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U.U.D.M. Project Report 2018:41

Infinity comes in all sizes

Joel Nissen

Examensarbete i matematik, 15 hp Handledare: Vera Koponen Examinator: Martin Herschend Oktober 2018

Department of Uppsala University

Infinity comes in all sizes

Joel Nissen

October 22, 2018

Introduction

This report will analyze and compare different cardinal . The cardinal numbers are sets of specific sizes and is often used as the measuring sticks with which we compare the sizes of other sets. Cardinals may be finite but for the most part this report will be focusing on infinite cardinals. We call the smallest infinite cardinal ω and it can easily be thought of as the of natural numbers (= {0, 1, 2,... }). Every in ω can be reached by up to it through finitely many elements, we therefore call sets of this size countable. Furthermore we have the infinite sets with size greater than ω and they are called uncountable. One example of such a set is the set of real numbers R. We usually use the symbol c to denote the cardinal corresponding to the size of R. However, this rises the question, is there a cardinal κ with the size ω < κ < c? If so, what kind of set would that correspond to? The answer to the first question is that neither the existence nor the absence of such a cardinal follows from our usual (ZFC). The statement that no such set exists is known as the hypothesis. As this report is mostly about answering the second question we will not be assuming the . The cardinals that we will discuss will be

• The smallest uncountable cardinal ω1 • The pseudo-intersection number p

• The dominating number d

• The bounding number b

• The splitting number s

• The reaping number r

• The disjoint number a

• The independence number i

• The homogeneity number hom

• The partition number par

1 • The shattering number h

• The continuum c

The order in which this will be discussed will be the same as the order in the book ’Combinatorial ’ by Lorenz J. Halbeisen. That is also the book that most of this report is based on. The final result will be a partial ordering of all these numbers as the ordering of these numbers is at least partially independent of ZFC, see page 193 in [Halbeisen, 2012]. This graph illustrates this partial ordering: a

ω1 p h par b r hom c

s d i Here a line between elements means that the leftmost element is smaller than or equal to the rightmost.

Basic definitions

The reader of this report is assumed to have some familiarity with set theory like the basics found in [Cunningham, 2016]. Some concepts may, however, need a refresher and some key parts may be new even to a reader familiar with many of the concepts. We start by defining a well ordering.

Definition 1. A well ordering of a set is a linear order where every non-empty has a smallest element.

A common example of a well ordered set is the natural numbers N. It is worth commenting that if we assume the of choice, as we will, it follows that all sets can be well ordered. Next we need

Definition 2. A A is a set with the property that if a ∈ A then we have a ⊂ A

And with this we can define the main objects of this report, ordinals, and cardinals.

Definition 3. An ordinal is a transitive set that is well ordered by the ’∈’ relation.

As the ordinals are said to be well ordered by the ’∈’ relation it is common to use the notation κ < η to denote κ ∈ η. Furthermore it can be noted that for any two ordinals κ, η we have that κ < η, κ = η or κ > η. To exemplify we have that the smallest ordinal

2 is the ∅ which is followed by {∅} < {∅, {∅}} and so on. One often choose to denote ∅ by 0, {∅} by 1 and so on. This gives an intuitive picture of how the finite ordinals and the natural numbers relates to each other.

Definition 4. A cardinal (or cardinal number) is an ordinal for which there exist no injective to a smaller ordinal.

A well known result is that for each set A there is a to a unique cardinal, this cardinal is called the of A and is denoted |A|. This is a common way of determining the size of a set and is an important notion in several branches of mathe- matics, not the least analysis. It is clear that if there is a bijection between two sets, then they have the same cardinality. In fact an important result worth mentioning is the Schr¨oder-Berstein theorem that states that

Theorem 1 (Schr¨oder-Berstein). If we have two sets A and B such that |A| ≤ |B| and |A| ≥ |B|, then |A| = |B|

For proof, see page 135 in [Cunningham, 2016]. Here it may be worth mentioning that not all ordinals and cardinals are finite, indeed this report will focus on infinite car- dinals. The smallest infinite ordinal is called ω and sets with cardinality ω = {0, 1, 2, ...} are called countable and conversely sets of cardinality greater than ω are called uncount- able.

We also need to establish some more notation. For two cardinals κ, λ we have that κ + λ denotes the cardinality of the set κ × {0} ∪ λ × {1} and κ · λ = |κ × λ|. It may be worth noting that for infinite cardinals κ and λ we have κ · λ = max(κ, λ). Finally, AB denotes the set of functions from A to B. An important case of this is the set of functions from a set A to the set 2 = {∅, {∅}} (denoted A2). This is because of the relation between A2 and the set of all of A (denoted P(A)). For every set a ∈ P (A) we have that A −1 there is an unique function fa ∈ 2 such that fa (1) = a. That fa is unique in this way is clear since if we assume that there is a function g ∈ A2 such that g−1(1) = a we note −1 −1 that fa (0) = g (0) and thus they all elements the same and we have fa = g. A We now take the function ψ : P (A) → 2 such that ψ(a) = fa. That this function is bijective has already been established. That means that we have a bijection from P (A) to A2 and thus they must have the same cardinality. Another important fact is that for an infinite cardinal κ we have that |κ2| = |κκ|. For proof, see Lemma 9.2.11 on page 222 in [Cunningham, 2016]. For a set A and a cardinal κ we will also use the notation [A]κ to denote the set of all subsets of A that have cardinality κ and fin(A) is the set of all finite subsets of A.A useful result here is

Lemma 1. Let A be a set of sets and let κ be a cardinal. If X ≤ κ for all X ∈ A then | S A| ≤ |A| · κ

For the proof of this, see Theorem 9.2.12 on page 223 in [Cunningham, 2016]. Fur- thermore we will often call a set of sets or functions a family. For a function f : A → B

3 and a subset S ⊂ A we use the notation f|S to denote the unique function from S to B such that it coincides with f on S. We say that f|S is the restriction of f to S.

The cardinals c and ω1

Another important cardinal is the so called continuum c. This is the cardinality of the real numbers and we have that ω < c. We even have c = |ωω|.

ω ω Proof. To prove that c = | 2| we use that c = |R| and | 2| = |N2| = |Q2| = |P (Q)| so ω c = | 2| ⇔ |R| = |P (Q)|

ω First of we prove c ≤ | 2| by taking an function f : R → P (Q) that sends each as follows x 7→ {q ∈ Q : q < x}. This is clearly injective since there is between every real number. Thus c ≤ |ω2|.

To show the converse we construct the function g : P (N) → R That acts like: X g(A) = 10−n n∈N For each set this generates a number consisting of only ones and zeros. Assume that A, B ∈ P (N) are different sets. Then there is a smallest element k that separates them so that in the kth position in the decimal expansion the generated numbers will differ, thus g(A) 6= g(B) and by the Schr¨oder-Bersteintheorem we have c = |ω2|. Then, by Cunningham, we know that c = |ωω|.

From our usual axioms (ZFC) one can neither prove nor disprove that there is a cardinal number between ω and c. To assume that c is the smallest infinite cardinal greater than ω is called the continuum hypothesis. The cardinals that will be discussed in this report will have c as an upper bound so to assume the continuum hypothesis would render the report rather short. Instead we say that the smallest uncountable cardinal is called ω1. This will be proven to be a lower bound of the rest of the cardinals in the report.

The pseudo-intersection number p

We define the property that a set A is almost contained in a set B if A\B is finite (denoted A ⊆∗ B). Furthermore, for a F ⊆ [ω]ω, we define a pseudo- intersection of F to be an infinite set that is almost contained in all sets in F . We also say that a family F has the strong finite intersection property (sfip) if all finite subfamilies of F have an infinite intersection.

Now assume that a family F has a pseudo intersection P . Then for a finite subfamily A ⊂ F we take the intersection of P and all the sets in A . Since P is infinite and for

4 each set we intersect it with we only remove finitely many elements the resulting set will be infinite. This set will clearly lie in the intersection of all sets in A and thus we have that if a set has an pseudo intersection it has the sfip. Definition 5. We define the pseudo intersection number p as the smallest cardinality of a family F ⊆ [ω]ω that has the sfip but lacks a pseudo intersection. Since c is the cardinality of the of ω it follows that p ≤ c. Furthermore to show that is uncountable we assume an enumeration of a family F that has the sfip and we will show that it has a pseudo intersection. ω First we assume that we have an enumerable family F = {Xi ∈ [ω] : i ∈ ω} that T has the sfip. We then define a0 = X0. By construction this makes a0 the smallest element of X0. We also define

0 \ an = Xi. i∈n Where n ∈ ω\{0}. Note that since F has the sfip we know that this is an infinite set. Now we take:

\ 0 an = (an\{ai : i ∈ n}). 0 This makes an the smallest element in an\{ai : i ∈ n}. Now we take Y = {an : n ∈ ω}. For each Xi we have that an ∈ Xi when i ∈ n and by this we conclude that Y is almost contained in Xi and thus a pseudo intersection of F .

The dominating number d

For two functions f, g ∈ ωω we say that f dominates g if f(n) > g(n) for all but finitely many n. We define a dominating family D ⊆ ωω to be a family of functions such that for each function h ∈ ωω there is a function f ∈ D such that f dominates h. With this we can define the dominating number d. Definition 6. The dominating number d is the smallest cardinality of any dominating family. It is a well known result that |ωω| = |ω2| = c so it follows that d ≤ c. To show that d is uncountable we will introduce a smaller cardinal and prove that that cardinal is uncountable.

The bounding number b

An unbounded family B ⊂ ωω is a family of functions for which there is no function that dominates all functions in B. Note that any dominating family must be unbounded since if there is a function that dominates all functions in a dominating family D then that function would not be dominated by any function in D.

5 Definition 7. The bounding number b is the smallest cardinality of any unbounded family. By earlier arguments we have b ≤ d. Now assume that we have an enumerable ω unbounded family B = {fn ∈ ω : n ∈ ω}. Now we define the function:

g(n) = max{fi(n): i ∈ n} + 1.

For all i, n ∈ ω : i < n we have that g(n) > fi(n) and thus we have found a function that dominates all functions in B, contradicting the assumption that B was an unbounded family. By this we conclude that there can be no countable unbounded family, so we have ω1 ≤ b.

The splitting number s

We say that a set X splits another set Y iff both Y ∩ X and Y \X are infinite. We also say that a family S ⊆ [ω]ω is a splitting family if for any infinite set Y ⊆ ω there exist X ∈ S s.t X splits Y . We now define: Definition 8. The splitting number s is the smallest cardinality of any splitting family.

It will be shown when discussing a later cardinal number that s ≥ ω1 by transitivity. First we will show that s ≤ d. Note that for a every dominating family D, there is a dominating family D0 such that |D| ≥ |D0| and all functions f ∈ D0 are strictly increasing. We know that if g dominates h and f(n) ≥ g(n) for all n ∈ ω, then f dominates h. Thus we create function ψ : ωω → ωω such that ψ(f(0)) = 1 and ψ(f)(n + 1) = max{f(n + 1)(f(n) + 1)}. For a dominating family D we then have that ψ(D) is a dominating family consisting of only strictly increasing functions and, since ψ clearly is surjective to its image, that |D| ≥ |ψ(D)|. Now for a strictly increasing function f, with f(0) > 0 we define the set

[  2n 2n+1  σf = { f (0), f (0) : n ∈ ω}, where we define f n+1(0) = f(f n(0)) and f 0(0) = 0. Furthermore we use [n, m) to denote {i ∈ ω : n ≤ i < m}. Now we let D be a dominating family consisting of only strictly increasing functions such that f ∈ D ⇒ f(0) > 0. We then define:

LD = {σf : f ∈ D}.

We note that |LD | ≤ |D| since if we have σf 6= σg then f 6= g. Furthermore we let ω fx be the function from ω to x ∈ [ω] defined as:

fx(n) = min(x\{fx(i): i ∈ n}). This can also be expressed as the unique increasing bijection from ω to x. Since we assumed that D was dominating we have that for every fx there is f ∈ D such that f

6 dominates fx. So for some N ∈ ω we have that for n ≥ N, f(n) > fx(n). Since f is strictly increasing we have that f n(0) ≥ n and thus we have

n n n n+1 ∀n ≥ N : f (0) ≤ fx(f (0)) < f(f (0)) = f (0). n n n+1 So we have fx(f (0)) ∈ [f (0), f (0)). Here we note the fact that we have S 2n 2n+1 S 2n+1 2n c n∈ω[f (0), f (0)) = σf while n∈ω[f (0), f (0)) = ω \ σf = σf . So for every n n c even n we have fx(f (0)) ∈ σf and for every odd n we have fx(f (0)) ∈ σf . Therefore we have |x ∩ σf | = |x \ σf | = ω so LD is splitting and consequently we have s ≤ d.

The reaping number r

A reaping family is a set R of infinite subsets of ω such that no set x ∈ [ω]ω splits all sets in R. Definition 9. The reaping number r is the smallest cardinality of any reaping family.

First of all we note that [ω]ω is a reaping family so this is well defined and clearly r ≤ c. A more specified upper bound for r will be discussed later. For now we will be focusing on the lower bound, namely that b ≤ r. To prove this we will show that that any family of cardinality κ < b can not be reaping. For this we assume that E = {xα : α ∈ κ < b} is a family of cardinality κ strictly lesser than b. We let gα be the unique strictly increasing bijection from ω to xα\{0} n ˜ andg ˜α(n) = gα(0). Now we use E to denote the set {g˜α, α ∈ κ}. By our assumption κ < b so E˜ is a bounded family. Thus we have a function f (which we may assume to be strictly increasing and f(0) > 0) such that it dominates all functions in E˜. Here we S 2n 2n+1 use the same construction as in the previous section and let x = n∈ω[f (0), f (0)). Then for each gα we have kα such that

n n n+1 ∀n ≥ kα : f (0) ≤ g˜α(f (0)) < f (0).

Thus we have that both xα\x and xα ∩ x are infinite and we can not have a reaping family of cardinality κ < b.

The almost disjoint number a

We say that two sets x, y ∈ [ω]ω are almost disjoint if their intersection is finite. A family A ⊆ [ω]ω is called almost disjoint if all sets in it are pairwise almost disjoint. Furthermore an almost disjoint family A is called maximal almost disjoint (mad) if A is not a proper subfamily to any almost disjoint family.

Definition 10. The almost disjoint number a is the cardinality of the smallest maximal almost disjoint family.

7 First we want to show that there exists a mad family of cardinality c. Since any almost disjoint family can be extended into a mad family (by Zorn’s lemma) it will suffice to find an almost disjoint family of size c. We use S = {si : i ∈ ω} to denote an enumeration of all S n ω the functions n∈ω ω. For a given f ∈ ω we define xf = {i ∈ ω : ∃n ∈ ω : f|n = si}. For two functions f 6= g we have that there exists N such that f(N) 6= g(N) so that for all n ≥ N we have f|n 6= g|n and thus |xf ∩ xg| = N (or N + 1 depending if you count ω the empty function). We can now let X = {xf : f ∈ ω} and note that there is a clear bijection from X to ωω and thus |X| = c. The next thing to show will be that from any mad family A one can create a bounded family of cardinality |A | and thus b ≤ a. We let A be a mad family of cardinality κ and index it by A = {xα : α ∈ κ}. S We then let z = ω\ α∈κ xα. If z would be infinite we note that it would have a finite (empty) intersection with all sets in A contradicting that A would be a mad family. 0 Thus z is finite. We furthermore define x0 = x0 ∪ z ∪ {0} and for n ∈ ω we define: 0 S 0 0 xn = (xn ∪ {n})\ k∈n xk. Here we note that x0 only differs from x0 with finitely many 0 0 elements. Furthermore for each n ≥ 1 we have that (xn\{n}) ⊆ xn and thus xn ∩ xα is 0 finite for n ∈ ω, α ∈ κ\{n}. Since we assumed that A is mad we have that each xn is S 0 S 0 infinite and since n ∈ k∈n+1 xk we have that n∈ω xn = ω. We also note that for a given ω set y ∈ [ω] there is a set α ∈ κ such that y ∩ xα is infinite and if α = n ∈ ω it follows 0 0 that y ∩ xn is infinite as well. From this we have that (A \{xk : k ∈ ω}) ∪ {xn : n ∈ ω} 0 0 is mad. Note also that for n, m ∈ ω where n 6= m we have that xn ∩ xm = ∅. We ω 0 use gn ∈ ω to denote the unique increasing bijection from ω to xn and the function h : ω → ω × ω as:

0 h(m) = hn, ki where m ∈ xn and k = gn(m). 0 S 0 Since all xk are disjoint and the fact that k∈ω xk = ω we know that for every 0 m ∈ ω there exist a unique n ∈ ω such that m ∈ xn. By construction we have that 0 0 h(xn) = {hn, ki : k ∈ ω}, for all n ∈ ω and since xω+ξ ∩ xn is finite ∀n ∈ ω, ξ ∈ κ, so is 0 ω h(xω+ξ ∩ xn). We now define the functions fξ ω, ξ ∈ κ as follows:

0 fξ(n) = max{k : hn, ki ∈ h(xω+ξ ∩ xn)}.

For ξ, χ ∈ κ we have that if fξ = fχ then fξ(ω) is an infinite subset of both xω+ξ and xω+χ, since A is an almost disjoint family we have that this implies ξ = χ. Now let B = {fξ : ξ ∈ κ}. By our previous arguments we have that |B| = |A | and if we assume that we have a function f that dominates all functions in B we get that the set H = h−1({hn, f(n)i, n ∈ ω}) would have an finite intersection with all members of A . To verify this we first note that H has exactly one element in common with all the 0 sets xn. Furthermore we note that for xω+ξ we have that f dominates fω+ξ so for some −1 0 0 N we have n ≥ N ⇒ f(n) > fω+ξ and thus y = h (hn, f(n)i) ∈/ xω+ξ ∩ xn but as y 0 clearly lies in xn we may conclude that y∈ / xω+ξ. As this holds for all n ≥ N we indeed have that H must have a finite intersection with all elements of A contradicting that it would be mad.

8 The independence number i

We call a family I ⊆ [ω]ω independent iff for any two numbers n, m ∈ ω and subsets X = {xi : i ∈ n} ⊂ I ,Y = {yj : j ∈ m} ⊂ I where X ∩ Y = ∅ we have that \ \ xi ∩ (ω\yj) is infinite. i∈n j∈m Assuming that we define T ∅ := ω, note that for A, B ∈ [ω]ω we have A ∩ B = A\Bc and T B = (S Bc)c (where Bc = ω\B) which yields \ \ \ [ xi ∩ (ω\yj) = xi\ yj. i∈n j∈m i∈n j∈m Furthermore, an independent family I is called maximal independent if there is no other independent family of which it is a proper subset.

Definition 11. The independence number i is the smallest cardinality of any maximal independent family.

First of all we want to show that there exist a maximal independent family I such that |I | = c. To do this we first note that it is equivalent to construct an independent family over the :

C = {hs, Ai : s ∈ fin(ω),A ⊆ P (s)} ω For any set x ∈ [ω] we define Px = {hs, Ai ∈ C : x ∩ s ∈ A}. We note that for x and y such that x 6= y and x\y 6= ∅ we may take s0 = {α, β} where, if x∩y = ∅, α ∈ x, β ∈ y and if x∩y 6= ∅, α ∈ x∩y, β ∈ x\y. By construction we have x∩s0 6= y ∩s0 so Px 6= Py. ω If we then take the set Ic = {Px : x ∈ [ω] } we have that it has cardinality c. Now we want to show that Ic is an independent family over C. We note that for any finite ω collection of sets xi ∈ [ω] , i ∈ {0, ..., m + n}, m ≥ 1 we may construct a finite subset s such that xi ∩ s 6= xj ∩ s for i 6= j. This is done in a similar fashion as previously. If x0\x1 = ∅ we take α0 ∈ x0, β0 ∈ x1\x0 and if x0\x1 6= ∅ we let α0 ∈ x0\x1, β0 ∈ x1 and define s0 = {α0, β0}. We note that x0 ∩ s0 6= x1 ∩ s0 and continue by defining α1 and β1 similarly so that x0 ∩ {α1, β1} 6= x2 ∩ {α1, β1} and let s1 = s0 ∪ {α1, β1}. We let s be the result we get after pairing all xi. By construction we get that xi ∩ s 6= xj ∩ s for all i 6= j. Furthermore we define A = {s ∩ xi : 0 ≤ i ≤ n} and sk = s ∪ {k}, ∀k ∈ ω\s. Finally we define Ak = A ∪ {t ∪ {k} : t ∈ A}.

We note that for i such that 0 ≤ i ≤ n, sk ∩xi ∈ Ak and therefore hsk,Aki ∈ Pxi , ∀i ∈

{0, ..., n} while for j such that n < j ≤ m + n we have hsk,Aki ∈/ Pxj . Thus we have \ [ {hsk,Aki : k ∈ ω\s} ⊆ Pxi \ Pxj . 0≤i≤n n

9 The next thing to show will be that max(r, d) ≤ i. We may start by showing r ≤ i. We let I ⊆ [ω]ω be a maximal independent family and we define: n\  [ o R = I \ J : I,J ∈ fin(I ),I ∩ J = ∅

S n S n Since |R| ≤ |2 · ( n∈ω[I ] )| = | n∈ω[I ] | we can apply lemma 1 to get |R| ≤ |I |. However we can let J = ∅ to get that I ⊆ R and thus |R| = |I |. Now we want to show that R is a reaping family. Assume that we have x ∈ [ω]ω such that it splits all elements of y ∈ R. This would mean that for every y ∈ R we would have that both \  [  y ∩ x = I ∩ x\ J

and \ [  y \ x = I\ J ∪ x

would be infinite. This would require that x∈ / I for otherwise we would find the set y = x\ S J ∈ R for some finite collection J that would not split with x as y \ x = x \ (S J ∪ x) = ∅ which is not infinite. But that in turn would imply that we have an element x∈ / I such that I ∪{x} would be an independent family contradicting our assumption that I was maximal. Thus we can not find a set x that splits all elements of R and we may conclude that R is a reaping family. From this we see that for any maximal independent family we may find a reaping family of the same cardinality. To show that d ≤ i we let the cardinal κ be such that κ < d and we will show that there is no maximal independent family of cardinality κ. For this we will need to define a few things. ω Let I = {Xξ : ξ ∈ κ} ⊆ [ω] be a independent family of cardinality κ and define 0 1 Iω = {Xn : n ∈ ω} ⊆ I . We also define Xn = Xn and Xn = ω\Xn, ∀n ∈ ω. For ω T g(k) every function g ∈ 2 we also define Cn,g = k∈n Xk . Note that Cn,g is infinite since I is an independent family. For a fixed g the set of all these sets will be denoted 0 Cg = {Cn,g : n ∈ ω}. Lastly (for a while) we define I = I \Iω and \ [ F = { I0\ J 0 : I0,J 0 ∈ fin(I 0),I0 ∩ J 0 = ∅}.

We now want to show that Cg has a pseudo-intersection P and that for every element y ∈ F we have that P ∩ y is infinite. We note that each element of Cg can be written on the form:

\ g(k) \ \ \ [ C = X = X ∩ (ω\X 0 ) = X \ X 0 . n,g k ki kj ki kj k∈n i∈S j∈T i∈S j∈T 0 where g(ki) = 0 for all i ∈ S and g(kj) = 1 for all j ∈ T . From this we conclude that each Cn,g ∈ Cg is infinite and that for each element x ∈ F we have

10 \ 0 [ 0 \ [  x ∩ C = I \ J ∩ X \ X 0 n,g ki kj \ 0 \  [ 0 [  = I ∩ X \ J ∪ X 0 i kj

0 0 and since all the sets I ,J {X : i ∈ S}, {X 0 : j ∈ T } ∈ fin( ) are disjoint we have ki kj I that every element of Cg has an infinite intersection with every element of F . We also note that F has the same cardinality as I 0 and by extension as I , namely κ < d. For each function h ∈ ωω we define:

h [ Yg = (Cn,g ∩ h(n)). n∈ω

We know that for every k ∈ n ∈ ω we have Cn,g ⊆ Ck,g. Now let ! h [ [ Yg,n = (Cm,g ∩ h(m)) \ (Ck,g ∩ h(k)). m∈ω k∈n h h h h We note that Yg,n ⊆ Cn,g and since Yg \Yg,n is finite we have that Yg is almost h contained in Cn,g for all n. Clearly, however, we have that Yg is not infinite for every h h. For instance Yg is finite for any bounded h. Our next goal will be to find a function h h such that Yg is infinite. For every A ∈ F and every n ∈ ω we note that A ∩ Cn,g is T infinite. Due to this we may define the function fA,n where fA,n(0) = (A ∩ Cn,g) and for k > 0:

fA,n(k) = min(A ∩ Cn,g\{fA,n(i): i ∈ k}).

or alternatively, the unique increasing bijection from ω to A ∩ Cn,g. We now use this function to define the function:

fA(n) = min(A ∩ Cn,g\{fA,n(k): k ∈ n}). This can also be defined using less notation as the function that given n returns the (n + 1) :th element of A ∩ Cn,g. As previously stated we know that |F | < d and from this fact it follows that the family {fA : A ∈ F } is not dominating. Thus we may find a function h0 such that it is not dominated by any function fA. Thus the set DA = {n ∈ ω : h0(n) > fA(n)} is infinite for each A ∈ F . For every n ∈ DA we thus have h0(n) ≥ fA(n) + 1 and by implication |Cn,g ∩ h0(n)| ≥ |Cn,g ∩ fA(n) + 1| ≥ n + 1 since fA(n) is the n + 1th element of A. In turn this tells us that:

[ (Ck,g ∩ h0(k)) ≥ n k∈n

h0 So it follows that the set Yg is infinite and therefore a pseudo-intersection of Cg that has an infinite intersection with all elements of F .

11 ω So in conclusion we have that for every g ∈ 2 there is a set Yg that is almost T g(k) T 0 T 0 contained in k∈n Xk for all n ∈ ω and such that Yg ∩ ( I \ J ) is infinite for any disjoint subfamilies I0,J 0 ∈ fin(I 0). We may deduce that whenever g, g0 ∈ ω2 are different functions it will follow that Yg and Yg0 are almost disjoint. Indeed, if we let 0 n0 ∈ ω be a number such that g(n0) 6= g (n0) we have that Yg is almost contained in 0 g(n0) g (n0) g(n0) c Xn0 while Yg0 is almost contained in Xn0 = (Xn0 ) . We now define:

ω Q0 = {g ∈ 2 : ∃n0 ∈ ω ∀k ≥ n0 (g(k) = 0)} and

ω Q1 = {g ∈ 2 : ∃n0 ∈ ω ∀k ≥ n0 (g(k) = 1)}.

Since Q0 ∪ Q1 is countable we may let {gn : n ∈ ω} be an enumeration of these 0 S functions. For a given n we define Ygn = Ygn \ {Ygk : k < n}. We also conclude that 0 the countable set {Ygn : n ∈ ω} is made up of pairwise . Now let

[ 0 0 [ 0 Z = Yg and Z = Yg . g∈Q0 g∈Q1

We remember that Iω = {Xn : n ∈ ω} ⊆ I and for any finite disjoint subfamilies 0 0 I,J ⊂ I we define I0 = I ∩ Iω,J0 = J ∩ Iω and I = I\Iω.J = J\Iω. Furthermore, pick m ∈ ω such that I0 ∪ J0 ⊆ {Xn : n ∈ m} and g ∈ Q0 such that

(Xn ∈ I0 ∪ J0 and g(n) = 0) ⇔ Xn ∈ I0. This leads us to:

\ [ \ 0 [ 0 \ g(n) ∗ \ 0 [ 0 I\ J ⊇ I \ J ∩ Xn ⊇ I \ J ∩ Yg. n∈m T 0 S 0 By our earlier results we know that the set ( I \ J ) ∩ Yg is infinite. Since g ∈ Q0 ∗ T 0 S 0 T S we know Yg ⊆ Z and ( I \ J ) ⊆ ( I\ J). This yields:

\ 0 [ 0 ∗ \ [  I \ J ∩ Yg ⊆ I\ J ∩ Z

In conclusion we have a set Z such that for any I,J ∈ fin(I ) such that I ∩ J = ∅ we have (T I\ S J)∩Z is infinite. Since we can make all the arguments above for Z0 ⊆ ω\Z we may conclude that I ∪ {Z} is an independent family. Thus we have no maximal independent family I such that |I | = κ < d.

Ramsey’s theorem

In order to describe the following cardinal number (the homogeneity number) we need to delve into the field of . More precisely, we need to be familiar with Ramsey’s theorem.

12 To understand Ramsey’s theorem it will be prudent to start with the pigeon hole principle. This states:

Theorem 2 (The pigeon hole principle). If n + 1 pigeons are roosting in n holes, then at least two pigeons must share a hole. In order to be more consistent with the following theorems, this may also be formu- lated as: + If n, k ∈ Z and n + k objects are colored with n colors, then at least two objects has the same color.

This theorem also has a infinite version that will be central to the upcoming theorem.

Theorem 3 (The infinite pigeon hole principle). If infinitely many objects are colored with finitely many colors, then infinitely many objects will have the same color.

Furthermore, by a coloring function we refer to a function π :[ω]n → r where n ∈ ω ω and r ∈ ω\{∅}. We call a set H ∈ [ω] homogeneous if π|[H]n is constant. We can also say that π is monochromatic on [H]n. Now we are ready to formulate and prove Ramsey’s theorem.

Theorem 4 (Ramsey’s theorem). For any number n ∈ ω, r ∈ ω\{∅} and any set S ∈ [ω]ω and for any coloring π :[S]n → r there is a set H ∈ [S]ω such that H is homogeneous for π.

In order to prove this we will use induction over n, our base case will be n = 2. We let S0 = S and a0 = min(S0). We also define the coloring τ0 : S0\{a0} → r as τ0(b) = π({a0, b}). We know from the infinite pigeon hole principle that there is a set ω S1 ∈ [S0] such that τ0|S1 is constant. We let ρ0 = τ0(b) for some b ∈ S1. Now we repeat this process for S1, let a1 = min(S1) and τ1 : S1\{a1} is a coloring defined as τ1(b) = π({a1, b}). ω Once again the infinite pigeon hole principle verifies that there is a set S2 ∈ [S1] such that τ1|S1 is constant. Again, we let ρ1 = τ1(b) for some b ∈ S1. We continue this process for all n ∈ ω in order to get an infinite set {an} and the not so infinite set {ρn}. We note that for n < k we have that π({an, ak}) = ρn. We define the coloring τ(an) = ρn. By the ω infinite pigeon hole principle we know that there is a set H ∈ {an} such that τ|H = ρ is constant. This in turn means that for any pair of elements {a, b} ∈ [H]ω, π({a, b}) = ρ and thus H is homogeneous. This concludes our base step. We now assume that Ramsey’s theorem holds for n ≥ 2 and let π :[ω]n+1 → r be an arbitrary coloring function. For a ∈ ω we define n πa :[ω\{a}] → r as

πa(x) = π(x ∪ {a}) By assumption we have that for any S0 ∈ [ω]ω and each a ∈ S0 we may find a set S0 0 n ω Ha ∈ [S \{a}] which is homogeneous with respect to πa. Now we fix a set S ∈ [ω] from which we define S0 = S and a0 = min(S). Furthermore we define:

13 Si Si+1 = Hai and ai+1 = min(Si+1)

By this we have an infinite chain of sets S0 ⊃ S1 ⊃ ... and elements a0 < a1 < . . . .

By construction the set {ak : k > i} is homogeneous with respect to πai . Let τ(ai) be a function such that τ(ai) = πai (ak) for some k > i. By the infinite pigeon principle we know that there is a set H ∈ {ai : i ∈ ω} such that τ is constant. Then for any n+1 {x0 < ··· < xn} ∈ [H] we have:

π({x0, . . . , xn}) = πx0 ({x1, . . . , xn}) = τ(x0) Thus we have that H is homogeneous with respect to π and by induction we have proven Ramsey’s theorem.

The homogeneity number hom

Now by Ramsey’s theorem we know that for every coloring function π :[ω]2 → 2 there is a set x ∈ [ω]ω such that π is monochromatic on [x]2. With this we define our next cardinal number:

Definition 12. The homogeneity number hom is the smallest cardinality of any family F ⊆ [ω]ω with the property that for any π :[ω]2 → 2 there is a set x ∈ F such that x is homogeneous for π.

We call such a family a homogeneous family. First of we can conclude that there exist a homogeneous family of cardinality c since [ω]ω is homogeneous by theorem 3. For a homogeneous family F such that |F | = hom we may define a function ϕ : ([ω]2 → 2) → F such that ϕ(π) = x where π|x is constant. Note that the image of ϕ is clearly a homogeneous family of cardinality hom and ϕ is by definition surjective to its image F 0. We also note that |[ω]2| = ω so we have

hom = |F 0| ≤ |[ω]2 → 2| = |ω → 2| = c and so we have that hom ≤ c. The next thing we want to show is that max{r, d} ≤ hom. We start with proving d ≤ hom. 0 Let F be a homogeneous family and let F = {fx : x ∈ F } be the family of functions such that fx is the unique strictly increasing bijection from ω to x. Furthermore, for any strictly increasing function f ∈ ωω such that f(0) = 0 we let {n, m} ∈ [ω]2 be such 2 that n < m and define πf :[ω] → 2 as follows:

πf ({n, m}) = 0 ⇔ ∃k ∈ ω : f(2k) ≤ n < m < f(2k + 2)

We know that there exist a set x ∈ F such that πf |[x]2 is constant, since F is homogeneous. Furthermore we note that for any n ∈ ω we have that there are only finitely many m so that πf ({n, m}) = 0. Thus we have that πf ({n, m}) = 1 for all 2 {n, m} ∈ [x] . Now we note that fx(0) ≥ f(0) = 0 and that fx(n) ≥ f(n) implies that

14 fx(n + 1) > f(n + 1) since πf ({fx(n), fx(n + 1)}) = 1 implies that fx(n + 1) ≥ f(2n + 2). ω Thus we have that fx dominates f. Since for every function g ∈ ω there is a strictly increasing function f with f(0) = 0 that dominates g this implies that F 0 is dominating and since |F 0| = |F | we have d ≤ hom. Next we want to show that r ≤ hom. We let F be a homogeneous family and ω 2 y ∈ [ω] . We define πy :[ω] → 2 as such:

πy({n, m}) = 0 ⇔ {n, m} ⊆ y or {n, m} ∩ y = ∅.

Then there is a set x ∈ F such that πy|[x]2 is constant. Assume that πy({n, m}) = 1 for some {n, m} ∈ [x]2. Then we have either n ∈ x\y and m ∈ x ∩ y or we have n ∈ y ∩ x and m ∈ y\x. Without loss of generality we may assume the former case. Now assume that there is an element k ∈ x such that n 6= k and m 6= k, for such a k we have either k ∈ y ∩ x or k ∈ y\x. In the former case we have that πy({m, k}) = 0 and for the latter case we have that πy({n, k}) = 0. Thus for a set x such that πy|[x]2 is constantly 1 we have |x| = 2. However, x was assumed to be infinite so we have that πy({n, m}) = 0 for {n, m} ∈ [x]2. This implies that for {n, m} ∈ [x]2 we have either {n, m} ∩ y = ∅ or {n, m} ⊂ y. Consequently we get that either x ⊆ y which would imply that x\y = ∅ or that x∩y = ∅. Either way, y does not split x and F is a reaping family. By this we conclude that every homogeneous family is also a reaping family and consequently r ≤ hom.

The partition number par

We call a set H ∈ [ω]ω almost homogeneous with respect to a coloring π if there is a finite set K ⊆ H such that H\K is homogeneous with respect to π. From this we will define our next cardinal number.

Definition 13. The partition number par is the smallest cardinality of any family P of colorings π :[ω]2 → 2 with the property that there is no H ∈ [ω]ω such that H is almost homogeneous with respect to all π ∈ P. We call such a family almost homogeneous. Our goal now will be to prove par = min{s, b} by showing first par ≤ min{s, b} and then, par ≥ min{s, b}. Then the Schr¨oder-Bersteintheorem will yield the equality we seek. We start with par ≤ s. Let S ⊆ [ω]ω be a splitting family and for each x ∈ S we 2 define the coloring πx :[ω] → 2 as:

πx({n, m}) = 0 ⇔ {n, m} ⊂ x ∨ {n, m} ∩ x = ∅ ω We let P = {πx : x ∈ S }. Since S is a splitting family we can, for every y ∈ [ω] find a x ∈ S such that y\x and y ∩ x are both infinite. By this we have that there are infinitely many sets {n, m}, {j, k} ⊂ y such that {n, m} ⊂ x and {j, k} ⊂ ω\x. Note that since y\x is infinite we have that ω\x is infinite. Thus y is not almost homogeneous with respect to πx and P is an almost homogeneous family.

15 To show par ≤ b we let B ⊆ ωω be a unbounded family such that each f ∈ B is strictly increasing. We know that this can be done without loss of generality by the 2 discussion in the bounding number. For every f ∈ B we define πf :[ω] → 2 as follows

πf ({n, m}) = 0 ⇔ f(n) < m for n < m. Now assume that there is a set H ∈ [ω]ω which is almost homogeneous with respect to all colorings in P = {πf : f ∈ B}. From this we construct the function g as

g(n) = min{m ∈ H : ∃k ∈ H : n < k < m} This can also be formulated as the function that maps n to the second element of H strictly greater than n. By this we have n < k < g(n) where k, g(n) ∈ H. By assumption

2 there is a set Kf ∈ fin(ω) for any arbitrary f ∈ B such that πf |[H\Kf ] is constant. For a given n there is only finitely many m > n such that g(n) ≥ m so a set T must be finite if it has the property that πf |[T ]2 is constantly 1. Indeed, assume that there exist ω 2 T ∈ [ω] such that for {n, m} ∈ [T ] we have πf ({n, m}) = 1 then we can find M ∈ T 2 so that f(n) < M and thus πf ({n, M}) = 0. We may thus conclude that πf |[H\Kf ] is constantly 0 for all f. However, since Kf is finite there exists a Nf ∈ ω such that Kf ⊆ Nf . Then we let {ni : i ∈ ω} be an enumeration of H\Nf . Let ni, ni+1 ∈ H\Nf . Then we have:

f(ni) < ni+1 < g(ni+1). Since all f ∈ B where assumed strictly increasing we know that for all k ∈ ω such that ni < k < ni+1 we have:

f(k) < f(ni+1) < ni+2 = g(k) Thus we have that g dominates all functions in B, contradicting that it would be a unbounded family. Finally we want to show that par ≥ min{s, b}. To do this we let κ be a cardinal such that κ < min{s, b} and let

[ω]2 P = {πξ ∈ 2, ξ ∈ κ} We intend to show that there exists a set H that is almost homogeneous with respect to every πξ ∈ P. For every ξ ∈ κ and n ∈ ω we define:

 π ({n, m}) if n 6= m f (m) = ξ ξ,n 0 otherwise −1 We note that |{fξ,n : ξ ∈ κ, n ∈ ω}| = κ · ω = κ. We now define Mξ,n to be fξ,n({0}) −1 if that set is infinite and fξ,n({1}) otherwise. By the infinite pigeon hole principle we know that all sets Mξ,n will be infinite. We also see that for any ξ ∈ κ there is at most one other ζ ∈ κ such that Mξ,n = Mζ,n. Thus we have |{Mξ,n : ξ ∈ κ, n ∈ ω}| = κ < s. ω Since {Mξ,n} is not a splitting family there is a set A ∈ [ω] such that A ∩ Mξ,n or

16 A\Mξ,n is finite for every ξ ∈ κ and n ∈ ω. We may thus conclude that for every ξ ∈ κ and n ∈ A there are constants Nξ,n ∈ ω and iξ,n ∈ {0, 1} such that for m > Nξ,n we have fξ,n(m) = iξ,n. From this we define the functions jξ : ω → 2 and gξ : ω → ω such that gξ(n) = Nξ,n and jξ(n) = iξ,n. Once again by the fact that κ < s we have that there exist a set B ⊆ A on which all functions jξ is almost constant. In other words, for every ξ ∈ κ there is a bξ ∈ B such that for all n ∈ B, jξ(n) = iξ. Since we also have that κ < b we have a function h : ω → ω that dominates all of the functions gξ for all ξ ∈ κ. Let cξ ∈ ω be the smallest number such that h(n) > gξ(n) for all ω n > cξ. Now let H = {xk : k ∈ ω} ∈ [B] be a set such that for all k ∈ ω, h(xk) < xk+1. Then for n, m ∈ H such that max{bξ, cξ} < n < m we have

gξ(n) < h(n) < m ⇒ πξ({n, m}) = fξ,n(m) = jξ(n) = iξ.

implying that H is almost homogeneous with respect to πξ for each ξ.

The shattering number h

We call a family H = {Aξ : ξ ∈ κ} shattering if it consists of mad families of cardinality c such that for every x ∈ [ω]ω there is a ξ ∈ κ such that x has an infinite intersection with at least two elements of Aξ. Definition 14. The shattering number h is the smallest cardinality of any shattering family.

We will prove that p ≤ h ≤ par. First we will show that one may create a shattering family from an almost homogeneous family. To do this we will use the following lemma:

2 Lemma 2. For every coloring π :[ω] → 2 there is a mad family Aπ of cardinality c such that every set A ∈ Aπ is almost homogeneous with respect to π.

Proof. To prove this we let A ⊂ [ω]ω be an almost disjoint family of cardinality c (The existence of such a family was proven in the section about the almost disjoint number) and let π :[ω]2 → 2 be a coloring function. By Ramsey’s theorem there exists an infinite set A0 ⊆ A for every set A ∈ A such that A0 is homogeneous for π and define A 0 = {A0 : A ∈ A }. Since A0 ⊆ A for all A we know that A 0 is still an almost disjoint ω 0 family of cardinality c. Now, for some cardinal κ ≤ c, let [ω] \A = {xξ : ξ ∈ κ}. Let 0 A0 = A and for every successor ordinal ξ ∈ κ we define:  Aξ ∪ {xξ} if xξ is homogeneous with respect to π   and for every A ∈ Aξ, |xξ ∩ A| < ω Aξ+1 =   Aξ othervise S Further, for limit ordinals ζ ∈ κ, we define Aζ = ξ∈ζ Aξ ∪{xζ } if xζ is homogeneous S S with respect to π and for every A ∈ ξ∈ζ Aξ we have |xζ ∩ A| < ω and Aζ = ξ∈ζ Aξ S otherwise. Finally we define Aπ = ξ∈κ Aξ. Since all sets that were added where almost

17 disjoint with all the already contained sets we know that Aπ is still an almost disjoint family of cardinality c. Furthermore we have that all elements of Aπ are homogeneous with respect to π. Now we just need to show that Aπ is mad. Assume that there is a ω set x ∈ [ω] such that x ∩ A is finite for all A ∈ Aπ. By Ramsey’s theorem we know that there exists an infinite set y ⊆ x such that it is homogeneous for π. However, the fact that |x ∩ A| < ω for all A ∈ Aπ tells us that the same holds for y. By this we ω 0 have that y ∈ [ω] \A . But then for some xξ0 ∈ Aξ0+1 we have that either y = xξ0 or

|y ∩xξ0 | = ω and since y ∩x is infinite we have that either of these cases would contradict the existence of such an x. Thus we have that Aπ is a mad family of cardinality such that every set A ∈ Aπ is homogeneous with respect to π. Thus we have concluded the proof for lemma 2.

With that we are ready to prove that h ≤ par. We let P be an almost homogeneous family and HP = {Aπ : π ∈ P} be a shattering family where Aπ is constructed as in ω the lemma. Then HP is shattering. To prove this we let H ∈ [ω] be some infinite subset of ω. Since P is an almost homogeneous family, there exist a π ∈ P such that H is not almost homogeneous for π. Then in Aπ ∈ HP we have a set A such that H ∩A is infinite (since Aπis mad). Since A is homogeneous the set H\A must be infinite as well, otherwise H\(H\A) would be homogeneous contradicting that H was not almost homogeneous with respect to π. However, since Aπ is still mad we have that H\A has an 0 0 infinite intersection with some A ∈ Aπ. By the fact that (H\A)∩A = ∅ we have A 6= A and thus H has an infinite intersection with two sets in Aπ and HP is subsequently shattering implying that h ≤ par. As we move on to prove that p ≤ h we need to establish a few definitions and lemmas. We say that a mad family A refines another mad family A 0 (both of cardinality c) if for each A0 ∈ A 0 there exist a set A ∈ A such that A0 ⊆∗ A. This is denoted A A 0. 0 Furthermore we call a shattering family H = {Aξ : ξ ∈ κ} refining if ξ > ξ implies that Aξ0 Aξ.

Lemma 3. Let κ be a cardinal such that κ < h, then for every family E = {Aξ : ξ ∈ κ} 0 of mad families of cardinality c there exists a mad family A which refines each Aξ ∈ E . Furthermore, A 0 has cardinality c.

Proof. In order to prove this we let κ < h and define E = {Aξ : ξ ∈ κ} to be a family of mad families of cardinality c. Then, by [Halbeisen, 2012, p.192], we have that for every ω 0 set x ∈ [ω] there is an infinite subset x ⊆ x such that for every Aξ ∈ E there is a set 0 ∗ A ∈ Aξ for which we have x ⊆ A. Let A 0 = {x0 : x ∈ [ω]ω}. If A 0 is a mad family it must be of cardinality c. If it was 0 0 0 of a cardinality ρ < c it would exist A ∈ A0 such that A ∩ x is finite for all x ∈ A and we would have that A 0 ∪ {A} would be an almost disjoint family containing A 0. 0 Furthermore we note that for every ξ ∈ κ we have that A Aξ.

Lemma 4. If H = {Aξ : ξ ∈ h} is a shattering family of cardinality h, then there 0 0 exists a refining shattering family H = {Aξ : ξ ∈ h} such that for each ξ ∈ h we have 0 Aξ Aξ.

18 Proof. This will be proven by transfinite induction. Our base case consist of picking 0 A0 = A0. Furthermore if the lemma holds for all Aξ : ξ ∈ η for some η ∈ h, then we 0 0 may apply lemma 3 on {Aξ : ξ ∈ η} ∪ {Aη} in order to get Aη . By transfinite induction 0 0 we have that H = {Aξ : ξ ∈ h} is indeed a refining shattering family of cardinality h and thus we have proven lemma 4.

Finally we are ready to show that p ≤ h. We know that there exist a refining shattering family H = {Aξ : ξ ∈ h} with cardinality h. We start by constructing a series of sets Fη for η ∈ h. We let F0 = {x0} for some x0 ∈ A0 and for η ∈ h we can choose an element from xη ∈ Aη such that it is a pseudo intersection to each Fξ S where ξ ∈ η and let Fη = ξ∈η Fξ ∪ {xη}, assuming we pick xξ in the same way. Let S F = ξ∈h Fξ. Now we have that F has the sfip but since H is shattering we know ω that for any set in [ω] it can not be almost contained in all xξ. Since H was assumed to be of cardinality h it follows that p ≤ h.

References

Lorenz J. Halbeisen, Combinatorial Set Theory, Springer, 2012

Daniel W. Cunningham, Set Theory: a first course, Cambridge University Press, 2016

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