
discrete topology, then any subset is closed) the set
∞
dk = ⋂ cm ∣k ≠ ∅.
m=0
Moreover dk is the restriction of dk+1 . By revealing this restriction we
get for any m ∈ N a coloring of Kk+m that does not contain a monochromatic graph Kn . Finally we arrive at the infinite complete graph with
vertex set N and a r-coloring that does not contain a monochromatic
57
complete subgraph with n vertices. This is a contradiction to Ramseys
Theorem 18.2.
Theorem 18.4 (Schur). For any r ∈ N there is a number s such that
any r-coloring of {1, 2, ⋯, s} has a monochromatic triple (n, m, n + m).
Proof. Let s = R(3, r) from the finite Ramsey Theorem 18.3. Let the
edges of Ks be colored by r ((k, l)) = r(∣k − l∣) where r∶ {1, ⋯, s} →
{1, ⋯, r} is a given r-coloring. By the finite Ramsey Theorem 18.3 there
is a monochromatic triangle (complete graph on 3 vertices) {a, b, c}, i.e.
for (w.l.o.g.) a < b < c
r (b − a) = r (c − b) = r (c − a).
Let n = b − a, m = c − b. We get n + m = c − a and
r(n) = r(m) = r(n + m).
Theorem 18.5 (Schur, ”Anti”-Fermat’s Theorem for Fp ). For any
n ∈ N there is a prime p(n) such that for all prime numbers p > p(n)
there is a non-trivial solution of Fermat’s equation
xn + y n = z n (mod p).
Proof. If p is a prime number then Fp is a field. The multiplicative
group F∗p is cyclic and has a generator q, i.e. ∀w ≠ 0 ∃m(w) ∈ {0, ⋯, p −
1} such that w = q m(w) (mod p). Let us write
m(w) = nj(w) + i(w)
i(w) ∈ {0, ⋯, m − 1}, j(w) ∈ N.
Let the coloring of w be given by r(w) = i(w). By Theorem 18.4, for
sufficiently large p there are u, v ∈ Fp such that r(u) = r(v) = r(u+v) = i.
Hence,
u = q nj(u)+i , v = q nj(v)+i ,
and u + v = q nj(u+v)+i
(mod p).
Finally we set
x = q j(u) , y = q j(v) and z = q j(u+v)
(mod p).
We arrive at
(xn + y n )q i = xn q i + y n q i = u + v = z n q i
(mod p)
and since Fp is a field
xn + y n = z n (mod p).
58
Definition 18.2. A subset n ⊂ N is called an IP set if there is an
infinite set {n1 , n2 , ⋯, nm , ⋯} ⊂ N such that
∀ m ∈ N, i1 < i2 < ⋯ < ik < ⋯ < im Ô⇒ ni1 + ni2 + ⋯ + nim ∈ N.
Theorem 18.6 (Hindman). Any finite coloring of N has a monochromatic IP-subset.
Proof. Let N = ⋃nk=1 Fk be a coloring of N, i.e. any set Fk corresponds
to one color. Let p = p + p ∈ βN be an idempotent ultrafilter. Then
N ∈ p ⇐⇒ N ∈ p + p ⇐⇒ {m ∈ N ∶ N − m ∈ p} ∈ p.
Now there is exactly one 1 ≤ k ≤ n such that Fk ∈ p and hence {m ∈ N ∶
Fk − m ∈ p} ∈ p. Therefore we can find
m1 ∈ Fk ∩ {m ∈ N ∶ Fk − m ∈ p}
such that Fk1 ∶= Fk ∩ (Fk − m1 ) ∈ p. Continuing this way we can find
m2 > m1 such that m2 ∈ Fk1 ∩ {m ∈ N ∶ Fk1 − m ∈ p} ∈ p and define
Fk2 ∶= Fk1 ∩ (Fk1 − m2 ) = Fk ∩ (Fk − m1 ) ∩ (Fk − m2 ) ∩ (Fk − (m1 + m2 )) ∈ p.
Hence, m1 , m2 , m1 + m2 ∈ Fk . By continuing ad infinitum we obtain an
increasing sequence (mk )k∈N such that for all n ∈ N
{mi1 + ⋯mil ∶ i1 < ⋯il , il < m} ∈ Fk .
Lemma 18.1. Let M ⊂ N be an IP set. Then there is an idempotent
ultrafilter p with M ∈ p.
Proof. Let for (ni )i∈N the sets
Nk ∶= {nj1 + ⋯ + njl ∶ k ≤ j1 < ⋯ < jl , } ∈ M
By definition Nk ∈ p for all p ∈ ⟨Nk ⟩βN . Let Nk denote its closure in βN,
i.e. the intersection of all closed sets containing it. The sets Nk are
closed and the family {Nk }k∈N has the finite intersection property and
hence, N ∶= ⋂k∈N Nk is a non-empty compact set in βN. We are going
to prove that N is a semi-group and hence, contains an idempotent
element.
Let p, q ∈ N . That means that Nk ∈ p, Nk ∈ q, ∀k. We need to show
that
{m ∈ N ∶ Nk − m ∈ p} ∈ q.
Let m ∈ Nk . Then
m = ni1 + ⋯ + nil , k ≤ i1 < ⋯ < il
and
Nk − m ⊃ {nj1 + ⋯ + njn ∶ il < j1 < ⋯ < jn } = Nil +1 ∈ p.
59
Therefore, since Nk ∈ q we get
Nk ⊂ {m ∈ N ∶ Nk − m ∈ p} ∈ q.
Definition 18.3. A subset N ∈ N is called an IP* set if N ∩ A ≠ ∅
for all IP sets A.
Lemma 18.2.
i) Let N ⊂ N be an IP* set then N ∩ A is an IP set for any IP set
A.
ii) N ⊂ N is an IP* set if and only if N ∈ p for any idempotent
ultrafilter p.
iii) A finite intersection of IP* sets is an IP* set.
Proof.
i) Let A be an arbitrary IP set. Then A ∈ p ∈ βN by Lemma 18.1.
Then either N ∩ A ∈ p or (N ∖ N ) ∩ A ∈ p. Hence, at least
one of the two sets contains an IP set (see proof of Hindman’s
Theorem 18.6). Since N is IP* it cannot be (N ∖ N ) ∩ A since
N ∩ ((N ∖ N ) ∩ A) = ∅.
ii) Assume that there is p = p + p and N ∉ p. ThenN ∖ N ∈ p and
there is an IP set A ⊂ N ∖ N (see proof of Hindman’s Theorem 18.6) - a contradiction.
On the other side if N ∈ p, ∀p = p + p and N is not IP*
then there is an IP set A ∩ N = ∅. Then A ⊂ N ∖ N and by
Lemma 18.1 there is q = q + q with A ∈ q. This implies N ∖ N ∈ q
- a contradiction.
iii) Since finite intersections of sets belonging to an ultrafilter belong to that ultrafilter ii) implies iii).
Lemma 18.3. Let X be a compact Hausdorff space and (xn )n∈N a
sequence in X. Then for any p, q ∈ βN:
(p + q) − lim xn = p − lim q − lim xk+l .
n
k
l
In particular for p = p + p
p − lim xn = p − lim p − lim xk+l .
n
k
l
Proof. Let (p + q) − limn xn = x and U (x) ∋ x. Then
{n ∈ N ∶ xn ∈ U (x)} ∈ p + q.
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That means:
q ∋ {k ∈ N ∶ {n ∈ N ∶ xn ∈ U (x)} − k ∈ p} =
= {k ∈ N ∶ {l ∈ N ∶ xl+k ∈ U (x)} ∈ p} ∈ q
or
{k ∈ N ∶ p − lim xk+l ∈ U (x)} ∈ q.
l
The latter implies the statement.
Let ∥x∥ ∶= d(x, Z) denote the distance to the nearest integer.
Theorem 18.7. For any > 0, n ∈ N and polynomials pk ∶ R → R with
pk (0) = 0, k = 1, ⋯, n the set
N ∶= {m ∈ N ∶ ∥pk (n)∥ < , k =, ⋯, n}
is an IP* set.
Proof.
● We start with n = 1 and g(x) = g1 (x) = αx (mod 1), α ∈ R+ .
Let xn = nα (mod 1). Choose an idempotent ultrafilter p =
p + p ∈ βN. Then by Lemma 18.3
∞ > c = p − lim xn = p − lim p − lim(k + l)α (mod 1)
n
k
l
= p − lim(kα (mod 1) + c)
k
= 2c.
Hence, c = 0.
● By induction on the degree m of the polynomial and linearity
of p − lim we have for xn = nm α (mod 1)
m−1
p − lim xn = p − lim p − lim (k m + ∑ pm−s (k)ps (l) + lm ) α (mod 1)
n
k
l
s=1
= p − lim (k α (mod 1)) + p − lim (lm α (mod 1))
m
k
l
= 2(p − lim xn ).
n
where ps , pm−s are polynomials of degree s, m − s respectively,
fixing the origin.
● The set
m
{n ∈ N ∶ ∥p(n) = ∑ αk nk ∥ < }
k=1
is thus contained in any idempotent ultrafilter and hence by
Lemma 18.2 an IP* set.
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● If we have several (finite number) of polynomials we use iii) of
Lemma 18.2.