Set Theory & Logic

8/3/2019 Set Theory & Logic

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1 Elementary Set Theory

Notation:{} enclose a set.{1, 2, 3} = {3, 2, 2, 1, 3} because a set is not dened by order or multiplicity.{0, 2, 4, . . .} = {x |x is an even natural number } because two ways of writing

a set are equivalent.is the empty set.x A denotes x is an element of A.N = {0, 1, 2, . . .} are the natural numbers.Z = {. . . , 2, 1, 0, 1, 2, . . .} are the integers.Q = { mn |m, n Z and n = 0 } are the rational numbers.R are the real numbers.

Axiom 1.1.Axiom of Extensionality Let A, B be sets. If (x)x A iff x Bthen A = B .

Denition 1.1 (Subset) . Let A, B be sets. Then A is a subset of B , written A B iff (x) if x A then x B .

Theorem 1.1. If A B and B A then A = B .

Proof. Let x be arbitrary.Because A B if x A then x BBecause B A if x B then x AHence, x A iff x B , thus A = B .

Denition 1.2 (Union) . Let A, B be sets. The Union A B of A and B is

dened by x A B if x A or x B .Theorem 1.2. A (B C ) = ( A B ) C

Proof. Let x be arbitrary.x A (B C ) iff x A or x B C iff x A or (x B or x C )iff x A or x B or x C iff (x A or x B ) or x C iff x A B or x C iff x (A B ) C

Denition 1.3 (Intersection) . Let A, B be sets. The intersection A B of Aand B is dened by a A B iff x A and x B

Theorem 1.3. A (B C ) = ( A B ) (A C )

Proof. Let x be arbitrary. Then x A (B C ) iff x A and x B C iff x A and ( x B or x C )iff (x A and x B ) or (x A and x C )iff x A B or x A C iff x (A B ) (A C )

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Denition 1.4 (Set Theoretic Difference) . Let A, B be sets. The set theoretic

difference A \ B is dened by x A \ B iff x A and x B .Theorem 1.4. A \ (B C ) = ( A \ B ) (A \ C )

Proof. Let x be arbitrary.Then x A \ (B C ) iff x A and x B C iff x A and not ( x B or x C )iff x A and ( x B and x inC )iff x A and x B and x inC iff x A and x B and x A and x inC iff x (A \ B ) and x (A \ C )iff x (A \ B ) (A \ C )

Provisional denition of function: Let A, B be sets. Then f is a function

from A to B written f : A B if f assigns a unique element b B to eacha A.But what is the meaning of assign?Basic idea is to consider f : R R dened by f (x) = x2 . We shall identify

f with its graph. To generalize this to arbitrary sets A and B we rst need theconcept of an ordered pair. IE, a mathematical object a, b satisfying

a, b = c, d iff a = c and b = d. In particular, 1, 2 = {1, 2}.

Denition 1.5 (Cartesian Product) . If A, B are sets, then their Cartesian Product A B = { a, b |a A, bB }

Denition 1.6 (Function) . f is a function from A to B iff f A B and for each a A, there exists a unique bB such that a, b f .

In this case, the unique value b is called the value of f at a, and we writef (a) = b.It only remains to dene a, b in terms of set theory.

Denition 1.7 (Ordered Pair) . a, b = {{a}, {a, b}}

Theorem 1.5. a, b = c, d iff a = c and b = d.

Proof. Clearly if a = c and b = d then a, b = {{a}, {a, b}} = {{c}, {c, d}} =c, d

1. Suppose a = b. Then {{a}, {a, b}} = {{a}, {a, a }} = {{a}, {a}} = {{a}}Since {{a}} = {{c}, {c, d}} we must have {a} = {c} and {a} = {c, d}. Soa = c = d, in particular, a = c and b = d.

2. Suppose c = d. Then similarly a = b = c so a = c and b = d.3. Suppose a = b and c = d.

Since {{a}, {a, b}} = {{c}, {c, d}} we must have {a} = {c} or {a} = {c, d}.The latter is clearly impossible, so a = c.Similarly, {a, b} = {c, d} or {a, b} = {c}. The latter is clearly impossible,and as a = c, then b = d.

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NB (Note Bene) - It is almost never necessary in a mathematical proof toremember that a function is literally a set of ordered pairs.

Denition 1.8 (Injection) . The function f : A B is an injection iff (a, a A) if a = a then f (a) = f (a )

Denition 1.9 (Surjection) . The function f : A B is a surjection iff (bB )(a A) such that f (a) = b

Denition 1.10 (Composition) . If f : A B and g : B C are functions,then their composition g f : A C is dined by (g f )(x) = g(f (x))

Theorem 1.6. If f : A B and g : B C are surjections, then g f : A C is also a surjection.

Proof. Let cC be arbitrary.Since g is surjective, bB such that g(b) = c.Since f is surjective, a A such that g(a) = b.Then, ( g f )(a) = g(f (a)) = c, hence g f is a surjection.

Denition 1.11 (Bijection) . A function f : A B is a bijection if f is an injection and a surjection.

Theorem 1.7. If f : A B and g : B C are bijections, then g f : A C is a bijection.

Proof. Composition of surjections is a surjection, and compositions of injectionsare injections.

Denition 1.12 (Inverse Function) . If f : A B is a bijection, then itsinverse, f 1 : B A is dened by f 1(b) = the unique a A such that f (a) = b.

Remarks - If f : A B is a bijection, it is easily checked that f 1 : B Ais a bijection.

In terms of ordered pairs, f 1 = { b, a : a, b f }

Denition 1.13 (Equinumerous) . Two sets, A and B , are equinumerous, writ-ten A B iff there exists a bijection f : A B .

Theorem 1.8 (Galileo) . Let E = {0, 2, 4, . . .} be the even natural numbers.Then, N E

Proof. We can dene a bijection f : N E by f (n) = 2 n

Remark 1.1. If is often extremely difficult to explicitly dene a bijection f :N A. However, suppose that f : N A is a bijection. For each n N let anbe f (n). Then, a0 , a 1 , . . . is a list of the elements of A such that every element occurs exactly once, and conversely, if such a list exists, then we can dene a bijection f : N A by f (n) = an

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Theorem 1.9. N Z

Proof. We can list the elements of Z : 0, 1, 1, 2, 2, . . .Theorem 1.10. N Q

Proof. We proceed in two steps.First, we prove that N Q + = {q Q : q > 0} and consider the following

innite matrix

11

21

31

21

22

32

1323

33

Proceed through the matrix along the indicated route adding rational num-bers to your list if they have not already occurred.

Second, we declare that N Q . In the rst part, we showed that there existsa bijection f : N Q + hence we can list Q by 0, f (1) , f (1) , . . . .

Denition 1.14 (Power Set) . If A is any set, then its power set is P (A) ={B : B A}, so P ({1, 2, . . . , n })is of size 2n .

Theorem 1.11 (Cantor) . N P (N)

Proof. This method of proof is called the diagonal argument.We must show that there does not exist a bijection f : N P (N).Let f : N P (N) be any function.So, we shall prove that f is not a surjection. Hence, we must nd a set

S N such that n N f (n) = S .We do this via a time and motion studyFor each n N we must make the n th decision. That is, if n S .We perform the n th task, that is, ensure f (n) = S .We make the n th decision so that it accomplishes the n th task, ie, n S iff

n f (n).Clearly, f (n) and S will differ on whether they contain n, thus, n

N f (n) = S , and so f is not a surjection.

In general, unattributed Theorems are due to Cantor.

Theorem 1.12. If A is any set, then A P (A)

Proof. Suppose that f : A P (A) is any function. Consider the set S = {a A : a f (a)}.

Then, a S iff a f (a), so f (a) = S . Hence, f is not a surjection.

Denition 1.15 ( ). Let A and B be sets.A B iff there exists an injection f : A B .A B iff A B and A B

Theorem 1.13. For any set A, A P (A)

Proof. We can dene an injection f : A P (A) by f (a) = {a}.Thus, A P (A)By Cantors Theorem, A P (A), thus A P (A).

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Corollary 1.14. N P (N) P (P (N)) . . .

Denition 1.16 (Countable) . A set A is countable iff either A is nite or A N . Otherwise, A is uncountable.

Theorem 1.15. Let A,B,C be sets.

1. A A

2. A B B A

3. A B and B C A C

Proof. 1. Let idA : A A be the function dened by idA (a) = a. Clearly,idA is a bijection, so A A

2. Suppose A B . Then there exists a bijection f : A B . Since f 1 : B Ais also a bijection, B A

3. Suppose that A B and B C . Then, there exist bijections f : A Band g : B C . As f, g are bijections, g f : A C is also a bijection, soA C

Theorem 1.16 (Cantor-Bernstein) . If A B and B A then A B .

The proof will be delayed.

Theorem 1.17 (Zarmelo) . If A, B are any sets, then either A B or B A.

This proof will be omitted, though the theorem is equivalent to the axiomof choice.

Axiom 1.2 (Axiom of Choice) . Suppose F is a set of nonempty sets. Then there exists a function f such that f (A) A for each A F . We say that f isa choice function for F

Theorem 1.18. N Q

Proof. We can dene an injection f : N Q by f (n) = n. Hence, N QWe next dene g : Q N as follows:If 0 = q Q , we can uniquely express q = ab where = 1 and (a, b) = 1,

and a,b > 0.

So, g(q) = 2+1

3a

5b

. We also dene g(0) = 0.Clearly, g is an injection, so Q NThus, by the Cantor-Bernstein Theorem, N Q

Lemma 1.19. (0, 1) R

Proof. From elementary claculus, we can dene a bijection f : (0, 1) R byf (x) = tan( x 2 )

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Theorem 1.20. R P (N)

Proof. By the lemma, it is eanough to show that (0 , 1) P (N).We make use of the fact that each r (0, 1) has a unique decimal expansion

r = 0 .r 1r 2 . . . such that 0 r n < 9 and the expansion doesnt end in an innitestring of nines. (this is to avoid two expansions such as 0 .500 . . . = 0 .4999 . . .)

First we dene a function f : (0, 1) P (N) as follows. Suppose thatr = 0 .r 0r 1r 2 . . . . Let p0 , p1 , . . . be the primes in increasing order. Then f (r ) ={ pr 0 +10 , p

r 1 +11 , . . .}. Clearly, f is an injection, hence (0 , 1) P (N)

Next, we dene a function g : P (N) (0, 1) as followes: If S N theng(S ) = 0 .S 0S 1S 2 . . . where S n = 7 if n S and S n = 5 if n S (note, 7 and 5are arbitrarily chosen), hence P (N) (0, 1)

By the Cantor-Bernstein Theorem, (0 , 1) P (N), hence R P (N)

Theorem 1.21. If S N then either S is nite or S N . That is, N has theleast innite size.

Proof. Suppose S N is innite. Let S 0 , S 1 , S 2 , . . . be the increasing enumera-tion of S . This list witnesses that N S

Hypothesis 1.1 (Continuum Hypothesis) . If X R , then either X is count-able or X R

Theorem 1.22 (Godel and Cohen) . If the axioms of set theory are consistent,then it is impossible to prove or disprove the Continuum Hypothesis from theseaxioms.

Denition 1.17 (Set of Finite Subsets) . Fin( N) is the set of nite subsets of

N.Clearly, N Fin( N) P (N)

Theorem 1.23. N Fin( N)

Proof. We can dene an injection f : N Fin( N) by f (n) = {n}.Next, we can dene g : Fin( N) N as follows.Suppose that S = {s0 , s1 , . . . , s n } = , where s0 < s 1 < . . . < s n . Then

g(S ) = ps 0 +10 . . . p s n +1n , with p0 , . . . , p n being the primes in increasing order,and also g() = 0. Clearly, g is an injection, so Fin( N) N

By Cantor-Bernstein, Fin( N) N

Denition 1.18 (The set of functions between sets) . If A, B are sets, then A B = {f |f : A B }Theorem 1.24. N N P (N)

Proof. We rst dene a function g : P (N) N N as follows:For each subset of N , the characteristic function is (x) : N {0, 1} dened

by S (n) = 1 if n S and S (n) = 0 if n /S . The set g(S ) = S N N .Clearly, g is an injection, and so P (N) N N

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Next we dene : N N P (N) by (f ) = { pf (0)+10 , . . . , p

f (n )+1n , . . .}, where

p0 , p1 , . . . are the primes in increasing order.Clearly, is an injection.Hence, N N P (N).So, by Cantor-Bernstein, N N P (N)

And now, a heuristic principle. If S is an innite set, then, in general, if each s S is determined by nitely many pieces of data, then S is countable,and if each s S is determined by innitely many independent pieces of data,then S is countable.

Denition 1.19 (EC( N)) . EC( N) is the set of eventually constant functionsf : N N, ie, functions f such that there exist a, b N such that f (n) =b n a

Theorem 1.25. N EC( N)

Proof. First, we dene a mat f : N EC( N) by f (n) = cn , where cn (t) =n t N .

Clearly, f is an injection, so N EC( N).Next, we dene a map : EC( N) N by (g) = pf (0)+10 . . . p

f (a )+1a , where

a is the least integer such that f (t) = f (a) t a. Clearly, is an injection,so EC( N) N

Thus, by Cantor-Bernstein, N EC( N)

We are almost ready to prove the Cantor-Bernstein Theorem. First, we willrequire a denition and a lemma.

Denition 1.20 (Image of Z ). If f : X Y and Z X , then f [Z ] = {f (z) :z Z } is the image of Z .

Lemma 1.26. If f : A B is an injection and C A, then f [A \ C ] =f [A] \ f [C ]

Proof. First suppose x f [A \ C ]. Then, a A \ C such that f (a) = x. Hence,x f [A].

Suppose x f [C ]. Then, c C such that f (c) = x. As a = c, thiscontradicts f being an injection, thus x f [A] \ f [C ]

Conversely, suppose x f [A] \ f [C ]. Then a A such that f (a) = x.Since x /f [C ], a /C , so a A \ C , therefore, x f [A \ C ]

And now, the proof of the Cantor-Bernstein Theorem.

Proof. Since A B and B A, there exist injection f : A B and g : B A.Let C = g[B ] = {g(b) : bB }.Claim 1 - B C .The map b g(b) is a bijection from B to C , so the claim is proved.Thus, it is enough to show A C , because then A C and C B , so

A B

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Let h = g f : A A, then h is an injection.

Dene by induction on n 0 A0 = A4, An +1 = h[An ] and C 0 = C , C n +1 =h[C n ].Dene a function k : A C by k(x) = h(x) if x An \ C n for some n and

k(x) = x otherwise.Claim 2 - k is an injection.Suppose that x = x are distinct elements of A.There are three cases:

1. Suppose x An \ C n , x Am \ C m for some n, m . Since h is an injection,k(x) = h(x) = h(x ) = k(x )

2. Suppose x, x /An \ C n n. Then k(x) = x = x = k(x )

3. Without loss of generality, suppose x An \ C n for some n and x /An \C n n. Then, k(x) = h(x) h[An \ C n ] = h[An ] \ h[C n ] = An +1 \ C n +1 .Hence, k(x) = h(x) = x = k(x )

Claim 3 - k is a surjection.Let cC be arbitrary. There are two cases.

1. Suppose c /An \ C n n, then k(c) = c

2. Suppose c An \ C n for some n. Since c C = C 0 , m such thatn = m +1, since h[Am \ C m ] = h[Am ]\ h[C m ] = An \ C n . So, a Am \ C msuch that k(a) = h(a) = c.

Therefore, k is a bijection, so A C

Theorem 1.27. R R R

Proof. Since R (0, 1), it is enough to prove that (0 , 1) (0, 1) (0, 1)We can dene an injection f : (0, 1) (0, 1) (0, 1) such that f (r ) = 12 , r .Next we dene an injection g : (0, 1) (0, 1) (0, 1) as follows.If r = 0 .r 0r 1 . . . r n . . . and s = 0 .s0s1 . . . s n . . . , then,g( r, s ) = 0 .r 0s0r 1s1 . . . r n sn . . . .By the Cantor-Bernstein Theorem, (0 , 1) (0, 1) (0, 1)

Denition 1.21 (Sym( N)) . Sym( N) = {f N N : f is a bijection }

Theorem 1.28. Sym( N) P (N)

Proof. Since Sym( N) N

N P (N), we have Sym( N) P (N).

Next, we dene a function f : P (N) Sym( N) by S f S where f S (2n) =2n + 1 and f S (2n + 1) = 2 n if n S and f S (2n) = 2 n and f S (2n + 1) = 2 n + 1otherwise.

This is an injection, and so P (N) Sym( N), so, by Cantor-Bernstein,P (N) Sym( N)

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Denition 1.22 (Finite Sequence) . Let A be a set. Then a nite sequence of

elements of A is an inject a0 , a 1 , . . . , a n , with a i A and n 0.a0 , a 1 , . . . a n = b0 , b1 , . . . bm if and only if n = m and i a i = bi

Denition 1.23 (Finite Sequence) . a0 , . . . a n = f : {0, . . . , n } A : f (i) =a i

Denition 1.24 (Finseq( A)) . Finseq( A) is the set of nite sequences of el-emetns of A.

Theorem 1.29. If A is a nonempty countable set, then N Finseq( A)

Proof. Fix some a A. Then, we can dene an injection f : N Finseq( A) :

n

n

a , . . . , aSo, N Finseq( A)Next, we dene an injection g : Finseq( A) N as follows:Since A is countable, there exists an injection e : A N . We dene

a0 , . . . , a n 2e (a 0 )+1 . . . pe(a n )+1n , where pi is the i th prime.

Thus, Finseq( A) N , and so, by the Cantor-Bernstein Theorem, N Finseq( A)

And now, we move on to Cardinal Arithmetic.

Denition 1.25 (Cardinality) . To each set A we associate an object, its car-dinality, denoted card( A) or |A| such that |A| = |B | iff A B

Examples

1. If A is a nite set, then |A| is its usual size.

2. |N | = 0

3. The innite cardinals begin 0 ,1 , . . . , ,+1 , . . .

Denition 1.26 (Cardinal Addition) . Let , be cardinal numbers, then + = card( A B ) where card( A) = , card( B ) = and A B =

Theorem 1.30. If A A and B B and A B = A B = , then A B A B

Proof. Since A A and B B , there exist bijections f : A A andg : B B , then we can dene a bijection h : AB A B by h(x) = f (x)if x A and h(x) = g(x) if x B .

Theorem 1.31. 0 + 0 = 0

Proof. Let E be the even natural numbers and O be the odd natural numbers.So 0 + 0 = card( E O ) = card N = 0

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In fact, if , are cardinal numbers, at least one of which is innite, then

+ = max {, }And it is clearly impossible to sensibly dene an operation of cardinal sub-traction.

Denition 1.27 (Cardinal Multiplication) . Let , be cardinals. Then =card( A B ) when |A| = and |B | = .

We have already checked that this is well dened.

Theorem 1.32. 0 0 = 0

Proof. Since N N N , we have 0 0 = card( N N) = card( N) = 0

In fact, if , are cardinals, at least one of which is innite, and neither iszero, then = max {, }

Denition 1.28 (Cardinal Exponentiation) . If , are cardinal numbers, then = card( A B ) where |A| = and |B | = .

Theorem 1.33. |R | = 2 0

Proof. We know R P (N) N {0,1} , hence 20 = card( N {0,1} = |R |

In fact, if is any cardinal number, then < 2 .This just restates that A P (A)

2 Relations

Denition 2.1 (Binary Relation) . A binary relation on a set A is a subset R A A. We usually write aRb instead of a, b R

Example 2.1. 1. The order relation on N is < = { n, m : n, m N , n < m }

2. divisibility relation on N+ is D = { n, m : n, m N+ , m divides n}

Observation: Thus P (N N) is the set of binary relations on N. SinceN N N, it follows that P (N N) P (N), in particular, P (N N) isuncountable.

Let R be a binary relation on A.

Denition 2.2 (Reexive Property) . R is reexive if (a A)aRa

Denition 2.3 (Symmetric Property) . R is symmetric if (a, b A)aRb bRa

Denition 2.4 (Transitive Property) . R is transitive if (a,b,c A)aRb bRc aRc

Denition 2.5 (Equivalence Relation) . R is an equivalence relation iff R isreexive, symmetric and transitive.

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Example: Consider the relation R dened on Z by aRb iff 3|a b

Theorem 2.1. R is an equivalence relation on Z .

Proof. Let a Z , then 3 |a a = 0. Hence, aRa , thus R is reexive.Let a, b Z . Suppose aRb, so 3|a b, so 3|b a = (a b), so bRa, thus,

R is symmetric.Let a,b,c Z . Suppose aRb and bRc, so 3|a b and 3 |b c, so, 3|(a b) +

(b c) = a c, so aRc , and so R is transitive.

Denition 2.6 (Equivalence Class) . Let R be an equivalence relation on A.For each x A, the corresponding equivalence class is [x] = {y A : xRy }

Example continued:The equivalence classes of R above are:

[0] = {. . . , 6, 3, 0, 3, 6, . . .}[1] = {. . . , 5, 2, 1, 4, 7, . . .}[2] = {. . . , 4, 1, 2, 5, 8, . . .}

Denition 2.7 (Partition) . Let A be a nonempty set. Then, a partition of Ais a collection {B i : i I } such that

1. (i I )B i =

2. (i, j I ) if i = j then B i B j =

3. (a A)(i I ) such that a B i . That is, A = iI B i

Theorem 2.2. Let R be an equivalence relation on A.Then, (a A)a [a] and If a, b A and [a][b] = , then [a] = [b]. Hence,

the set of equivalence classes {[a] : a A} forms a partition of A.

Theorem 2.3. Let {B i : i I } be a partition of A. Dene the binary relation E on A by aEb iff (i I ) such that a, b B i . Then, E is an equivalencerelation and the E equivalence classes are precisely {B i : i I }

Example: How many equivalence classes are there on {1, 2, 3}?Answer: There are exactly ve equivalence relation, corresponding to{{1, 2, 3}}, {{1}, {2, 3}}, {{2}, {1, 3}}, {{3}, {1, 2}}, {{1}, {2}, {3}}

Denition 2.8 (Irreexive Property) . R is irreexive iff (a) a, a /R

Denition 2.9 (Trichotomy Property) . R satises the Trichotomy Property if (a, b A) exactly one of the following holds, aRb,a = b,bRa.

Denition 2.10 (Linear Order) . A, R is a linear order if R is irreexive,transitive, and satises the trichotomy property.

Denition 2.11 (Partial Order) . Let R be a binary relation on A, then A, Ris a partial order if R is irreexive and transitive.

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Denition 2.12 (Isomorphism) . Let A, < and B, be partial orders. A

map f : A B is an isomorphism if the following conditions are satised:1. f is a bijection

2. (a, b A)a < b iff f (a) f (b)

In this case, we say that A, < an B, are isomorphic, and write A, < =B,

Theorem 2.4. Z , < = Z , >

Proof. Dene f : Z Z by f (z) = z, then a < b iff a > b iff f (a) > f (b).Thus, f is an isomorphism

Notation - Let D be the strict divisibility relation on N+ . that is, aDb

a < b&a |bTheorem 2.5. N+ , D = P (N),

Proof. N+ P (N), so no bijection exists.

Theorem 2.6. R \ { 0}, < = R , r 2 > . . . r k > . . . f ( 1)Let s be the greatest lower bound of {r n : n 1}.t R \ { 0} such that f (t) = sClearly, t < 0, thus t2 > t

f (t2 ) > s , hence n 1 such that r n < f (

t2 ).But then t2 1 + . . . + 1

n t} with T Th( N) is a

nite subset and n1 , . . . , n t 1Let m > max {n1 , . . . , n t }. Let s : V N satisfy s(x) = m.Then, N, + , ,


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By compactness, is satisable.

Then,A

satises Th(N

) and yet s(x) = c > 1A

+ . . . + 1A

n for any numbernSo, c is an innite, or nonstandard, natural number. Furthermore, we can

assume A is countable, and so we get the structure N < Q Z .Thus, the statements that are true about N in rst order logic do not uniquely

identify N.

Denition 4.12 (Isomorphism) . Let A , B be structures for the rst order language L

A function f : A B is an isomorphism iff the following conditions aresatised:

1. f is a bijection

2. for each n-ary predicate symbol P and a1 , . . . , a n A, then a1 , . . . , a n P A iff f (a1), . . . , f (an ) P B

3. for each constant symbol c, f (cA ) = cB

4. For each n-ary function symbol h and a1 , . . . , a n A, f (hA (a1 , . . . , a n )) =

hB (f (a1), . . . , f (an ))

Theorem 4.4. Suppose : A B is an isomorphism if is any sentence,then A |= iff B |=

We will actually prove the more general statement:

Theorem 4.5. Suppose : A B is an isomorphism, and s : V A. If isany wff, then A |= [s] iff B |= [ s]

We shall need the following technical lemma:

Lemma 4.6. With the above hypotheses, for each term t, (s(t)) = s(t)

Proof. We argue by induction on the complexity of t.First suppose that t is a variable x.Then, (s(x)) = (s(x)) = ( s)(x) = (s(x)) = s(x)Next, suppose that t is a constant symbol c.Then, (s(c)) = (cA ) = cB = s(c)Suppose that t is f t 1 , . . . , t n , by the inductive hypothesis, we know (s(t i )) =

s(t i ).Thus, (s(f t 1 , . . . , t n )) = (f A (s(t1), . . . , s (tn ))) = s(f t 1 , . . . , t n ))

And now we can prove the theorem

Proof. We argue by induction on complexity of a wff .Assume is atomic, say, P t 1 , . . . , t nA |= P t 1 , . . . , t n [s] iff s(t1), . . . , s (tn ) P A

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iff (s(t1)) , . . . ,(s(tn )) P B

iff s(t1), . . . , s(tn ) P B

Next, suppose = , which is similar, as is = Finally, suppose = v . Then, A |= v [s]iff for all a A, A |= [s(v|a)]iff for all a A, B |= [ s(v|a)]iff for all a A, B |= [( s)(v|(a))]As is surjective, for all bB , B |= [( s)(v|b)]

Denition 4.13 (Models) . Let T be a set of sentencesA is a model of T iff A |= for all T Mod( T ) is the class of models of T .

Abbreviation: if E is a binary predicate symbol, then we write xEy insteadof Exy .

Example 4.2. Let T be the following:(x)xEx(x)(y)xEy yExThen, Mod( T ) is the class of graphs

Denition 4.14 (Axiomatizable) . Let C be a class of structures.C is axiomatizable iff there exists a set of sentences T such that C = Mod( T )If there exists a nite set of sentences T , such that = Mod( T ) then is

nitely axiomatizable.

Theorem 4.7. C is nitely axiomatizable iff C can be axiomatized by a singlesentence.

Example 4.3. The class of graphs is nitely axiomatizableThe class of innite graphs is axiomatizableIs the class of nite graphs axiomatizable? Is the class of innite graphs

nitely axiomatizable?

Theorem 4.8. Let T be a set of sentences in a rst order language.If T has arbitrarily large nite models, then T has an innite model.

Proof. For each n 1, let n be the sentence which says There exist at leastn elements.

Consider the set of sentences = T {n : n 1}We claim that is nitely satisableSuppose 0 is any nite subset. Then, 0 = T 0 {n 1 , . . . , n

t}

Let m = max {n1 , . . . , n t }, then T has a nite models A of size greater thanm.

Clearly, A is a model of 0 .By compactness, there exists a model B of . Thus, B is an innite model

of T .

Corollary 4.9. The class F of nite graphs is not axiomatizable.

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Corollary 4.10. The class C of innite graphs is not nitely axiomatizable.

Denition 4.15 (Semantically Implies) . Let be a set of wffs and let be a wff.

Then, logically/semantically implies , written |= iff for every struc-ture A for L and every s : V A, if A satises with s, then A satises with s.

Denition 4.16 (Valid) . The wff is valid iff |=

Now we will return to the development of syntax. will denote the set of logical axioms, dened later.

Denition 4.17 (Deduction) . Let be a set of wffs and let be a wff. Adeduction of from is a nite sequence of wffs 1 , 2 , . . . , n such that n = and for each 1 i n either i or there exist k, j < i such that k is j i

In the latter case, we say that i follows from j and j i by modusponens (MP).

Denition 4.18 (Theorem) . is a theorem of , written, iff there existsa deduction of from .

And now we nally arrive at the two main theorems of the course.

Theorem 4.11 (The Soundness Theorem) . If then |=

Theorem 4.12 (Godels Completeness Theorem) . If |= then .

Denition 4.19 (The Logical Axioms, ) . is a generalization of iff for some n 0 and variables x1 , . . . , x n we have that is x1 . . . xn . When n = 0 we see that is a generalization of .

The logical axioms are all generalizations of all wffs of the following forms:

1. Tautologies

2. (x xt ), where xt means to substitute t for all instances of x in when t is substitutable for x in .

3. (x( )) (x x )

4. ( x ) where x does not occur free in .

5. x = x

6. (x = y ( )) where is atomic and is the result of replacing some, possibly none, of the occurrences of x with y.

Explanations. A tautology is a wff which can be obtained from a propositional tautology by substituting wffs.

xt is the result of substituting t for each free occurrence of x in . We say t is substitutable for x in if no variable of t becomes bound by a quantier in xt .

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Now, we will prove the Soundness theorem, which is easy.

We will make use of the following resultLemma 4.13. Each logical axiom is valid.

And now, the proof of Soundness.

Proof. We argue by the minimum length n 1 of a deduction of from , that |= .

If n = 1, then .Clearly, if , then |= Suppose . Then is valid, and so |= .Next suppose n > 1.Let 1 , . . . , n be a deduction of from . Then, must follow by modus

ponens, and so there are earlier wffs and ( ).Since initial segments of deductions are also deductions, we have that

and ( ) by deductions of length less than n.Let A be any structure and s : V A. Suppose A |= [ s]. By the inductive

hypothesis, A |= and A |= ( )Hence, |=

Denition 4.20 (Inconsistent) . A set of wffs is inconsistent iff there existsa wff such that and .

Otherwise, is consistent.

Corollary 4.14. If is satisable, then is consistent.

Proof. Suppose that is satisable, say, A is a structure, s : V A andA

|= [ s].Suppose is inconsistent, then and By Soundness, |= and |= Then, |= [s] and |= [s], which is impossible.

Now, we shall begin working toward Completeness.First, we must prove a number of meta-theorems.

Theorem 4.15 (Generalization Theorem) . If and x does not occur freein then x

Proof. We argue by induction on the minimum length of a deduction of from.

Suppose that n = 1.Then, . If , then x , so done.If , then, as x is not free in , x does not occur free in . So x

is in . And so a deduction from of x is 1-, 2- x and 3-x, bymodus ponens.

Next, suppose that n > 1.Then, in a proof of minimum length n, follows from earlier wffs and

.

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By the induction hypothesis, x and x( ).

So, a deduction of x from would proceed by rst deducing x , thenx( ). Then, by the logical axioms, we have x( ) (x x).And so, by modus ponens, we get x x, and another modus ponens givesus x.

Denition 4.21 (Tautologically Implies) . { 1 , . . . , n } tautologically implies iff ( 1 ( 2 (. . . ( n ))) . . .) is a tautology.

Theorem 4.16 (Rule T) . If 1 , . . . , n and { 1 , . . . , n } tautologically implies , then .

Proof. Obvious, by repeated application of modus ponens.

Theorem 4.17 (Deduction Theorem) . If { } then ( )

Proof. We argue by induction on the minimal length of a deduction of from { }.

Suppose n = 1.First, suppose . Then, the following is a deduction of . 1-,

2-( ( )) and 3- Second, assume = .Then, is a tautology, and so of course ( )Suppose n > 1.Then, in a proof of minimal length n, we must have that follows from

earlier wffs and ( ) by modus ponens.By induction hypothesis, ( ) and ( ( ))Clearly, {( ), ( ( ))} tautologically implies .By Rule T, ( )

Theorem 4.18 (Contraposition) . {} iff {}

Proof. Suppose {} By deduction theorem, ( )By rule T, ( )Hence, {} , so ( )Hence, {} The other direction is similar.

Theorem 4.19 (Reductio Ad Absurdum) . If {} is inconsistent, then

Proof. Since {} is inconsistent, there exists a wff such that {} and {}

By deduction theorem, ( ) and ( )Clearly, {( ), ( )} tautologically implies By rule T,

Remark 4.2. If is inconsistant, then for every wff

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Proof. Suppose that and . Clearly ( ( )) is a tautology.

By Rule T , .Remark 4.3. Thus, to prove the consistancy of ZFC, it is enough to show that ZFC 0 = 1 .

Applications: Some theorems about equality:

1. x(x = x)

Proof. x(x = x)

2. xy(x = y y = x)

Proof. x = y (x = x y = x) by axiom 6. x = x by axiom 5,x = y y = x by Rule T and axioms 1,2. y(x = y y = x) by

Generalization

3. xyz(x = y (y = z x = z))

Proof. We have y = x (y = z x = z) by axiom 6. Then x =y y = x by the previous statement, so we have by Rule T and 1,2

(x = y (y = z z = x)), and so by Generalization three appliedthree times, we have xyz(x = y (y = z z = x)).

Theorem 4.20 (Generalization on Constants) . Assume that and c is a constant symbol which doesnt occur in . Then there exists a variable y which

doesnt occur in scuh that ycy . Furthermore, there exists a deduction of

ycy from in which c doesnt occur.

Remark 4.4. This says that if we suppose (c) and c does not occur in ,then for a suitable variable y, we have that y(y).

More importantly, suppose that is a consistant set of formulas in the lan-guage L . Let L + = L {c} where c is a new constant symol. Then remainsconstant in L + .

Proof. Suppose that (*)= 1 , . . . , n is a deduction of from . Let y be avariable with any i . We claim that (**)= ( 1)cy , . . . , ( n )cy is a deduction of

cy from . We will check that for all i n, either ( i )cy or follows by

MP. We will break into cases.

1. Suppose that i . Then i doesnt involve c, and so ( i )cy = i .

2. Suppose i . Then it is easily checked that ( i )cy .

3. Suppose that there exist j,k < i such that k is ( j i ). Then ( k )cy is(( j )cy ( i )cy ). So ( i )cy follows by Modus Ponens from ( j )cy and ( k )cy .

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Let be the elements of which actually occur in (**). Then cy . As

y does not occur free in , By generalization, ycy .Hence ycy . Finally note that c certainly doesnt occur in (**).

Recall the proof of generalization, and we see that c also doesnt occur inthe proof of ycy from .

Corollary 4.21. Suppose that xc where c is a constant that doesnt occur in {}. Then x via a deduction in which c does not occur.

Proof. By the above theorem, y(xc )cy for some variable y which doesntoccur in xc . Since c doesnt occur in , we have that ( xc )cy = xy . Thus, yxy .

By the exercise, teh following is a logical axiom: yxy . Thus xxy .Since x doesnt occur free in yxy , generalization igves yxy x.By deduction, (yx

y x), and since yx

yby Modus Ponens, wE

get x.

Theorem 4.22 (Existence of Alphabetic Variants) . Let be a wff, t a term and x a variable. Then there exists a wff (which differs from only in thequantied variables) such that

1. and

2. t is a substitute for x in

Proof. Omitted, see Enderton

4.2 The Completeness Theorem

Finally, we are ready to begin a proof of the completeness theorem.

Theorem 4.23 (The Completeness Theorem) . If |= then .

We will make use of the following:

Lemma 4.24. The following are equivalent:

1. The Completeness Theorem

2. If is consistant then is satisable.

Proof. 1 2: Suppose that is consistant. Then there exists a wff suchthat . By completeness, |= . Hence there exists a structure A and afunction s : V A such that A satises with s, but A |= [s]. In particular,A satises with s.

2 1: Suppose that . By reductio ad absurdum, {} is consistent.Thus, it is satisable, and so |= .

We now prove the following version of the completeness theorem:

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Theorem 4.25 (Completeness) . If is a consistent set of wffs in a countable

languageL

, then there exists a countable structureA

for L

and a function s : V A such that A satises with s.

We will proceed in several lemmas.

Lemma 4.26. Expand L to a language L + by adding a countably innite set of new constant symbols. Then remains consistant in L + .

Proof. Suppose not. Then there exists a wff of L + such that ( ) inL + .

Suppose that c1 , . . . , c n include the new constants, if any, which appear in . By generalization on constants, there exist variables y1 , . . . , yn such that y1 , . . . , yn , in L where is the result of replacing each ciwith yi . Since yi is substitutable for yi in , in L . This is acontradiction.

Lemma 4.27 (Add Witnesses) . Let 1 , x1 , . . . , n , xn enumerate all pairs, x where si a wff of L + and x is a variable. Let 1 be the wff x11

(1)x 1c1 where c1 is the rst new constnat which doesnt occur in . If n > 1then n is the wff xnn (n )x ncn where cn is the rst new constant symbol which doesnt occur in 1 , . . . ,n , 1 , . . . , n 1 . Let = {n |n 1}. Then is consistant.

Proof. Suppose not. Then let n 0 be the least integer such that {1 , . . . , n +1 }is inconsistant. Then by reductio ad absurdum, {1 , . . . , n } n +1 . Recallthat n +1 has the form x ()xc . By rule T, we have {1 , . . . , n } x adn {1 , . . . , n } xc .

Since c doesnt occur in {1 , . . . , n } {}, we have {1 , . . . , n } x. This contradicts n +1 being the rst contradiction, or the consistency of if n = 0.

Lemma 4.28. We can extend to a consistent set of wffs such that for every wff of L + , either or .

Proof. Let 1 , . . . , n , . . . be an enumeration of the wffs of L + . We shall deneinductively an increasing sequence 0 1 . . . of consistent sets of wffs. Let 0 = .

Suppose inductively that n has been dened. If n { n +1 } is consitent,then let n +1 = n { n +1 }. If it is inconsistent, then by reductio ad absur-dum, n n +1 , and so we let n +1 = n { n +1 }. Clearly = nsatises our requirements.

Notice now that is deductively closed. That is, if , then .Otherwise if /, then , and so and , contradicting theconsistency.

Lemma 4.29. For each of the following wffs, and so :

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1. x(x = x)

2. (x)(y)(x = y y = x)

3. (x)(y)(z)(( x = y y = z) x = z)

4. For each n-ary preducate symbol P , x1 , . . . , xn ,y1 , . . . , yn ((x1 = y1 . . . xn = yn ) P x 1 . . . x n = P y1 . . . yn )

5. For each n-ary function symbol f , x1 , . . . , xn ,y1 , . . . yn ((x1 = y1 . . . xn = yn ) fx 1 . . . x n = fy 1 . . . yn )

Similarly, since is deductively closed and xy(x = y y = x) , wehave that if t1 , t2 are terms, then (t1 = t2 t2 = t1) .

Lemma 4.30. We construct a structure A for our language L + as follows:Let T be the set of terms of L + and dene a binary relation E on T by t1Et 2iff (t1 = t2) .

Then E is an equivalence relation.

Proof. Let t T . Then ( t = t) , and so tEt for all t, so E is reexive.Suppose that t1Et 2 . Then ( t1 = t2) , and so ( t2 = t1) , so t2Et 2 , so

E is symmetric.Transitivity is similar.

For each t T , we dene [t] = {s T |tEs }, then we dene A = {[t]|t T }.For any n-ary predicate symbol, we dene an n-ary relation P A on A by

[t1], . . . , [tn ] P A iff P t 1 , . . . , t n .

Lemma 4.31. P A

is well dened.Proof. Suppose that [ s1] = [t1], . . . , [sn ] = [tn ]. We must show that P s 1 , . . . , s n iff P t 1 , . . . , t n . By hypothesis ( s i = t i ) for each i.

Since ((s i = t i . . . sn = tn ) (Ps 1 , . . . , s n P t 1 , . . . , t n )) , theresult follows.

For each constant symbol c, we dene cA = [c], and for each n-ary func-tion symbol f , we dene an n-ary operation f A on A by f A ([t1], . . . , [tn ]) =[f t 1 , . . . , t n ]. As above, f

A is well-dened.Finally, we dene s : V A by s(x) = [x].

Lemma 4.32. For every term t, s(t) = [ t].

Proof. By denition this is true when t is a constant symbol or a variable.Suppose that t = f t 1 , . . . , t n . By induction, we assume that s(t i ) = [ t i ]

for each i. Thus s(f t 1 , . . . , t n ) = f A (s(t1), . . . , s(tn )) = f

A ([t1], . . . , [tn ]) =[f t 1 , . . . , t n ].

Lemma 4.33 (Substitution Lemma) . If the term t is substitutable for x in ,then A |= xt [s] iff A |= [s(x | s(t))].

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And our nal claim:

Lemma 4.34. For each wff of L + , A |= [s] iff

Proof. We argue by deduction on the complexity of . First suppose that isatomic. Assume that is t1 = t2 . Then A |= ( t1 = t2)[s] iff s (t1) = s(t2), iff [t1] = [t2], iff (t1 = t2) .

Now suppose that is P t 1 , . . . , t n . Then A |= ( P t 1 , . . . , t n )[s] iff s(t1), . . . , s(tn ) P A , iff [t1], . . . , [tn ] P A iff P t 1 , . . . , t n .

Next we consider the cases where is not atomic.Suppose that is . Then A |= [s] iff A |= [s] iff /, iff A .The case is is similar.Finally, suppose is x . By construction, for some constant c, (x

xc ) . Call this (*).First suppose that A |= x[s], then in particular, A |= [s(x |[c])], that is,

A |= [s(x | s(x))].By the substitution lemma, A |= xc [s], and hence by the induction hypoth-

esis, xc . Thus xc /, so (*) implies that x /. Thus x .Conversely, suppose that A |= x[s]. Then there exists a term t T such

that A |= [s(x |[t])], that is, A |= [s(x | s(t ))].Let be an alphabetic variant of such that t is substitutable for x in .

Then A |= [s(x | s (t))]. By the substitution lemma, A |= ( )xt [s], and by theinduction hypothesis, ( )ct . Since ( x ( )xt ) , we must havex /. Thus x /.

Finally, let A 0 be the structure for L obtained by forgetting the interpre-tations of the new constants. Then A 0 satises with s.

Thus, the completeness theorem holds.Corollary 4.35. |= .

Theorem 4.36 (Compactness Theorem) . Let be a set of wffs in a countable rst order language. If is nitely satiable, then is satisable in somecountable structure.

Proof. Suppose that is nitely satisable. Let 0 be any nite subset,then 0 is satisable. Hence, by soundness, 0 is consistent. Since every nitesubset of is consistent, is consistent. By completeness, is satisable in acountable structure.

Theorem 4.37. Let T be a set of sentences in a rst order language. If the

class C = Mod(T ) is nitely axiomatizable, then there exists a nite subset T 0 T such that C = Mod(T 0)

Proof. Suppose that C = Mod(T ) is nitely axiomatizable. Then there existsa sentence such that C = Mod(), since Mot (T ) = Mod(), we have T |= .Hence by completeness, T . Thus, there is a nite set T 0 T such that T 0. By soundness T 0 |= . Hence, C = Mod(T ) Mod(T 0) Mod() = C .

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Denition 4.22. Let A , B be structures for the rst order language L . Then A

and B

are elementarily equivalent, written A

B

iff for every sentence L , A |= iff B |= .

Remark 4.5. A B implies A B , but the converse is false (eg, nonstan-dard arithmetic.)

Denition 4.23 (Complete Set) . A consistent set of sentences is said to becomplete iff for every sentence , either T or T .

Denition 4.24 (Theory) . A theory is a set of sentences. A consistant theory is complete if it is as a set of sentences.

Theorem 4.38. Let T be a complete theory in a rst order langague L . If A and B are models of T , then A B .

Proof. Let be any sentence. Then T or T . Suppose that T . Bysoundness if T is true then is true. So A |= and B |= .

On the other hand, if T , then by soundness we have A |= andB |= .

Theorem 4.39 (Los-Vaught Test) . Let T be a consistent theory in a countablelanguage L . Suppose

1. T has no nite models

2. If A are countably innite models of T , then A B .

Then T is complete.

Proof. Suppose that T satises the two conditions. Assume for contradictionthat T is not complete. Then there is a sentence such that T and T .By reduction ad absurdum, T {} and T {} are both consistent. BYcompleteness, there exist countable models A and B of T {} and T {}respectively. By the rst property, they are countably innite, and by thesecond, A B , contradiction.

Corollary 4.40. T DLO is complete.

Proof. T DLO has no nite models. Also we have seen by the back and forthargument that any two countable dense linear orders are isomorphism, and soby L-V, we are done.

Corollary 4.41. Q , < R ,

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