A SET-THEORETIC APPROACH to OBTAINING INFINITY by MATTHEW MICHAEL JONES B.A., University of Colorado, Colorado Springs, 2013

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A SET-THEORETIC APPROACH TO OBTAINING INFINITY by MATTHEW MICHAEL JONES B.A., University of Colorado, Colorado Springs, 2013

A thesis submitted to the Graduate Faculty of the University of Colorado Colorado Springs in partial fulﬁllment of the requirements for the degree of Master of Science Department of Mathematics 2016 ii

This thesis for the Master of Science degree by Matthew Michael Jones has been approved for the Department of Mathematics by

Greg Oman

Gene Abrams

Greg Morrow

Date: 12/14/2016 iii

Jones, Matthew Michael (M.S., Mathematics) A Set-Theoretic Approach To Obtaining Inﬁnity Thesis directed by Assistant Professor of Mathematics, Dr. Greg Oman

ABSTRACT

Many set-theoretic axioms have been formulated over the years which imply (along with the other axioms for set theory) the existence of what one would intuitively regard as an inﬁnite set. In this thesis, we identify a property common to many such axioms, and then we obtain a more general class of “inﬁnity axioms.” iv

TABLE OF CONTENTS

CHAPTER I. INTRODUCTION 1 II. PRELIMINARIES 6 III. RESULTS 16 REFERENCES 25 1

CHAPTER I

INTRODUCTION

The concept of “infinity” is ubiquitous in mathematics. Indeed, the collection of real numbers, the collection of prime numbers, and the collection of real-valued, continuous functions are all examples of infinite collections of objects. This is a notion we intuitively understand: a collection of objects is infinite if, in some sense, its elements “go on forever.” In modern mathematics, this is hardly precise enough to be a formal definition. But how can one define infinity? There are many answers in the literature. The most common context in which such answers appear is in the mathematical field of set theory. Informally, a set is a collection of objects. For instance, {1, 2, 3} denotes the collection (set) of numbers 1, 2, and 3. It is generally recognized that the theory of sets is robust enough to encapsulate all of mathematics. Hence it is natural to investigate notions of infinity within this theory. A brief history follows. Mathematician Richard Dedekind produced many classical writings between the years 1871 and 1888 which promoted a set-theoretic formulation of mathematical ideas. In 1888 he published a presentation of the basic elements of (a somewhat primitive) set theory, making a bit more explicit the operations on sets and mappings he had been using since 1871. His definition of an infinite set is given below.

Definition 1. (Dedekind Infinite Set) Let X be a set. Then X is Dedekind Infinite if and only if there exists a bijection between A and X for some proper subset A of X. 2

One may argue that Dedekind’s definition of an infinite set conforms to our informal notions of infinity. Indeed, let us consider some intuitively “infinite” set X. We will informally argue there is a bijection between A and X for some proper subset A of

X. Indeed, since X is in some sense inﬁnite, we can pick x1,x2,x3,... ∈ X. We now deﬁne f : X → X \{x1} by

x if x∈ / {x1,x2,x3,...}, and  f(x)=  . xi+1 if x = xi.  

It is easy to see that f is a bijection between X and X\{x1}. Thus X is Dedekind Infinite. Conversely, one can give an argument (which is necessarily informal at this point since we have yet to introduce the ZFC axioms) that if X is not infinite, then X is not Dedekind Infinite. Another great mathematician, Paul Stäckel, also made contributions to set theory. While at the University of Hannover in 1907, Stäckel gave us his definition of an infinite set.

Definition 2. (StäckelInfinite Set) Let X be a set. Then X is StäckelInfinite if and only if there exists a well-order on X such that the reversal order is not a well-order on X.

In the next chapter we discuss well-orders in more detail and prove the Well- Ordering Theorem which states that every set can be well-ordered. Assuming (for now) the Well-Ordering Theorem, we note how Stäckel’s definition also seems to capture our informal notion of what an infinite set should be. Indeed, let us consider some set X which is, informally, “infinite.” By the Well-Ordering Theorem, there exists a well-order < on X. Thus there must exist a least element of X, call it x1. 3

There must also exist a least element of X \{x1}, call it x2. We can continue this construction indefinitely as X is “infinite” to create the set S := {x1,x2,...}. It is clear that S has no greatest element, and thus in the reversal order, there is no least element. Hence X is a Stäckel Infinite Set. Another contributor to the foundations of set theory was Alfred Tarski. He was a brilliant mathematician; in fact, he was the youngest person ever to complete a doctorate at Warsaw University. Tarski’s mathematical interests were exceptionally broad, yet his first paper, published in 1920 when he was 19 years old, was on set theory. In what follows, we let P(X) denote the power set of X, that is, the set of all subsets of X. We now give Tarski’s definition of an infinite set.

Definition 3. (Tarski Infinite Set) Let X be a set. Then X is a Tarski Infinite Set if there exists a nonempty S ( P(X) such that for all Y ∈S there exists Z ∈S such that Y ( Z.

In some sense, Tarski had appropriately defined what one may want an infinite set to be. We now argue that any Tarski Infinite set must be, informally, an infinite set and conversely. Indeed, let X be a Tarski Infinite Set. Then there exists a nonempty S⊆P(X) such that for all Y ∈ S there exists Z ∈ S such that Y ( Z. Intuitively then, S itself must be “infinite”. To wit, there exist s0,s1,s2, ··· ∈ S such that s0 ( s1 ( s2 ( ··· . Thus we can choose x0 ∈ s0, x1 ∈ s1 \ s0, x2 ∈ s2 \ s1, and so on.

It follows that these objects must be distinct. Since x0,x1,x2, ··· ∈ X, we see that X must be “infinite.” Conversely, now let X be (intuitively) an “infinite” set. Since X is “infinite” we can pick x0,x1,x2, ···∈ X such that xi =6 xj for i =6 j. Define

S := {{x0}, {x0,x1, }{x0,x1,x2}, ···}. 4

Then S =6 ∅ and S ⊆ P(X) such that for all x ∈ S there exists a y ∈ S such that x ( y. Several years later, John von Neumann introduced the now standard axiom pos- tulating the existence of an infinite set. Before we give von Neumann’s definition, we must define what is contemporarily know as an inductive set.

Deﬁnition 4. For a set x, let x+ := x ∪ {x}. Now deﬁne a set I to be inductive if

(1) ∅ ∈ I, and (2) n+ ∈ I whenever n ∈ I.

Von Neumann’s Axiom of Infinity postulates the existence of an inductive set. One may argue that any inductive set is informally infinite. Indeed, let I be an inductive set. From (1) of the definition, ∅ ∈ I. It follows from (2) that ∅+ ∈ I. Applying (2) again we find that ∅++ ∈ I. Continuing in the fashion we see ∅, ∅+, ∅++, ∅+++, ··· ∈ I. Since these members of I are distinct (this is easy to show) and because we can continue the construction of the above members indefinitely, we find that I satisfies our informal notion of an infinite set. There are some sets, however, that appear to be infinite yet fail to be Inductive sets. Indeed, let I be and inductive set and choose y∈ / I. Next, set

I′ := {{y, i}: i ∈ I}.

It is easy to see that F : I → I′ by F (i) = {y, i} is a bijection between I and I′. Intuitively then, one may feel that I and I′ must be the same “size”, yet we now show that neither properties (1) or (2) from Deﬁnition 5 holds for I′. Indeed, ∅ ∈/ I′ as for all i′ ∈ I′, y ∈ i′. Furthermore, for any i′ ∈ I′, i′ contains exactly two sets while 5 i′+ contains exactly three sets. Thus i′+ cannot be in I′. Thus the set I′, although seemingly “inﬁnite”, is not an inductive set. 6

CHAPTER II

PRELIMINARIES

We now state the standard axioms for ZC − I set theory in which we will be working. We refer the reader to the references for two standard texts on set theory.

Axiom 1. (Extensionality Axiom) Let A and B be sets. If for all sets x: x ∈ A if and only if x ∈ B, then A = B .

In other words, if two sets have exactly the same members, then they are equal.

Axiom 2. (Empty Set Axiom) There exists a set ∅ such that for all sets x, x∈ / ∅.

Intuitively, there is a set having no members. Moreover, Extensionality implies that there is a unique set with no members.

Axiom 3. (Pairing) Let A and B be sets. There exists a set C such that for all sets x: x ∈ C if and only if x = A or x = B.

More informally, for any sets A and B, there is a set having as members just A and B.

Axiom 4. (Union Axiom) Let A be a set. Then there is a set B such that for all sets x: x ∈ B if and only if there is some y ∈ A such that x ∈ y.

The Axiom of Union asserts that if A is any set, then the members of the members of A form a set.

Axiom 5. (Power Set Axiom) Let A be a set. There exists a set B such that for all sets x: x ∈ B if and only if x ⊆ A. 7

Thus the collection of subsets of a given set is again a set.

Axiom 6. (Subset Axiom Schema) Let ϕ(x) be a formula in the language of set theory (possibly with parameters) in which no variable other than x occurs free. Then for any set A, there is a set B such that for all sets s: s ∈ B if and only if s ∈ A and ϕ(s).

The above axiom schema is also known as the Axiom Schema of Speciﬁcation. It simply asserts that the collection of members of a set A with a certain property forms a (sub)set (of A).

Axiom 7. (Axiom of Choice) Suppose that A is a family of nonempty sets. Then there is a function F (a “choice function” for A) with domain A such that F (x) ∈ x for all x ∈ A.

In other words, this “choice function” F picks out a member from each member of A.

Axiom 8. (Replacement Axiom Schema) For any formula φ(x,y) not containing the letter B, the following is an axiom:

∀A[(∀x ∈ A)∀y1∀y2(φ(x,y1)& φ(x,y2) =⇒ y1 = y2)

=⇒ ∃B ∀y(y ∈ B ⇐⇒ (∃x ∈ A) φ(x,y))].

Since this axiom will play little to no role in what follows, we omit clarifying comments as they will be largely irrelevant to this thesis. Now that the axioms are in place, we move on to state some fundamental deﬁni- tions. First note that it follows from the Extensionality Axiom that {x,y} = {y,x}. 8

One may find this less than desirable and wish for the order in which sets are listed to matter (more on this soon). It turns out (possibly surprisingly) that this is indeed possible with the list of axioms we presented above. The following definition is a simplified version of one given by Kuratowski in the early 1920s.

Deﬁnition 5. (Ordered Pair) Let x and y be sets. Deﬁne hx,yi := {{x}, {x,y}}.

We now show that this deﬁnition lives up to its name by showing hx,yi captures that x is “ﬁrst” and y is “second.”

Theorem 1. Let u,v,x,y be sets. Then hu,vi = hx,yi if and only if u = x and v = y.

Proof. We begin with the easy direction. Indeed, assume u = x and v = y. Then of course hu,vi = hx,yi and we are done. Now assume hu,vi = hx,yi. Thus

{{u}, {u,v}} = {{x}, {x,y}}.

Since the two sets are equal they must have the same members (Extensionality Ax- iom). Thus

(1) {u} ∈ {{x}, {x,y}}, and (2) {u,v} ∈ {{x}, {x,y}}.

From (1), we know that either

(a) {u} = {x} or (b) {u} = {x,y}, 9 and from (2), either

First suppose (b) holds. Then u = x = y. Now (c) and (d) are equivalent, and tell us that u = v = x = y. Similarly if (c) holds, we have the same situation. The ﬁnal case to consider is when (a) and (d) hold. From (a), we have u = x. From (d), we have either u = y or v = y. In the ﬁrst case, (b) holds, and we have already handled that case. In the second, v = y and we are done.

We are now able to deﬁne a relation.

Deﬁnition 6. A relation is a set of ordered pairs.

For a relation R, we sometimes write xRy in place of hx,yi∈ R. Before proceeding, we pause to remind the reader of several well-studied properties of certain relations.

Deﬁnition 7. Let R be a relation on a set A. Then R is transitive if for all x,y,z ∈ A: if xRy and yRz, then xRz.

Deﬁnition 8. Let R be a relation on a set A (that is R ⊆ A × A). Then R is said to satisfy trichotomy on A provided that for any x,y ∈ A: exactly one of xRy, yRx, x = y holds.

We are now able to deﬁne a linear ordering on a set A, which will later help us deﬁne a well-order.

Deﬁnition 9. R is a linear ordering on A if and only if R is a relation on A that is transitive and satisﬁes trichotomy on A. 10

Note the usual ordering relation < on the set R of real numbers is a linear ordering.

Moreover, if C := {Ai : i ∈ I} is a family of sets with the property that for distinct i, j ∈ I: either Ai ⊆ Aj or Aj ⊆ Ai, then ⊆ is a linear ordering on C. In what follows, x ≤ y abbreviates x

Deﬁnition 10. Let < be a linear ordering on a set A, and S ⊆ A with S 6= ∅. We call m ∈ S a least element of S with respect to < (if it exists) if and only if m ≤ x for all x ∈S.

Note that a least element of a set, if one exists, is automatically unique. Indeed, if m1 and m2 are both least elements, then m1 ≤ m2 and m2 ≤ m1, which implies m1 = m2. We are now equipped to introduce the notion of a well-order.

Deﬁnition 11. A well-ordering on a set A is a linear ordering < on A with the further property that every nonempty subset of A has a least element with respect to <.

We are almost ready to show that any set can be well-ordered (that is, for any set A, there is a well-order on A). We first must give one more definition, and also state and prove the Transfinite Induction Principle for future use.

Deﬁnition 12. If < is a linear ordering on A and t ∈ A, then the set

seg t = {x: x

is called the initial segment up to t. We sometimes write segA t to make explicit that the order to which we are referring is on the set A. 11

Theorem 2. (Transﬁnite Induction Principle) Assume that < is a well-order on a set A. Assume further that B is a subset of A with the property that for every t in A: if seg t ⊆ B, then t ∈ B.

Then B = A.

Proof. Let < be a well ordering on A, and let B be a subset of A be such that for every t in A:

if seg t ⊆ B, then t ∈ B.

Assume by way of contradiction that B =6 A. Then B ( A, so A \ B =6 ∅. Since < is a well ordering on A, it follows that A \ B has a least element m. Hence, for any y

We are now able to state and prove one of the most controversial and fascinating results in set theory: the Well-Ordering Theorem. The theorem was originally proved by none other than Zermelo himself.

Theorem 3. (The Well-Ordering Theorem) Every set can be well-ordered.

Proof. Let S be a nonempty set (it is trivial to verify that ∅ is a well-order on ∅), and let γ be a choice function for P(S) \{∅}. Suppose now that X ⊆ S and that

γ(S \ segX (x)) = x.

We shall require the following lemma. 12

Lemma 1. Suppose (X,

segX (α)= segY (β).

Then α = β.

This is easy to see since if (X,

segX (α) = segY (β), then

α = γ(X \ segX (α)) = γ(X \ segY (β)) = β.

We will now show that if (X,

X = Y or X =6 Y . We ﬁrst consider the case X = Y , and we will show that

In other words, we will show that for any a, b ∈ X: a

(0.1) segX (y) = segY (y) for all y ∈ Y.

Indeed, ﬁx y ∈ Y and suppose

′ ′ segX (y ) = segY (y )

′ ′ for all y ∈ Y with y

We claim segY (y) is “closed downward” in (X,

(0.2) if α ∈ segY (y), β ∈ X, and β

To see this, let α ∈ segY (y), β ∈ X and β

Next, observe that segY (y) =6 Y as y∈ / segY (y), but y ∈ Y . Thus segY (y) =6 X.

Let x be the least element of X (relative to the order

X (relative to the order

Hence segY (y) = segX (x) and it then follows from Lemma 1 that x = y and thus segX (y) = segY (y) as desired. We have established (0.1).

Now let a, b ∈ X. Then observe that a

(∗) if (X,

′ Next, let G := {(Xi,

(0.3) < is transitive on S′.

To see this, suppose x,y,z ∈ S′ with x

(Xi,

(0.4) < satisﬁes trichotomy on S′.

′ To wit, let x,y ∈ S . Since the γ-sets are chained, there is some γ-set (Xi,

∗ not. Then there is y ∈ X such that y

(0.5) < is a well-order on S′.

′ ′ Now, let x ∈ S . By deﬁnition of S , x ∈ Xi for some γ-set (Xi,

′ (0.6) segS (x) = segXi (x).

′ It is clear that segXi (x) ⊆ segS (x). The other inclusion follows immediately from another application of (∗). Our next claim is that

(0.7) (S′,<) is a γ-set.

′ To see this, we let x ∈ S be arbitrary. Then x ∈ Xi for some γ-set (Xi,

′ γ(S\segS (x)) = γ(S\segXi (x)) = x (the ﬁnal two equalities follow from (0.6) and the fact that (Xi,

CHAPTER III

RESULTS

We have been studying how to formalize the notion of “infinite” in the theory ZC − I. For now, we change gears and consider the analogous problem for “finite.” Certainly, we would like for ∅ to be “finite.” Further, if X is a “finite” set and we simply add one element to X, then we want the new set to remain finite. This motivates the following definition.

Deﬁnition 13. Let P (X) be a property of sets. Say that P (X) is k-ﬁnite if the following hold:

(1) P (∅) is true, and (2) For all X,Y : if P (X) is true, then P (X ∪ {Y }) is true.

So informally, a k-finite property (formula) P (X) in the language of set theory is a property that holds for all finite sets. We now recall Definition 1 from the introduction and state its negation for future use.

Deﬁnition 14. (Dedekind Finite Set) Let X be a set. Then X is Dedekind-ﬁnite if and only if there does not exist a bijection between between A and X for any proper subset A of X.

We now show that “Dedekind Finiteness” is a k-ﬁnite formula.

Proposition 1. Let P (X) be “X is Dedekind Finite.” Then P (X) is k-ﬁnite.

Proof. First note there does not exist a proper subset of ∅ and thus P (∅). Next assume there exists an X such that P (X). Then there does not exist a set A with 17 a bijection F : A → X where A is a proper subset of X. Let Y be any set and consider the set X ∩ {Y }. If Y ∈ X the problem is trivial. If Y∈ / X assume by way of contradiction there exists b ( X ∩ {Y } with a bijection f : b → X ∩ {Y }. Either Y ∈ b or Y∈ / b. First consider Y∈ / b. Then b ⊆ X. Let b∗ = b \{f −1(Y )}.

∗ ∗ Then f|b∗ : b → X is a bijection where b is a proper subset of X, and we reach a contradiction.

Now consider Y ∈ b. Let b1 = b \{Y }. Then b1 ( X. If f(Y ) = Y consider f|b1 : b1 → X which is a bijection where b1 ( X, and thus we reach a contradiction.

If f(Y ) =6 Y , consider g : b1 → X deﬁned by

f(α) when α =6 f −1(Y ), and  g(α)=  f(Y ) when α = f −1(Y ).   Note g is a bijection, and again we reach a contradiction. Therefore we conclude there does not exist a b ( X ∩ {Y } with a bijection f : b → X ∩ {Y }. In other words P (X ∪ {Y }).

As another example, let us negate the definition of “Stäckel Infinite Set”:

Definition 15. (StäckelFinite Set) Let X be a set. Then X is StäckelFinite if and only if for every well-order on X, the reversal order is also a well-order on X.

Now we show that “X is Stäckel-finite” is also a k-finite formula.

Proposition 2. Let P (X) be “X is St¨ackelFinite.” Then P (X) is k-ﬁnite.

Proof. First note that the only well order on ∅ is ∅ and its reversal order (again ∅) is also a well order. Thus P (∅). 18

We now assume P (X) for some set X and let Y be any set. Let R be a well order on X ∪ {Y }. Deﬁne

RX = {hα, βi∈ R: α =6 Y, β =6 Y }.

∗ Then RX is a well order on X, and thus the reversal-order RX is also a well-order on ∗ X. Hence, for every nonempty A ⊆ X there exists m ∈ A such that mRX α for all α ∈ A. Let B =6 ∅ with B ⊆ X ∪ {Y }. Either Y ∈ B or Y∈ / B. Consider Y ∈ B. Deﬁne A = B \{Y }. If A = ∅ then B = {Y } and thus Y is the least element of B.

∗ If A =6 ∅ then there exists m ∈ A such that mRX α for all α ∈ A. Since R is a well order on X ∪ {Y } it follows that it is also a linear ordering on X ∪ {Y }, and thus the reversal order R∗ is also a linear ordering on X ∪ {Y }. Therefore R∗ is transitive and satisﬁes the trichotomy on X ∪ {Y }. Thus we must have either

mR∗Y or YR∗m.

The ﬁrst case gives m as the least element while the second, using transitivity, gives Y as the least element. Now consider Y∈ / B. Then B ⊆ X from which it follows that there exists m ∈ B

∗ ∗ such that mRX α for all α ∈ B. This implies mR α for all α ∈ B. Thus m is a least element. In either case we see that any nonempty subset of X ∪ {Y } has a least element in reference to the reversal order R∗. In other words R∗ is a well-order on X ∪ {Y }, or P (X ∪ {Y }).

Next, we present another example of a k-ﬁnite formula.

Proposition 3. Let P (X) := for all Y : if Y∈ / X, then there is no surjection f : X → X ∪ {Y }. 19

Proof. First we will show P (∅). Note for any Y we have Y∈ / ∅. We proceed using the method of proof by contradiction. Indeed, assume there exists a surjection F : ∅ → {Y }. If this were true then we need to have F (X) = Y for some X ∈ ∅, but there is nothing in the empty set. Therefore no such X exists and we reach a contradiction. Thus no such surjection F can exist. In other words, P (∅). We will now show P (X ∪ {α}) when P (X) and α is any set. Indeed, assume P (X) for some set X and let α be any set. Let Y be any set such that Y∈ / X ∪ {α}. Assume by way of contradiction there exists a surjection f : X∪{α} → X∪{α}∪{Y }.

Our plan is to show there exists a surjection fX : X → X ∪ {Y } with Y∈ / X, thus contradicting P (X). We have two cases. Either α ∈ X or α∈ / X.

Case 1: α ∈ X. Let fX : = f. Since α ∈ X it follows that X ∪ {α} = X, and that X ∪ {α} ∪ {Y } = X ∪ {Y }. Since we have assumed that f is a surjection it follows that fX is also a surjection and we have reached a contradiction. Thus we conclude P (X ∪ {α}).

Case 2: α∈ / X. Deﬁne fX : X → X ∪ {Y } by

f(β) when f(β) =6 α  fX (β)=  Y when f(β)= α.   First we will show that the codomain of fX is X ∪ {Y }. Indeed, for any β ∈ X we have f(β) ∈ X ∪ {α} ∪ {Y } from which it follows that fX (β) ∈ X ∪ {α} ∪ {Y }. It follows that fX (β) ∈ X ∪ {Y }, as desired.

We now show that fX is onto X ∪ {Y }. Indeed let b ∈ X ∪ {Y }. We will show there exists β ∈ X such that fX (β) = b. We will consider the sub-cases f(α) ∈/ X and f(α) ∈ X. Case 2a: f(α) ∈/ X. We consider more sub-cases where either b ∈ X or b = Y . 20

Case 2a1: b ∈ X. Since b ∈ X there exists β ∈ X such that f(β) = b (since f is a surjection). Now, β =6 α since then f(α) = b ∈ X, contradicting that f(α) 6∈ X. It follows that fX (β) = b. Thus fX is a surjection which contradicts P (X). Therefore no such surjection f can exist. In other words P (X ∪ {α}).

Case 2a2: b = Y . Since b = Y , there exists β ∈ X ∪ {α} such that f(β) = Y as f is a surjection. Yet again we consider sub-cases. Either β ∈ X or β = α.

Case 2a2(i): β ∈ X. Since β ∈ X, Y =6 α, and f(β) = Y , we must have fX (β) = f(β) = Y = b. Thus fX is a surjection which contradicts P (X). Therefore no such surjection f can exist. In other words P (X ∪ {α}).

Case 2a2(ii): β = α. Since f(α)= f(β)= Y = b where f : X ∪{α} → X ∪{α}∪{Y } is a surjection with Y∈ / X ∪ {α}, there must exist βy ∈ X such that f(βy) = α.

It follows that fX (βy) = Y = b. Thus fX is a surjection which contradicts P (X). Therefore no such surjection f can exist. In other words P (x ∪ {α}). Case 2b: f(α) ∈ X. We again consider more sub-cases where either b ∈ X or b = Y .

Case 2b1: b ∈ X. If b ∈ X then there exists β ∈ X ∪ {α} such that f(β) = b as f is a surjection. Either β ∈ X or β = α.

Case 2b1(i): β ∈ X. Since β ∈ X and f(β)= b ∈ X it follows that fX (β)= b. Thus fX is a surjection which contradicts P (X). Therefore no such surjection f can exist. In other words P (X ∪ {α}).

Case 2b1(ii): β = α. Since f(α) = b ∈ X, where f : X ∪ {α} → X ∪ {α} ∪ {Y } is a surjection, we know there exists βb ∈ X such that f(βb) = α. It follows that fX (βb) = f(α) = b. Thus fX is a surjection which contradicts P (X). Therefore no such surjection f can exist. In other words P (X ∪ {α}). 21

Case 2b2: b = Y . Since b = Y and f(α) ∈ X there must exist β ∈ X such that f(β)= Y . It follows that fX (β)= Y = b. Thus fX is a surjection which contradicts P (X). Therefore no such surjection f can exist. In other words P (X ∪ {α}). In every case we see P (X ∪ {α}). Thus P (X) is k-ﬁnite.

As a final illustration, we negate “Tarski-Infinite Set” and show that the resulting formula is k-finite.

Deﬁnition 16. (Tarski Finite) Let X be a set. Then X is Tarski-ﬁnite if and only if for every nonempty S⊆P(X), S has a maximal element with respect to containment.

Proposition 4. Let P (X) be “X is Tarski Finite.” Then P (X) is a k-ﬁnite formula.

Proof. Indeed, let P (X) be as above. First we will show P (∅). Note that P(∅) = {∅}, and thus the only nonempty subset S of P(∅) is S = {∅}. Since ∅ 6( ∅ it follows that ∅ is a maximal element. Thus P (∅). We will now show P (X ∪ {Y }) when P (X) and Y is any set. Indeed assume P (X) for some set X and let Y be any set. Let S⊆P(X ∪{Y }) with S= 6 ∅. We will show that S has a maximal element with respect to containment. We begin by deﬁning

Sx = {β \{Y }: β ∈ S}.

Since S= 6 ∅ it follows that Sx =6 ∅. Since β ⊆ X ∪ {Y } for all β ∈S it follows that

β \{Y }⊆ X for all β ∈ S. In other words Sx ⊆ P(X). Since P (X) we know there exists M ∈Sx such that

M 6( α ∀α ∈Sx. 22

Noting that Y is not a member of α for any α ∈ Sx and adding Y to both sides we see

(0.8) M ∪ {Y }( 6 α ∪ {Y } ∀α ∈Sx.

It follows by the construction of Sx that for each β ∈S there exists α ∈Sx such that β ⊆ α ∪ {Y }. Using this fact it then follows from equation (0.8) that

(0.9) M ∪ {Y }( 6 β ∀β ∈S.

If M ∪ {Y }∈S it is a maximal element by equation (0.9) and thus P (X ∪ {Y }). Thus we now consider M ∪ {Y } ∈/ S, from which it follows that M ∈ S. We will show M is maximal in S. Assume by way of contradiction there exists Z ∈ S such that M ( Z.

As Z =6 M ∪ {Y }, it follows that

M ⊆ Z \{Y } since Y∈ / M. Assume

(0.10) M = Z \{Y }.

It follows that Z \{Y } ( Z from which it follows that Y ∈ Z. Therefore adding {Y } to both sides of (0.10) we have M ∪ {Y } = Z 23 and we reach a contradiction. Hence,

M ( Z \{Y }.

But Z \{Y }∈Sx thus contradicting M being the maximal element of Sx. Therefore we conclude there does not exist a Z ∈S such that M ( Z. In other words M is a maximal element of S, or P (X ∪ {Y }).

We are finally ready to present the main result of this thesis. Recall from the introduction that many mathematicians have presented notions of “infinite set.” We argued that some notions are more natural than others. We now present a vast class of “infinity axioms” which, we will argue, are strong enough to be adopted as axioms for infinity.

Theorem 4. Suppose that P (X) is a k-ﬁnite formula in the language of set theory. Then (ZC −I)+∃X¬P (X) proves the existence of a non-empty well-ordered set with no largest element.

Proof. We work in the theory (ZC − I)+ ∃X¬P (X). Let X0 be a set such that

¬P (X0). Since P (X) is k-finite, by definition, P (∅). Hence X0 =6 ∅. Next, let < be a well-order on X0. Now let S := {x ∈ X0 : P (seg(x))}. We claim that S is nonempty. Indeed, X0 is non-empty, so there is a least element x0 ∈ X0 (relative to <). Observe that seg(x0) = ∅. Since P (X) is k-finite, we have (by definition)

P (seg(x0)). Hence x0 ∈ S, so S =6 ∅. Now suppose by way of contradiction that S has a largest element, say ℓ. We consider two cases. 24

Case 1 ℓ is the largest element of X0. Then X0 = seg(ℓ) ∪ {ℓ}. Now, since ℓ ∈ S,

P (seg(ℓ)). Since P (X) is k-ﬁnite, P (seg(ℓ) ∪ {ℓ}). But then P (X0), contradicting

¬P (X0).

′ ′ Case 2 ℓ is not the largest element of X0. Then there is ℓ such that ℓ<ℓ . Without loss of generality, ℓ′ is least with this property. Then seg(ℓ′) = seg(ℓ) ∪ {ℓ′}. Recall again that P (seg(ℓ)). Since P (X) is k-ﬁnite, we see that P (seg(ℓ′)). But then ℓ<ℓ′ and ℓ′ ∈ S, contradicting our assumption that ℓ is the largest element of S.

We now reflect on the above theorem in light of our previous results. Recall that we have given plausibility arguments that both the notions of a Dedekind Infinite Set and a Tarski Infinite Set are “correct” ways to rigorously formalize infinity (more specifically, an infinite set). We now prove that if instead of assuming either of these axioms, we simply assume the existence of a set X for which ¬P (X) for some k-finite formula P (X), then we can prove the existence of both types of sets. Though this will take us too far afield, one can also prove the existence of an inductive set.

Corollary 1. The theory (ZC − I)+ ∃X¬P (X), where P (X) is a k-finite formula proves the existence of both a Dedekind Infinite Set and a Tarski Infinite Set.

Proof. As shown above, we have a nonempty well ordered set S with no greatest element. For s ∈ S, let s+ be the least element of S larger than s. Now deﬁne F : S → S by F (s) := s+. Then it is easy to see that F is one-to-one, and the least element of S is not in the range of F . Hence F is a bijection between S and F [S]. As for the latter assertion, we observe that the set X := {seg(s): s ∈ S} implies that S is Tarski Inﬁnite. This concludes the proof. 25

REFERENCES

Jech, Thomas. 2006. Set Theory. Berlin: Springer-Verlag. Enderton, Herbert. 1977. Elements Of Set Theory. New York: Academic Press.