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

The vocabulary V of D consists of a n-ary relation for each table where n is the number of columns in the table. That is, the vocabulary contains a unary relation F and binary relations P and B corresponding to the Female, Parent, and Birthday tables.

The V-structure D interprets these relations as rows of the tables. For example, D |= B(a, b) if and only if “ab” is a row of the Birthday table. This completely describes the V-structure D. For example, we see that

D |= F (Dot), D |= P (Zelda, M ax), and

D |= ¬B(Zelda, July 28).

In addition to B, F , and P , we can define first-order formulas expressing various other relations in D. For example, the formula ¬F (x) says that x is male. The formula z(P (x, z) P (z, y)) says that x is a grandparent of y. The conjunction of this formula with F (x) says that x is a grandmother of y. The formula yB(x, y) says that x is a date and the negation of this formula says that x is a person. The formula z(B(x, z) B(y, z)) asserts that x and y share the same birthday. There is no end to the relations that can be defined (see Exercise 2.2).

We return to this example at the end of Chapter 3 where we discuss Prolog. Prolog is a programming language based on first-order Horn logic that can be used to present and search any relational database.

2.4.3 Linear orders. Next, we look at some structures in the vocabulary V< consisting solely of the binary relation <. Rather than use the notation “< (x, y)” we use the more familiar “x < y” to express that the binary relation < holds for the ordered pair (x, y). As our choice of symbols indicates, each of the structures we consider interprets < as “less than.”

We consider four V<-structures denoted by N<, Z<, Q<, and R<. To define each structure, we must state what the underlying set is and how the symbols are to be interpreted. The underlying sets of the above structures are, in order, the natural numbers, the integers, the rational numbers, and the real numbers.

Each of these structures interprets < in the usual way. We can present these structures more concisely as follows:

•N< = (N| <)

•Z< = (Z| <)

•Q< = (Q| <)

•R< = (R| <).

These four structures have a lot in common. They are all V<-structures and each of them models the following V<-sentences:

x y((x < y) → ¬(y < x))

x(¬(x < x))

x y((x < y) (y < x) (x = y))

x y z(((x < y) (y < z)) → (x < z)).

Taken together, these sentences say that < linearly orders the underlying set. Each of the four structures models each of these four sentences. However, this is not true for all V<-sentences. Let ϕ be the sentence x y(y < x), saying that there is no smallest element. Clearly, R<, Q<, and Z< are models of ϕ. However, N< does have a smallest element, namely 1. So N< does not model ϕ, rather N< models the sentence x y¬(y < x), asserting that there is a smallest element. Call this sentence θ. Note that θ is equivalent to ¬ϕ. The sentence θ distinguishes N< from the other three models.

Next let us find a first-order sentence distinguishing Z< from the other three structures. Observe that Z< has no smallest element and it is not dense. A linearly ordered set is dense if between any two elements, there is another element. This property can be expressed in first-order logic by the following V<-sentence

x y((x < y) → z((x < z) (z < y))).

Call this sentence δ. Both Q< and R< model δ. Between any two rational numbers a and b there exist infinitely many rational numbers [(a + b)/2 for one]. The same is true for the real numbers. However, the integers are not dense. Between 1 and 2 there are no other integers. So Z< |= ¬δ. The V<-sentence ϕ ¬δ distinguishes Z< from the other three structures.

Now suppose that we want to distinguish between Q< and R<. We may use the fact that R< is bigger than Q<. (In the next section, we discuss the size of a structure in detail and show that, is some precise sense, there are more real

72 Structures and first-order logic

numbers than rational numbers.) Another distinguishing characteristic is ordercompleteness. A linear order is order-complete if it cannot be split into two open

intervals. The set of rational numbers, for example, is the union of the intervals

√ √

(−∞, 2) and ( 2, ∞). The parentheses “(” and “)” indicate that the intervals

√

do not contain the end points (this is what we mean by “open”). Since 2 is not a rational number, every rational number is in one of these two intervals. So Q< does not have order-completeness. The structure R<, on the other hand, does have order-completeness. This is a distinguishing characteristic of the real numbers.

However, if we attempt to find a V<-sentence that distinguishes R< from Q<, we will fail. For every V<-sentence ϕ, R< |= ϕ if and only if Q< |= ϕ. We give an elementary proof of this in Section 5.2. Our first-order language is too weak to express any di erence in these structures. We noted that R< is order-complete whereas Q< is not, but we cannot express this with a firstorder sentence (try it). Rather, order-completeness is a second-order concept. In second-order logic we can express things like “there do not exist two subsets such that. . . ” We also noted that R is bigger than Q, but, as we will see in Chapter 4, first-order logic can not distinguish between one infinite number and another. Both Q< and R< are infinite, and that is all first-order logic can say. From the point of view of V<-sentences, the structures R< and Q< are identical.

2.4.4 Number systems. Although first-order logic cannot tell the di erence between the V<-structures Q< and R<, it can tell the di erence between the real numbers and the rational numbers in vocabularies other than V<. Consider the vocabulary of arithmetic {+, ·, 0, 1} having binary functions + and ·, and constants 0 and 1. Let Var denote this vocabulary and consider the following Var-structures:

•A = (Z|+, ·, 0, 1),

•Q = (Q|+, ·, 0, 1),

•R = (R|+, ·, 0, 1),

•C = (C|+, ·, 0, 1).

The underlying sets of these structures are, in order, the integers, the rational numbers, real numbers, and the complex numbers. Each of these structures interprets the symbols of Var in the usual way.

Rather than use the formal notation “+(x, y) = z” we use the more conventional “x + y = z.” Likewise, we write x · y instead of ·(x, y). We let 2 abbreviate (1 + 1), x2 abbreviate x · x, and so on. Any polynomial having natural numbers

as coe cients is a Var-term. Equations such as (for example)

x5 − 9x + 3 = 0

are Var-formulas. Again, 3 and x5 are not symbols in Var, they are abbreviations for the Var terms (1 + (1 + 1)) and x · (x · (x · (x · x))), respectively.

We still cannot express order-completeness in this vocabulary, but we can

distinguish between the structures R and Q. The Var-sentence x(x2 = 2) asserts

√

the existence of 2. It follows that R models this sentence and Q does not. Likewise, the equation 2x + 3 = 0 has a solution in Q but not in A. So Q |=x(2x + 3 = 0), whereas A |= ¬ x(2x + 3 = 0).

To progress from N to Z to Q, we add solutions for more and more polyno-

mials. We reach the end of the line with the complex numbers C. The complex

√

numbers are obtained by adding to the reals, the solution i = −1 of the equation x2 + 1 = 0. The Var-sentence x(x2 + 1 = 0) distinguishes C from the other structures in our list. The set C consists of all numbers of the form a + bi where a and b are both real numbers. The Fundamental Theorem of Algebra states that for any nonconstant polynomial P (x) having coe cients in C, the equation P (x) = 0 has a solution in C (this is true even for polynomials of more than one variable). So there is no need to extend to a bigger number system. By virtue of adding a solution of x2 + 1 = 0 to R, we have added a solution for every polynomial.

The names of these number systems reflect historical biases. The counting numbers 1, 2, 3, . . . are the “natural” numbers to consider in mathematics. Negative numbers are not natural, the square root of 2 is irrational, and the square root of −1 is imaginary. The names suggest that things get more complicated as we progress from “natural” numbers to “complex” numbers. From the point of view of first-order logic, however, this is backwards. The structure C is the most simple. The structure R is not simple like C, but it does have many desirable properties. We will discuss the properties of these two structures in Chapter 5. The structure A is not so nice. The “A” stands for arithmetic, which sounds quite elementary. However, from the point of view of first-order logic, A is most complex. We investigate the structure A in Chapter 8.

2.5 The size of a structure

For any set U , |U | denotes the number of elements in U . For a V-structure M , |M | means |UM |, the number of elements in the underlying set UM of M . We refer to |M | as the size of M . For example, if M2 is Graph 2 from Section 2.1, then |M2| = 5. If the underlying set of M is infinite, then we could just write

|M | = ∞ and say no more, but this oversimplifies the situation. It implies that any two infinite sets have the same size. This is not the case. To explain this, we need to say precisely what we mean by “same size.”

Let A and B be two finite sets. Picture each set as a box of ping pong balls. Imagine reaching into box A with your left hand and box B with your right hand and removing one ball from each. Repeat this process. Reach in to the boxes and simultaneously remove a ball from each, and again, and again. Eventually, one of the boxes is emptied. If box B is emptied first, then we conclude that box A must have contained at least as many balls as box B at the outset. That is, |B| ≤ |A|. Since A and B are finite, this is elementary. For infinite sets we take this idea as definition of “|B| is less than or equal to |A|.”

Definition 2.33 Let A and B be sets. We define “|B| ≤ |A|” as follows: |B| ≤ |A| if there exists a one-to-one function f from B into A.

The function in this definition plays the same role as our right and left hands in the preceding discussion. The definition requires that f is one-to-one and has domain B. Given any element b in B, the function “picks out” an element f (b) from A. If such a function exists, we conclude that |B| ≤ |A|.

Example 2.34 Let P be the set of all prime natural numbers and let E be the set of all even natural numbers. Let f : P → E be defined by f (p) = 2p. This function is one-to-one. We conclude that |P | ≤ |E|. That is, there are at least as many even numbers as there are prime numbers.

Example 2.35 Recall that N × N denotes the set of all ordered pairs (m, n) of natural numbers. Let f : N → N × N be defined by f (n) = (n, 1) for all n N. This function is one-to-one. We conclude that |N| ≤ |N × N|. The reader should not be surprised by this fact. Less obvious is the fact that the opposite is true. Consider the function g : N × N → N defined by g(m, n) = 2m3n. This too is a one-to-one function. So not only is |N| less than or equal to |N × N|, but also |N × N| is less than or equal to |N|. Naturally, we conclude that these two sets have the same size.

Definition 2.36 Let A and B be sets. We say A and B have the same size and write |A| = |B| if both |B| ≤ |A| and |A| ≤ |B|. We write |A| < |B| if both |A| ≤ |B| and it is not the case that |A| = |B| .

So to show that two sets A and B have the same size we must demonstrate a one-to-one function from A to B and a one-to-one function from B to A. It su ces to show there exists a function f from A to B (or from B to A) that is both one-to-one and onto (since f −1 is also one-to-one and onto). Such a function is called a one-to-one correspondence or a bijection.

Example 2.37 Let N be the natural numbers and again let E denote the even natural numbers. In some sense there are “more” natural numbers than even numbers (since E N). However, these two sets have the same size. This is witnessed by the function f (x) = 2x defining a bijection from N onto E.

Example 2.38 Let R be the real numbers and let I be (0, 1), the set of all reals between 0 and 1. The function f : R → I defined by f (x) = (2/π) arctan x is a bijection from R onto I. So |R| = |I|.

If sets A and B can be put into one-to-one correspondence with each other, then they must have the same size. The following theorem states that the converse is also true. If |A| ≤ |B| and |B| ≤ |A|, then there must exist a bijection between A and B. This provides an alternative definition for “same size.”

Theorem 2.39 Sets A and B have the same size if and only if there exists a bijection from A onto B.

Proof Only one direction requires proof. As we previously remarked, if there exists a bijection between A and B, then A and B must have the same size. We now prove the opposite: if |A| = |B|, then such a bijection necessarily exists.

Suppose A and B have the same size. By the definition of “same size” there exist one-to-one functions f : A → B and g : B → A. Our goal is to demonstrate a bijection h : A → B. Before defining h, we define some sequences.

Given any a A, we define a (possibly finite) sequence sa as follows. Let a1 = a. Now suppose am A has been defined for some m N. Take bm B such that g(bm) = am. If no such bm exists, then the sequence ends. Otherwise, if bm does exist, the sequence continues. Take am+1 A such that f (am+1) = bm. Again, if no such am+1 exists, the sequence terminates. Note that the sequence alternates between elements of A and elements of B. The sequence sa can be depicted as follows:

There are three possibilities for the sequence sa. Either it terminates with some element ai A, or it terminates with some element bi B, or it never terminates. These three possibilities partition the set A into three subsets.

•Let AA be the set of all a A such that sa terminates in A.

•Let AB be the set of all a A such that sa terminates in B.

•Let AN be the set of all a A such that sa never terminates.

Similarly, we can define sequences sb that begin with b B and partition B as follows:

• Let BA be the set of all b B such that sb terminates in A.

•Let BB be the set of all b B such that sb terminates in B.

•Let BN be the set of all b B such that sb never terminates.

The function f , when restricted to AA, is a bijection f : AA → BA. We know that f is one-to-one. To see that it is onto, take any b BA. Since the sequence sb terminates in A, there must exist a AA such that f (a) = b. (Otherwise, sb would be the one-element sequence b). Likewise g, when restricted to BB , forms a bijection g : BB → AB . Finally, AN and BN are in one-to-one correspondence by either g or f . A bijection h : A → B can now be defined by putting these three parts together.

For finite sets, Theorem 2.39 is elementary. To determine how many ping pong balls are in a given box, we put the ping pong balls into one-to-one correspondence with the set {1, 2, 3, . . . , k} for some k N (that is, we count them). We say that two boxes contain the same number of ping pong balls if each can be put into one-to-one correspondence with the same set {1, 2, 3, . . . , k} and, hence, with each other. If A and B are infinite, we may have di culty visualizing them as boxes of ping pong balls. We extrapolate our definitions for infinite sets from the corresponding definitions for finite sets. Furthermore, we employ the following assumption.

Assumption: If A and B are sets, then |A| ≤ |B| or |B| ≤ |A|.

For finite A and B, this assumption is a fact that can be proved. If we remove ping pong balls one at a time from each of two given boxes, eventually one (or both) of the boxes will be emptied. We must be careful, however, when handling boxes containing infinitely many ping pong balls (see Exercise 2.43). For infinite A and B, we accept this assumption without proof. It is equivalent to an axiom of mathematics known as the Axiom of Choice.

It follows from this assumption that, for any infinite set A, |N| ≤ |A|. This leads to a crucial dichotomy of infinite sets: either |N| = |A| or |N| < |A|.

Definition 2.40 A set A is denumerable if there exists a bijection between A and N.

Definition 2.41 A set A is countable if it is either finite or denumerable. Otherwise, A is uncountable.

Proposition 2.42 The set of rational numbers Q is countable.

Proof Clearly, |N| ≤ |Q| (since N Q). Conversely, each nonzero element in Q can be written in a unique way as a reduced fraction of natural numbers times (−1)m for m = 1 or 2. Let f : Q → N be defined by f ( ab (−1)m) = 2a3b5m where ab is reduced. Further, let f (0) = 0. Now f is a one-to-one function from Q into N. By definition, |Q| ≤ |N|. Hence Q and N have the same size.

In a similar manner, we showed in Example 2.35 that N×N has the same size as N. So N × N is a countable set. We use this to prove the following useful fact.

Proposition 2.43 The union of countably many countable sets is countable.

Proof For each n N, let An be a countable set. Let U denote the union of these sets. If the Ans are each denumerable and are disjoint from one another, then U is as big as possible. Suppose this is the case. So each An can be enumerated as {a1, a2, a3, . . .}. Let f (m, n) denote the mth element in the enumeration of An. This defines a bijection f : N × N → U . We conclude that U has the same size as N × N. Since N × N is countable, so is U .

An example of an uncountable set is provided by the set of all subsets of N. For any set A, the set of all subsets of A is called the power set of A, denoted by P(A). We show that |P(A)| is always strictly bigger than |A|.

Proposition 2.44 For any set A, |A| < |P(A)|.

Proof To show that |A| < |P(A)| we must show that both |A| ≤ |P(A)| and

|A| = |P(A)|.

The one-to-one function f : A → P(A) defined by f (a) = {a} (for each a A) shows that |A| ≤ |P(A)|.

To show that |A| = |P(A)|, we must show that there does not exist a bijection between A and P(A). Let g be an arbitrary one-to-one function from A to P(A). We show that g is necessarily not onto. (Note that the above one-to-one function f is not onto.) For each element a in A, either a is in the set g(a) or a is not in g(a). Let X be the set of those elements a in A for which a is not in g(a). Then a X if and only if a g(a). For each a A, it cannot be the case that g(a) = X (otherwise we would have a X if and only if a X which is absurd). Since X is not in the range of g, g is not onto. Since g was arbitrary, we conclude that no one-to-one function from A to P(A) is onto.

Corollary 2.45 Any denumerable set has uncountably many subsets.

In particular, there are uncountably many subsets of N. We use this fact to show that there are uncountably many real numbers.

Proposition 2.46 The set of real numbers R is uncountable.

78 Structures and first-order logic

Proof We define a one-to-one function f from P(N) into R.

Let X be an element of P(N). Then, as a subset of the natural numbers, X contains at most 10 single-digit numbers, at most 90 two-digit numbers, at most 900 three-digit numbers, and so forth. Let rX be the real number between 0 and 1 described as follows. The first two digits following the decimal point represent the number of single-digit numbers in X. These are succeeded by each of the single-digit numbers in X listed in ascending order. The next two digits in the decimal expansion of rX represent the number of two-digit numbers in X. These are followed by the list of the two-digit numbers in X. The next three digits state how many three-digit numbers are in X, and so forth.

For example, let X = {2, 4, 5, 6, 7, 8, 9, 10, 24, 213, 3246}. There are 07 singledigit numbers in X (namely 2, 4, 5, 6, 7, 8, and 9), there are 02 two-digit numbers (namely 10 and 24), there is 001 three-digit number (213), and 0001 four-digit number (3246). So we have

rX = 0.0724567890210240012130001324600000000 . . .

The number rX contains a complete description of the set X. It follows that the function f : P(N) → R defined by f (X) = rX is a one-to-one function. Hence |P(N)| ≤ |R|. Since P(N) is uncountable, so is R.

We next show that there are only countably many V-formulas for any countable vocabulary V.

Proposition 2.47 If the vocabulary V is countable, then so is the set of all V-formulas.

Proof We define a one-to-one function f from the set of all V-formulas into N. Since V is countable, we can assign a di erent natural number to each symbol occurring in a V-formula. Then to each V-formula, there is an associated finite sequence of natural numbers. Suppose that a given V-formula ϕ has a1, a2, . . . , am as its associated sequence of natural numbers. Define f (ϕ) as the

product

2a1 · 3a2 · 5a3 · · · · · pann

where pn denotes the nth prime number. We recall two basic facts about the natural numbers: there are infinitely many primes and there is a unique way to factor any given natural number into primes. So we can factor the natural number f (ϕ) to recover the sequence a1, . . . , an and the formula ϕ. It follows that f is a one-to-one function as was required.

By Proposition 2.47, most subsets of N are not definable in any countable vocabulary. The same idea used to prove Proposition 2.47 can be used to show that there are countably many sentences in English or any other natural language. So there exist uncountably many real numbers that elude description in