Hedman. A First Course in Logic, 2004 (Oxford)

150 Properties of first-order logic

Put yet another way T + is the limit of the sequence T0, T1, T2, . . . If we continue this process forever, then T + is the end result.

We must verify that T + has all of the desired properties. First of all,

a contradiction from T +. Since formal proofs are finite, we could derive a contradiction from a finite subset of T +. Since every finite subset of T + is consistent, so is T +.

So T + is a theory. We next show that T + is a complete theory. Let ϕ be an arbitrary V+-sentence. Then ϕ is ϕi for some i. Since either ϕi or ¬ϕi is in Ti T +, T + is complete.

Finally, we must show that T + has Properties 1 and 2. Since T0 = Γ, every sentence of Γ is in T +. So T + has Property 1. To show that T + has Property 2, let xθ(x) be a V+-sentence in T +. This sentence is ϕm+1 for some m. Since this sentence is in T +, Tm {¬ϕm+1} is inconsistent. In this case, Tm+1 is defined as Tm {ϕm+1} {θ(ci)} for some constant ci. So θ(ci) is in T + and T + has Property 2.

Having successfully defined T +, we next define a V+-structure M + that models T +. The underlying set U + of M + is a set of variable-free V+-terms. Let t1 and t2 be two V+-terms that do not contain variables. We say t1 and t2 are the “same” if T + says they are. That is, t1 and t2 are the same if and only if T + t1 = t2. (Note that, since T + is complete, either T + t1 = t2 or T + ¬(t1 = t2).) Let U + be such that every variable-free V+-term is the same as some term in U + and no two terms of U + are the same. So if t1 and t2 are the same, then U + does not contain both of them,it contains exactly one term that is the same as these terms.

To complete our description of M +, we must say how M + interprets V+. Since the elements of U + are V+-terms, there is a natural interpretation. For any constant c V+, there exists a unique term t U + such that T + t = c. The structure M + interprets the constant c as the element t in its underlying set. Moreover, M + interprets the relations and functions of V+ in the manner described by T +. More precisely,

This completes our description of M +.

Claim For any V+-sentence ϕ, M + |= ϕ if and only if T + ϕ.

Proof Since every first-order sentence is equivalent to a sentence that uses only the fixed symbols ¬, , , and = (and neither , ←, ↔, nor ), we may assume with no loss of generality that these are the only fixed symbols occurring in ϕ. We proceed by induction on the total number of occurrences of ¬, , and in ϕ.

If ϕ has no occurrences of these three fixed symbols, then ϕ must be atomic. In this case, M + |= ϕ if and only if T + ϕ by the definition of M +.

Suppose now that ϕ has a total of m + 1 occurrences of ¬, , and . Our induction hypothesis is that the claim holds for any sentence having m or fewer occurrences of these symbols.

If ϕ has the form ψ θ, then T + ϕ if and only if both T + ψ and T + θ (since T + is a complete theory). By our induction hypothesis, this happens if and only if M + models both ψ and θ and, therefore, ϕ as well.

If ϕ has the form ¬ψ, then T + ϕ if and only if ψ is not in T + (since T + is a complete theory). By our induction hypothesis, this happens if and only if M + does not model ψ. By the semantics of ¬, M + does not model ψ if and only if M + |= ϕ.

Lastly, suppose that ϕ has the form xθ(x). By Property 2 and our definition of U +, T + ϕ if and only if T + θ(s) for some term s in U + (since T + is a complete theory). By our induction hypothesis, T + θ(s) if and only if M + |= θ(s). Finally, by the semantics of , M + |= θ(s) for some term s U + if and only if M + |= ϕ.

This completes the proof of the claim.

It follows from this claim that M + |= T +. Hence, we have demonstrated a model for Γ as was required.

Corollary 4.3 Let Γ be a countable set of formulas. If Γ is consistent, then Γ has a countable model.

Proof The structure M + from the proof of Theorem 4.2 is countable.

The following corollary is a countable version of the Compactness theorem for first-order logic.

Corollary 4.4 A countable set of formulas is satisfiable if and only if every finite subset is satisfiable.

Proof Let Γ be a countable set of formulas. We prove that Γ is unsatisfiable if and only if there exists a finite subset of Γ that is unsatisfiable. Clearly, if there exists a finite subset of Γ that is not satisfiable, then Γ is not satisfiable either. So suppose Γ is not satisfiable. By Theorem 4.2, Γ is inconsistent. That is, Γ

152 Properties of first-order logic

for some contradiction . Since formal proofs are finite, ∆ for some finite subset ∆ of Γ. By Theorem 3.4, ∆ |= and ∆ is unsatisfiable.

The following corollary is a countable version of the Completeness theorem for first-order logic.

Corollary 4.5 For any countable set of formulas Γ, Γ ϕ if and only if Γ |= ϕ.

Proof If Γ ϕ, then Γ |= ϕ by Theorem 3.4. Conversely, suppose that Γ |= ϕ. Then Γ {¬ϕ} is unsatisfiable. By Theorem 4.2, Γ {¬ϕ} is inconsistent. That is,

Γ {¬ϕ}

for some contradiction . By Contrapositive,

Γ {T } ¬¬ϕ,

where T is the tautology ¬ . Finally,

Γ ϕ

by the Tautology rules and Double negation.

All of the results of this section can be extended to include uncountable sets of sentences. We state and prove both the Compactness theorem and the Completeness theorem in their full generality in Section 4.3. This requires familiarity with cardinal numbers.

4.2 Cardinal knowledge

We return to our discussion of infinite sets. In Section 2.5, we defined what it means for two sets to have the “same size.” We now introduce numbers to represent the size of a set. These numbers are called cardinals and the size of a set is called the cardinality of the set. If a set is finite, then its size is some natural number (or zero if the set is empty). So each natural number is a cardinal. The Hebrew letter (aleph) is used with subscripts to denote infinite cardinals. The smallest infinite cardinal is 0. This is the cardinality of the set N and, therefore, of every countably infinite set.

The cardinality of set A is denoted |A|. In Section 2.5 we made the assumption that for any sets A and B, either |A| ≤ |B| or |B| ≤ |A|. This assumption allows us to list the cardinals in ascending order as follows:

0, 1, 2, 3, . . . , 0, 1, 2, . . .

The cardinals 1, 2, and beyond are uncountable cardinals. We showed that the set of real numbers is uncountable in Section 2.5. This raises a new question: where in the above list does |R| fall? Is it equal to 1 or some other uncountable cardinal? We address this question and state some surprising results at the end of the present section. As we shall see, it is possible that the cardinality of the reals is bigger than n for each natural number n. The above list of cardinals is only a partial list. To extend this list we must discuss ordinal numbers.

4.2.1 Ordinal numbers. There are two types of numbers: cardinals and ordinals. Whereas cardinals regard quantity, ordinals regard the length of an ordered list. The di erence between cardinals and ordinals is the di erence between 7 and 7th. This distinction is mere pedantry for finite numbers. For infinite numbers, however, the distinction between cardinals and ordinals is essential.

Example 4.6 Consider the following ordered lists of natural numbers:

A = {1, 2, 3, . . .}

B = {2, 3, 4, . . .} {1}

C = {3, 4, 5, . . .} {1, 2}

D = {1, 3, 5, 7, . . .} {2, 4, 6, 8, . . .}.

As sets, each of these is identical to N. The cardinality of each of these sets is 0. However, the order in which these sets are listed di ers. In B, the number 1 follows infinitely many numbers. In this sense, B is longer than A. Likewise C is longer than B and D is the longest of the four lists. Ordinal numbers recognize this distinction. The ordinal number ω describes the length of the natural numbers with the usual order. So ω describes the ordered set A. The length of B is denoted ω +1. Likewise, the ordinal ω +2 describes C. Finally, the ordinal representing the length of D is ω + ω.

Whereas every set has a cardinality, not every set has an ordinality. Ordinality is defined only for sets that are well ordered. A linearly ordered set is a set X with a binary relation < so that

1.for all a and b in X, exactly one of the following hold: either a < b, b < a, or a = b, and

2.for all a, b and c in X, if a < b and b < c then a < c.

That is, a linearly ordered set is a set equipped with a notion of “less than” by which any two nonequal elements can be compared. A well-ordered set is a linearly ordered set that is ordered in such a way that every nonempty subset has a least element.

Example 4.7 The natural numbers N with the usual ordering is a well ordered set. Any given set of natural numbers must contain a smallest number. The rational numbers Q with the usual ordering is a linearly ordered set that is not well ordered. To see this, consider the set {1/n | n N}. This subset of the rational numbers does not contain a smallest element.

Any finite linearly ordered set is well ordered. The ordinality of a finite set does not depend on the particular order of the set. If ten people are standing in a queue, then, regardless of their arrangement, one thing is certain: the tenth person is last. As Example 4.6 demonstrates, the same cannot be said for infinite sets.

Definition 4.8 Let A be a finite well ordered set. The ordinality of A is the same as the cardinality of the set.

So the ordinals, listed in ascending order, begin with the finite ordinals 0, 1, 2, 3, . . . . To continue the list we apply the following rule.

Given any nonempty set of ordinals, there exists a least ordinal greater than each ordinal in that set.

All ordinals are generated by repeated application of this single rule. The least ordinal greater than each finite ordinal is denoted by the Greek letter ω (omega). So ω is the smallest infinite ordinal. This is the ordinality of N with the usual ordering. The least ordinal greater than ω is denoted ω + 1. The least ordinal greater than ω + 1 is ω + 2. These ordinals were illustrated in Example 4.6.

For any ordinal α, the least ordinal greater than α is called the successor of α and is denoted α + 1. Let A be a well ordered set having ordinality α. Then α + 1 is the ordinality of the well ordered set A {b} where b is a new element (not in A) that is greater than each element of A. Every ordinal has a successor, but not every ordinal has an immediate predecessor. An ordinal that has an immediate predecessor is called a successor ordinal. A nonzero ordinal that is not the successor of any ordinal is called a limit ordinal. For example, ω is the smallest limit ordinal.

The ordinals have a natural order. For any ordinal α, the successor of α and all subsequent ordinals are greater than α. We let < denote this order and refer to this as the usual order for the ordinals.

Proposition 4.9 Any set of ordinals with the usual order is a well ordered set.

Proof Let A be a set of ordinals. It is clear that A is a linearly ordered set. To show that it is a well ordered set, we must show that any nonempty subset X of A contains a least element. If X happens to contain 0, then X certainly has a least element. So suppose that 0 X. Let L be the set of all ordinals that are

less than every ordinal in X. Since 0 X, L is nonempty. So there exists a least ordinal greater than each ordinal in L. This is the least ordinal in X.

Since the set of all ordinals less than α is well ordered, it has an ordinality. We naturally define the ordinality of this set to be α. For example, the set {0, 1, 2, 3} of ordinals less than 4 has ordinality (and cardinality) 4. More generally, we now define the ordinality of an arbitrary well ordered set.

Definition 4.10 The well ordered set A has ordinality α if there exists a one-to- one correspondence f from A onto the set {β | β < α} that preserves the order. By “preserves the order” we mean that, for any x and y in A, x < y if and only if f (x) < f (y).

This is an unambiguous definition of ordinality that agrees with all of the facts we have previously stated about ordinality. (In particular, the reader can verify the ordinalities stated in Example 4.6.)

The ordinal α is identified with the set {β | β < α}. Clearly, any ordinal α uniquely determines the set {β | β < α}. Conversely, given {β | β < α}, we can define α as the least ordinal greater than each ordinal in this set. In light of this association, we consider α and {β | β < α} to be interchangeable entities. So the ordinal 4 is the set {0, 1, 2, 3}. The purpose of this is to facilitate our notation. In particular, we write |α| to denote the cardinality of the set {β | β < α}. We refer to α as being countable or uncountable depending on whether |α| is countable or uncountable.

Whereas there is only one countably infinite cardinal, there are many countably infinite ordinals (see Exercise 4.21). We proceed now to list some countable ordinals. The first ordinal is 0. After 0, we have the successors 1,2,3. . . followed by the limit ordinal ω. This is then followed by ω + 1, ω + 2, ω + 3, and so forth. The least ordinal greater than each ordinal in the set {ω + n | n N} is the limit ordinal ω + ω also known as ω · 2. This has successor ω · 2 + 1 which has successor ω · 2 + 2. Continuing in this manner we arrive at the limit ordinals ω · 3, ω · 4, and so forth. The least ordinal greater than each ordinal in the set {ω · n | n N} is the ordinal ω · ω also known as ω2. Likewise, ω3, ω4, and the limit ωω are each

ordinals as are ωωω and

ωωωωωω .

Each of these ordinals is countable. The least ordinal greater than each countable ordinal is denoted ω1. The cardinal 1 is defined as |ω1|. Likewise, ω2 denotes the least ordinal greater than each ordinal of cardinality 1. We define 2 as |ω2|. Whereas 2 is the cardinal immediately following 1, ω2 does not immediately follow ω1. Rather, ω1 is followed by ω1 + 1, ω1 + 2, and so forth.

The list of ordinals cannot be exhausted. Given any set of ordinals, there exist ordinals greater than all of those in that set. So it is nonsense to speak

of the totality of all ordinals. When we refer to the list of ordinals, it should be understood that this is not a complete list. There necessarily exist ordinals beyond those in any list, no matter how extensive. In particular, we forbid ourselves from referring to the set (or list) of all ordinals. Although it is alluring terminology, “the set of all ordinals” does not make sense.

We conclude our discussion of ordinal numbers by introducing the Well Ordering Principle. Consider the set Q. With its usual order, this set does not have an ordinality. As demonstrated in Example 4.7, this is not a well ordered set. With another order, however, Q is a well ordered set. Since Q has the same size as N, we can enumerate Q as {q1, q2, q3, . . .}. The rational numbers with this order has ordinality ω. As Example 4.6 shows, Q may have di erent ordinalities when arranged in a di erent order. Likewise, we can impose a well ordering on any set. This is the Well Ordering Principle

Proposition 4.11 (Well Ordering Principle) Any set X can be enumerated as

{xβ | β < α} for some ordinal α. Moreover, we may require that α be the least ordinal such that |α| = |X|.

Proof First we show that there exists an order < that makes X a well ordered set. Since there exist arbitrarily large ordinals, there exists an ordinal γ with |X| ≤ |γ|. By the definition of |X| ≤ |γ|, there exists a one-to-one function f from X into {β | β < γ}. For any x and y in X, we define x < y to mean f (x) < f (y). Since {β|β < γ} is well ordered, so is X with this order. Let α be the ordinality of this well ordered set.

Now consider the set of all ordinals δ with |δ| = |X|. Since it contains α , this is a nonempty set of ordinals. By Proposition 4.9, there exists a least ordinal α in this set. Since |α| = |X|, there exists (by Theorem 2.39), a one-to-one correspondence g from {β|β < α} onto X. For each ordinal β < α, let xβ denote g(β). This provides the required enumeration {xβ |β < α} of X.

The Well Ordering Principle is in fact equivalent to the statement that every set has a cardinality. It is also equivalent to our earlier assumption that, for any sets A and B, either |A| ≤ |B| or |B| ≤ |A|. Each of these statements is equivalent to an axiom of mathematics known as the Axiom of Choice. This axiom can be stated as follows: the Cartesian product of nonempty sets is nonempty. We view this as a reasonable axiom and employ it without further comment.

4.2.2 Cardinal arithmetic. The list of cardinal numbers begins with

0, 1, 2, 3, . . . , 0, 1, 2, . . .

We extend this list indefinitely by using ordinal numbers as subscripts.

We define the infinite cardinals by induction on the ordinals. Prior to stating this definition, we discuss what we mean by “induction on the ordinals.”

This version of induction, known as transfinite induction, can be used to show that some property P holds for each ordinal α. Like other forms of induction, transfinite induction consists of two steps: the base step and the induction step. First, we show that P holds for 0. This is the base step. Second, we show that if P holds for all β < α, then it holds for α as well. This is the induction step. If we successfully complete these two steps, then we can rightly conclude that P does, in fact, hold for each ordinal α (since there is no least ordinal for which property P does not hold).

Definition 4.12 We define the infinite cardinals by transfinite induction. First we define (again) 0 to be |N|. Let α be a nonzero ordinal. Suppose that ι has been defined for each ι < α. Let γ be the least ordinal such that |γ| > ι for all ι < α. We define α to be |γ|.

Having defined the cardinal numbers, we now define arithmetic operations for these numbers. Cardinal arithmetic must not be confused with ordinal arithmetic. Previous reference was made to ω + ω, ω · 2, and ωω. Since ω is an ordinal, these are expressions of ordinal arithmetic (each represents a countable ordinal). We turn now to cardinal arithmetic.

Definition 4.13 Let κ and λ be cardinals. Let A and B be disjoint sets with |A| = κ and |B| = λ.

•Addition: κ + λ = |A B|.

•Multiplication: κ · λ = |A × B|.

•Exponentiation: κλ = |F (B, A)| where F (B, A) is the set of all functions f : B → A having B as a domain and a subset of A as a range.

Note that these definitions are independent of our choice of A and B. The requirement that A and B are disjoint is needed only for adding finite cardinals.

If κ and λ are finite cardinals, then these definitions correspond to the familiar notions of addition, multiplication, and exponentiation. We demonstrate (but do not prove) this fact with an example.

Example 4.14 Let A = {a1, a2, a3} and let B = {b1, b2, b3, b4}.

Addition. We have A B = {a1, a2, a3, b1, b2, b3, b4}. Clearly |A| + |B| = 3 + 4 = 7 = |A B|.

Multiplication. Recall that A × B is the set {(ai, bj )|1 ≤ i ≤ 3, 1 ≤ j ≤ 4}.

Observing the above arrangement of the elements in A × B, we see that the size of A × B is 3 · 4 = 12. So |A| · |B| = 3 · 4 = 12 = |A × B|.

Exponentiation. The set F (B, A) consists of all functions f : B → A. Each function is determined by the values of f (b) for b B. For each of the four elements in B, there three possible values for f (b) in A. It follows that there are 3 · 3 · 3 · 3 = 34 functions in F (B, A). We see that |A||B| = 34 = 81 = |F (B, A)|.

So for finite cardinals, addition, multiplication, and exponentiation are nothing new. We now consider these operations for infinite cardinals. It turns out that adding and multiplying two infinite cardinals are remarkably easy tasks (easier than adding and multiplying finite cardinals). In contrast, exponentiation for infinite cardinals is remarkably hard. We deal with the two easier operations first. All there is to know about the addition and multiplication of infinite cardinals stems from the following result.

Theorem 4.15 Let κ be an infinite cardinal. Then κ · κ = κ.

Proof We prove that this holds for κ = α by transfinite induction on α. If

α= 0, then this follows from Example 2.35 where it was shown that |N×N| = |N|. Suppose now that κ = α for α > 0. Our induction hypothesis is that λ·λ =

λ for all infinite cardinals λ smaller than κ.

Let δ be the least ordinal such that |δ| = κ. We regard δ as the set of ordinals less than δ. By the definition of cardinal multiplication, κ · κ is the cardinality of the set δ × δ of ordered pairs of ordinals less than δ. We show that |δ × δ| = |δ| by arranging the elements of δ × δ into a well ordered set having ordinality δ.

Now, δ × δ is well ordered by the lexicographical order defined as follows: (β1, β2) precedes (γ1, γ2) lexicographically if and only if either β1 < γ1 or both β1 = γ1 and β2 < γ2 where < is the usual order for ordinals. This order is analogous to the alphabetical order of words in a dictionary. The ordinality of this well ordered set is δ2 which is bigger than δ.

We now impose a new order on δ × δ. We claim that the new order makes δ × δ a well ordered set having ordinality δ. We denote this order by and define it as follows:

(β1, β2) (γ1, γ2)

if and only if

either (β1, β2) precedes (γ1, γ2) lexicographically

OR

γ2 is larger than both β1 and β2.

The set δ×δ with the order is a well ordered set. We leave the verification of this as Exercise 4.18. This is also true of the lexicographical order. The crucial feature of is that, with this order, each element of δ × δ has fewer than κ predecessors. This is not true of the lexicographical order.

Let (β1, β2) be an arbitrary element of δ × δ. To see that this element has fewer than κ predecessors, first note that (β1, β2) (β, β) where β is the larger of β1 and β2. Further, (γ1, γ2) does not preceed (β, β) if either γ1 or γ2 is larger than β. Because of this, the predecessors of (β1, β2) are contained in (β + 1) × (β + 1). For example, suppose (β1, β2) = (1, 3). Then (β1, β2) (3, 3). The set of all elements of δ × δ that preceed (3, 3) are contained in the following square:

Note that this is the set 4 × 4 (recalling that the ordinal 4 is identified with {0, 1, 2, 3}). So, with the order , there are fewer than |4 × 4| = 16 predecessors of the ordered pair (3, 3). Likewise, for any β < δ, there are fewer than |(β + 1) ×(β + 1)| elements of δ × δ that preceed (β, β) in the order .

Since δ is least such that |δ| = κ and β < δ, we have |β + 1| = |β| < κ. By our induction hypothesis, |(β + 1) × (β + 1)| = |β + 1|. It follows that each element of δ × δ has fewer than κ predecessors in the order as was claimed.

Let γ denote the ordinality of δ × δ with . If γ were larger than δ, then there would necessarily exist elements with κ = |δ| predecessors. Since we have shown that this is not the case, we conclude that γ ≤ δ. It follows that κ · κ = |δ × δ| = |γ| ≤ |δ| = κ. Since it is clear that κ ≤ κ · κ, we have κ · κ = κ as was desired. By induction, this holds for κ = α for each ordinal α.

Corollary 4.16 Let κ and λ be nonzero cardinals. If either κ or λ is infinite, then λ · κ is the larger of κ and λ.

Proof Suppose that κ is infinite and λ ≤ κ. We have κ ≤ λ · κ ≤ κ · κ. Since, κ · κ = κ by Theorem 4.15, we conclude that λ · κ = κ. Likewise, if λ is infinite and κ ≤ λ, then λ · κ = λ.

Proof If one of κ and λ is infinite and the other is finite, then this corollary follows from Exercise 2.36. So suppose that κ and λ are both infinite. If λ ≤ κ,