Solutions Manual to Naive Set Theory by Paul Halmos

Naive Set Theory: Solutions Manual George Mplikas September 2015

Contents 2 Misc. Exersizes 2.1 Logical Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Logical Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 3 4

3 Unordered Pairs

5

4 Unions and Intersections

6

5 Complements and Powers

10

6 Ordered Pairs

16

7 Relations

20

8 Functions

22

9 Families

23

10 Inverses and Composites

27

11 Numbers

31

12 Peano Axioms

32

13 Arithmetic

33

14 Order

37

15 Axiom of Choice

38

16 Zorn’s Lemma

39

17 Well Ordering

42

18 Transfinite Recursion

44

19 Ordinal Numbers

46

20 Sets Of Ordinal Numbers

47

21 Ordinal Arithmetic

48

22 The Schr¨ oder-Bernstein Theorem

49

23 Countable Sets

51

24 Cardinal Arithmetic

52

1

25 Cardinal Numbers

54

26 Extra Exercises 56 26.1 Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

2

2 2.1

Misc. Exersizes Logical Operators

Exercise 2.1. We show that ∧, ∨ are commutative. We do this via truth table. Proof. Consider the logical sentences p, q and the following truth table: p T T F F

q T F T F

p∨q T T T F

q∨p T T T F

Thus, p ∨ q ⇐⇒ q ∨ p Proof. Consider the logical sentences p and q and the following truth table: p T T F F

q T F T F

p∧q T F F F

Thus, p ∧ q ⇐⇒ q ∧ p.

3

q∧p T F F F

2.2

Logical Rules

Exercise 2.2. We show that p ∨ p ⇐⇒ p and p ∧ p ⇐⇒ p, via truth table. Proof. Consider the logical sentence p and the following truth table: p∨p T F

p T F Thus, p ∨ p ⇐⇒ p

Proof. Consider the logical sentence p and the following truth table: p∧p T F

p T F Thus, p ⇐⇒ p ∧ p.

Exercise 2.3. We show that p ⇒ p ∨ q via truth table. Proof. Consider the logical sentences p and q and the following truth table: p T T F F

p∨q T T T F

q T F T F

Thus, if p is true, p ∨ q is true, i.e. p ⇒ p ∨ q.

4

3

Unordered Pairs

Exercise 3.1. Consider the sets ∅, {∅}, {{∅}}, etc.; Consider all the pairs such as {∅, {∅}}, formed by two of them; consider the pairs formed by any such two pairs, or else mixed pairs formed by a singleton and any pair; and proceed ad infinitum. Are all the sets obtained in this way distinct from one another? Yes, all sets obtained in this way are distinct. Proof. By the axiom of pairing, we can form A and B as given, and by the axiom of specification, they are the only such sets which contain their elements. Suppose, to the contrary, that A = B. If A = B, then for all a ∈ A, a ∈ B, and for all b ∈ B, b ∈ A. Now, either A and B are singletons, or they are pairs. • A and B are singletons: Say, A = {a}, and B = {b}. Thus, by the Axiom of Extension, we have a = b. Implying that A and B were formed in the same way. • A and B are pairs: Let A = {a, a0 }, B = {b, b0 }. Without loss of generality, suppose that a = b, and a0 = b0 . If this is the case, then by the Axiom of Extension, again, we have that a, b were obtained in the same fashion, as well as a0 , b0 . But in this case, A and B were constructed in the same manner. Either way, we have a contradiction. Thus, A and B are formed distinctly.

5

4

Unions and Intersections

Throughout these exercises we consider a arbitrary non-empty collection C, of sets. For simplicity we will assume that C = {A, B, C} for arbitrary non-empty sets (unless specified) A, B, C and will make liberal use of the notation {x : S S(x)}, in-place of the more precises yet cumbersome notation {x ∈ C : S(x)}. Exercise 4.1. We show the following: A∪∅=A

(1)

A∪B =B∪A

(2)

A ∪ (B ∪ C) = (A ∪ B) ∪ C

(3)

A∪A=A

(4)

A ⊂ B ⇐⇒ A ∪ B = B Proof. ( ⇐⇒ ) Since ∅ has no elements, for all a ∈

(5) S

C, a ∈ / ∅ and we have,

x ∈ A ∪ ∅ = {x : x ∈ A ∨ x ∈ ∅} ⇐⇒ x ∈ A ∨ x ∈ ∅ = {} ⇐⇒ x ∈ A Thus, A ∪ ∅ = A Proof. ( ⇐⇒ )

A ∪ B = {x : x ∈ A ∨ x ∈ B} = {x : x ∈ B ∨ x ∈ A} = B ∪ A

Proof. ( ⇐⇒ )

x ∈ A ∪ (B ∪ C) ⇐⇒ x ∈ A ∨ x ∈ B ∪ C ⇐⇒ x ∈ A ∨ x ∈ B ∨ x ∈ C ⇐⇒ x ∈ {x : x ∈ A ∨ x ∈ B} ∨ x ∈ C ⇐⇒ x ∈ A ∪ B ∨ x ∈ C ⇐⇒ x ∈ (A ∪ B) ∪ C

Proof. ( ⇐⇒ )

6

A ∪ A = {x : x ∈ A ∨ x ∈ A} = {x : x ∈ A} = A

Proof. (⇒) We note that A ⊂ B iff for all x ∈ A, x ∈ B. Thus, we have that x ∈ A ∪ B = {x : x ∈ A ∨ x ∈ B} ⇒ x ∈ A ∨ x ∈ B ⇒ x ∈ B by the comments above. And likewise, x ∈ B ⇒ x ∈ B ∨ x ∈ A ⇒ x ∈ {x : x ∈ A ∨ x ∈ B} ⇒ x ∈ A ∪ B That is, A ∪ B = B. (⇐) Suppose A ∪ B = B. Then, x ∈ A ⇒ x ∈ {x : x ∈ A ∪ x ∈ B} = A ∪ B = B That is, A ⊂ B. Exercise 4.2. We show the following: A∩∅=∅

(1)

A∩B =B∩A

(2)

A ∩ (B ∩ C) = (A ∩ B) ∩ C

(3)

A∩A=A

(4)

A ⊂ B ⇐⇒ A ∩ B = A

(5) (6)

Proof. (1) Suppose that A 6= ∅, then

(⇒) x ∈ A ∩ ∅ = {x : x ∈ A ∧ x ∈ ∅} ⇒ x ∈ ∅ ∧ x ∈ A ⇒x∈∅ (⇐) (∅ ⊂ A) ∧ (∅ ⊂ ∅) ⇒∅⊂A∩∅

Proof. (2) ( ⇐⇒ ) 7

A ∩ B = {x : x ∈ A ∧ x ∈ B} = {x : x ∈ B ∧ x ∈ A} = B ∩ A

Proof. (3) ( ⇐⇒ ) x ∈ A ∩ (B ∩ C) ⇐⇒ x ∈ A ∧ x ∈ B ∩ C = {x : x ∈ B ∧ x ∈ C} ⇐⇒ x ∈ A ∧ x ∈ B ∧ x ∈ C ⇐⇒ x ∈ {x : x ∈ A ∧ x ∈ B} ∧ x ∈ C ⇐⇒ x ∈ A ∩ B ∧ x ∈ C ⇐⇒ x ∈ (A ∩ B) ∩ C

Proof. (4) ( ⇐⇒ ) A ∩ A = {x : x ∈ A ∧ x ∈ A} = {x : x ∈ A} = A

Proof. (5) (⇒) Suppose that A, B 6= ∅. We note that A ⊂ B iff for all x ∈ A, x ∈ B; then, x∈A⇒x∈B ⇒x∈A∧x∈B ⇒x∈A∩B And likewise, x ∈ A ∩ B = {x : x ∈ A ∧ x ∈ B} ⇒ x ∈ A ∧ x ∈ B ⇒ x ∈ A That is, A ∩ B = A. (⇐)Suppose that A ∩ B = A. Then, x ∈ A ⇒ x ∈ A ∩ B = {x : x ∈ A ∧ x ∈ B} ⇒ x ∈ A ∧ x ∈ B ⇒ x ∈ B That is, A ⊂ B. Exercise 4.3. We wish show that (A ∩ B) ∪ C = A ∩ (B ∪ C) ⇐⇒ C ⊂ A: We first show (1) in the following, noting that (2) was proved in the text; From page fifteen, A ∩ (B ∪ B) = (A ∩ B) ∪ (A ∩ C) (1) and, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) Proof. (⇒)

8

(2)

Suppose that not all of A, B, C are the emptyset. Then, x ∈ A ∩ (B ∪ C) ⇒ x ∈ A ∧ x ∈ B ∪ C Now, either x ∈ B or x ∈ C: x ∈ B ∧ x ∈ A ⇒ x ∈ A ∩ B ⇒ (A ∩ B) ∪ (A ∩ C) x ∈ C ∧ x ∈ A ⇒ x ∈ A ∩ C ⇒ (A ∩ B) ∪ (A ∩ C) Either way, x ∈ (A ∩ B) ∪ (A ∩ C). To prove the reverse inclusion, we note that similar logic holds; either x ∈ A ∩ B or x ∈ A ∩ C.

Now, we show that (A ∩ B) ∪ C = A ∩ (B ∪ C) ⇐⇒ C ⊂ A: Proof. (⇐)

Suppose that C ⊂ A. Then, for all x ∈ C, x ∈ A; By Exercise 4.1.5 and the proof above, we have the following: (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C) = A ∩ (B ∪ C)

(⇒) Suppose that (A ∩ B) ∪ C = A ∩ (B ∪ C). Then, x ∈ C ⇒ x ∈ {x : x ∈ A ∩ B ∨ x ∈ C} = (A ∩ B) ∪ C = A ∩ (B ∪ C) by the assumption. Thus, x ∈ C ⇒ x ∈ A ∩ (B ∪ C) ⇒x∈A∧x∈B∪C ⇒x∈A That is, C ⊂ A.

9

(3)

5

Complements and Powers

Throughout these exercises we assume that ”all sets are subsets of one and the same set E, and that all complements (unless otherwise specified) are formed relative to that E” (Halmos). Likewise, we will often use the more convenient, yet less accurate notation {x : S(x)} in place of {x ∈ E : S(x)} Exercise 5.1. We prove the following: A − B = A ∩ Bc A ⊂ B ⇐⇒ A − B = ∅

(1) (2)

A − (A − B) = A ∩ B

(3)

A ∩ (A − C) = (A ∩ B) − (A ∩ C)

(4)

c

(5)

A ∩ B ⊂ (A ∩ C) ∪ (B ∩ C ) c

(A ∪ C) ∩ (B ∪ C ) ⊂ A ∪ B

(6)

Proof. (1) ( ⇐⇒ )

A − B = {x ∈ E : x ∈ A ∧ x ∈ / B} ⇐⇒ x ∈ A ∧ x ∈ /B ⇐⇒ x ∈ A ∧ x ∈ B c ⇐⇒ x ∈ A ∩ B c

Proof. (2) (⇒) Suppose that A − B 6= ∅, but A ⊂ B; since A ⊂ B, if x ∈ A, x ∈ B. Thus, if x ∈ A − B = {x ∈ E : x ∈ A ∧ x ∈ / B}, x ∈ / B. However x ∈ A, which implies x ∈ B: A contradiction.

(⇐) Suppose that A − B = ∅ and that there exists some x ∈ A which is not in B, i.e. A 6⊂ B. However this is a contradiction since x∈A∧x∈ / B ⇒ x ∈ {x ∈ E : x ∈ A ∧ x ∈ / B} = A − B and yet A − B = ∅. Proof. (3) ( ⇐⇒ ) 10

By the proof above, A − (A − B) = A − (A ∩ B c ) = {x ∈ E : x ∈ A ∧ x ∈ / A ∩ Bc} ⇐⇒ x ∈ A ∧ x ∈ / A ∩ Bc ⇐⇒ x ∈ A ∧ x ∈ / Bc ⇐⇒ x ∈ A ∧ x ∈ B ⇐⇒ x ∈ {x ∈ E : x ∈ A ∧ x ∈ B} = A ∩ B

Proof. (4) ( ⇐⇒ )

x ∈ A ∩ (B − C) ⇐⇒ x ∈ A ∧ x ∈ (B − C) ⇐⇒ x ∈ A ∧ x ∈ {x ∈ E : x ∈ B ∧ x ∈ / C} ⇐⇒ x ∈ A ∧ x ∈ B ∧ x ∈ /C ⇐⇒ x ∈ / {x ∈ E : x ∈ A ∧ x ∈ C} ∧ x ∈ A ∩ B ⇐⇒ x ∈ / A∩C ∧x∈A∩B ⇐⇒ x ∈ {x ∈ E : x ∈ / A ∩ C ∧ x ∈ A ∩ B} ⇐⇒ x ∈ (A ∩ B) − (A ∩ C)

Proof. (5) (⇒) Consider C ⊂

S

C, as stated above.

x ∈ A ∩ B ⇒ x ∈ {x ∈ E : x ∈ A ∧ x ∈ B} ⇒x∈A∧x∈B

(1)

⇒ (x ∈ A ∧ x ∈ B) ∧ (x ∈ C ∨ x ∈ C c )

x ∈ C ∧ (1) ⇒ x ∈ A ∩ C ∧ x ∈ B ∩ C

(2)

x ∈ C c ∧ (1) ⇒ x ∈ A ∩ C c ∧ x ∈ B ∩ C c

(3)

Thus, either way, x ∈ (A ∩ C) or x ∈ (B ∩ C c ), i.e. x ∈ (A ∩ C) ∪ (A ∩ C c ) Proof. (6) (⇒)

11

x ∈ (A ∪ C) ∩ (B ∪ C c ) ⇒ x ∈ A ∪ C ∧ x ∈ B ∪ C c ⇒ (x ∈ A ∨ x ∈ C) ∧ (x ∈ B ∨ x ∈ C c )

(1)

We note that either x ∈ C, or x ∈ C c : x ∈ C ∧ (1) ⇒ x ∈ / C c ∧ (x ∈ B ∨ x ∈ A) ⇒x∈A∪B x ∈ C c ∧ (1) ⇒ x ∈ / C ∧ (x ∈ A ∨ x ∈ B) ⇒x∈A∪B

(2)

(3)

Either way, x ∈ A ∪ B. Exercise 5.2. We show that P(E) ∩ P(F ) = P(E ∩ F ): Proof. Let E, F be sets such that E, F ⊂ E ; via Axiom of Powers, P(E), P(F ); specifically, from the discussion in Halmos1 , P(E) = {X : X ⊂ E}, and P(F ) = {X : X ⊂ F }. Then, we have the following: X ∈ P(E) ∩ P(F ) ⇐⇒ X ∈ P(E) ∧ X ∈ P(F ) ⇐⇒ X ∈ {Y : Y ⊂ E ∧ Y ⊂ F } ⇐⇒ X ⊂ E ∧ X ⊂ F ⇐⇒ ∀x(x ∈ X ⇒ (x ∈ E ∧ x ∈ F )) ⇐⇒ ∀x ∈ X, x ∈ E ∩ F ⇐⇒ X = {x ∈ E : x ∈ X} ⊂ E ∩ F ⇐⇒ X ∈ {Y : Y ⊂ E ∩ F } = P(E ∩ F ) Exercise 5.3. We show that P(E) ∪ P(F ) ⊂ P(E ∪ F ): Proof. Let E, F be given as in the discussion of proof 5.2.0. We first claim that if A, B are sets, then (x ⊂ A ∨ x ⊂ B ⇒ x ⊂ A ∪ B). Indeed; if x ⊂ A, then x ⊂ A ∪ B and likewise, x ⊂ B implies x ⊂ B ∪ A. With this in mind, we have the following: X ∈ P(E) ∪ P(F ) = {X : X ∈ P(E) ∨ X ∈ P(F )} ⇐⇒ X ∈ P(E) ∨ X ∈ P(F ) ⇐⇒ X ∈ {Y : Y ⊂ E} ∨ X ∈ {Y : Y ⊂ F } ⇐⇒ X ∈ {Y : Y ⊂ E ∨ Y ⊂ F } ⇒ X ∈ {Y : Y ⊂ E ∪ F } = P(E ∪ F ) 1 Halmos,

P. ”Naive Set Theory”. pg 19

12

T

Exercise 5.4. We show that

X∈C

P(X) = P(

T

X∈C

X):

Proof. Let C be a non-empty collection of sets. (⇒) z∈

\

P(X) ⇒ (∀X ∈ C)(z ∈ P(X))

X∈C

⇒ (∀X ∈ C)({z} ⊂ P(X)) ⇒ (∀X ∈ C)(z ∈ X) \ ⇒z∈ X X∈C

⇒ {z} ⊂

\

X

X∈C

⇒ z ∈ P(

\

X)

X∈C

(⇐) z ∈ P(

\

X) ⇒ z ∈ {Y : Y ⊂

X∈C

\

X}

X∈C

⇒z⊂

\

X

X∈C

⇒ (∀X ∈ C)(z ⊂ X) ⇒ (∀X ∈ C)(∀v ∈ z)(v ∈ X) ⇒ (∀X ∈ C)(∀v ∈ z)({v} ⊂ P(X)) ⇒ (∀X ∈ C)(∀v ∈ z)(v ∈ P(X)) ⇒ (∀X ∈ C)(z ∈ P(X)) \ ⇒z∈ P(X) X∈C

And this completes the proof. Exercise 5.5. We show that

S

X∈C

P(X) ⊂ P(

S

X∈C

X):

Proof. Let C be a collection of sets. A common fact used in the following proof is that if x ∈ P(X), then x ⊂ X for any set X. This is indeed true, since

13

P(X) = {Y : Y ⊂ X}. With this in mind, we have [ z∈ P(X) ⇒ (∃Y ∈ C)(z ∈ P(Y )) X∈C

⇒z⊂Y ∈C ⇒ z ⊂ {x : x ∈ X(∀X ∈ C)} =

[

X

X∈C

⇒ (∀s ∈ z)(s ∈

[

X)

X∈C

⇒ (∀s ∈ z)(s ∈ P(

[

X))

X∈C

⇒ z ∈ P(

[

X)

X∈C

Which completes the proof. Exercise 5.6. We show that

T

X∈P(E)

X = ∅:

Proof. Let E be a set and consider P(E) = {Y : Y ⊂ E}. We note that this is non-empty because ∅ ∈ P(E), T since ∅ is a subset of every set. We also note that the statement (x ∈ ∅ ⇒ x ∈ X∈P(E) X) is vacuously satisfied since ∅ has no elements. Thus, we only show the forward inclusion: (⇒) x∈

\

X ⇒ (∀X ∈ P(E))(x ∈ X)

X∈P(E)

⇒ x ∈ ∅ ∈ P(E) And this completes the proof. Exercise 5.7. We show that (E ⊂ F ⇒ P(E) ⊂ P(F )): Proof. Suppose that E ⊂ F for sets E, F . Then, we have the following: e ∈ P(E) ⇒ e ⊂ E ⇒ (∀x ∈ e)(x ∈ E) ⇒ (∀x ∈ e)(x ∈ F )

(by Assumption)

⇒e⊂F ⇒ e ∈ P(F )

(by Definition)

Which completes the proof. Exercise 5.8. We show that E =

S

P(E) =

S

X∈P(E)

Proof. Let E be a set. Then, we have the following: 14

X:

(⇐) x∈

[

P(E) ⇒ (∃Y ∈ P(E))(x ∈ Y ) ⇒x∈Y ⊂E ⇒x∈E

(⇒) x ∈ E ⇒ {x} ∈ P(E) ⇒ x ∈ {x} ∈ P(E) [ ⇒ x ∈ {X : X ∈ P(E)} [ = P(E)

Which completes the proof. ´S ¯ E is a set that includes E, typically, as a Exercise 5.9. We show that P proper subset. ´S ¯ E = P(E) = {X ∈ E : X ⊂ E}. If E is non-empty, Proof. Indeed, P then it is clear that E Ĺ P(E). However, if E = ∅, then P(E) = ∅ and E = P(E).

15

6

Ordered Pairs

Throughout this section, we will consider A, B, C, X, Y to be any sets, and will use the notation {x : S(x)} in place of the more accurate, yet cumbersome S notation {x ∈ C : S(x)}. Exercise 6.1. We show the following: (A ∪ B) × X = (A × X) ∪ (B × X)

(1)

(A ∩ B) × (X ∩ Y ) = (A × X) ∩ (B × Y )

(2)

(A − B) × X = (A × X) − (B × X)

(3)

(A = ∅ ∨ B = ∅) ⇐⇒ (A × B = ∅) (A ⊂ X) ∧ (B ⊂ Y ) ⇐⇒ (A × X ⊂ X × Y )

(4) (5)

Proof. (1) (⇒) (x, y) ∈ (A ∪ B) × X ⇐⇒ (x, y) ∈ {(a, b) : a ∈ A ∪ B ∧ b ∈ X} ⇐⇒ x ∈ A ∪ B ∧ y ∈ X ⇐⇒ (x ∈ A ∨ x ∈ B) ∧ y ∈ X (by Distributivity)

⇐⇒ (x ∈ A ∧ y ∈ X) ∨ (x ∈ B ∧ y ∈ X) ⇒ (x ∈ A ∪ X ∧ y ∈ A ∪ X) ∨ (x ∈ B ∪ X ∧ y ∈ B ∪ X)

(by Remarks in Text)

⇒ {x, {x, y}} ⊂ P(A ∪ X) ∨ {x, {x, y}} ⊂ P(B ∪ X) ⇐⇒ (x, y) ∈ P(P(A ∪ X)) ∨ (x, y) ∈ P(P(B ∪ X))

(by Remarks in Text)

⇐⇒ (x, y) ∈ (A × X) ∪ (B × X)

(⇐) (x, y) ∈ (A × X) ∪ (B × X) ⇐⇒ (x, y) ∈ (A × X) ∨ (x, y) ∈ (B × X) (by Definition)

⇒ (x ∈ A ∧ y ∈ X) ∨ (x ∈ B ∧ y ∈ X) ⇐⇒ (x ∈ A ∨ x ∈ B) ∧ y ∈ X ⇐⇒ x ∈ A ∪ B ∧ y ∈ X ⇒ x ∈ (A ∪ B) ∪ X ∧ y ∈ (A ∪ B) ∪ X ⇒ {{x}, {x, y}} ⊂ P((A ∪ B) ∪ X) ⇐⇒ {{x}, {x, y}} ∈ P(P((A ∪ B) ∪ X)) ⇒ (x, y) ∈ (A ∪ B) × X

Which completes the proof. 16

Proof. (2) (⇒) (c, d) ∈ (A ∩ B) × (X ∩ Y ) ⇐⇒ c ∈ A ∩ B ∧ d ∈ X ∩ Y ⇐⇒ (c ∈ A ∧ c ∈ B) ∧ (d ∈ X ∧ d ∈ Y ) ⇐⇒ (c ∈ A ∧ d ∈ X) ∧ (c ∈ B ∧ d ∈ Y )

(by Commutativity)

⇒ (c ∈ A ∪ X ∧ d ∈ A ∪ X) ∧ (c ∈ B ∪ Y ∧ d ∈ B ∪ Y ) ⇒ {{c}, {c, d}} ⊂ P(A ∪ X) ∧ {{c}, {c, d}} ⊂ P(B ∪ Y ) ⇐⇒ (c, d) ∈ P(P(A ∪ X)) ∧ (c, d) ∈ P(P(B ∪ X)) ⇐⇒ (c, d) ∈ A × X ∧ (c, d) ∈ B × X ⇐⇒ (c, d) ∈ (A × X) ∩ (B × X)

(⇐) (c, d) ∈ (A × X) ∩ (B × X) ⇐⇒ (c, d) ∈ (A × X) ∧ (c, d) ∈ (B × X) ⇐⇒ (c ∈ A ∧ d ∈ X) ∧ (c ∈ B ∧ d ∈ X) (by Commutativity)

⇐⇒ (c ∈ A ∧ c ∈ B) ∧ d ∈ X ⇐⇒ (c ∈ (A ∩ B)) ∧ d ∈ X ⇒ c ∈ (A ∩ B) ∪ X ∧ d ∈ (A ∩ B) ∪ X ⇒ {{c}, {c, d}} ⊂ P((A ∩ B) ∪ X) ⇒ (c, d) ∈ P(P((A ∩ B) ∪ X)) ⇐⇒ (c, d) ∈ (A ∩ B) × X

Which completes the proof. Proof. (3) (⇒) (c, d) ∈ (A − B) × X ⇐⇒ c ∈ (A − B) ∧ d ∈ X ⇐⇒ (c ∈ A ∧ c ∈ / B) ∧ d ∈ X ⇐⇒ (c ∈ A ∧ d ∈ X) ∧ (c ∈ / B ∧ d ∈ X) (since c ∈ / B)

⇒ (c, d) ∈ A × X ∧ (c, d) ∈ / B×X ⇐⇒ (c, d) ∈ (A × X) − (B × X)

17

(⇐) (c, d) ∈ (A × X) − (B × X) ⇐⇒ (c, d) ∈ (A × X) ∧ (c, d) ∈ / (B × X) ⇒ (c ∈ A ∧ d ∈ X) ∧ (c ∈ / B∨d∈ / X) (since (d ∈ X ∧ d ∈ / X, →←))

⇒ (c ∈ A ∧ d ∈ X) ∧ (c ∈ / B) ⇐⇒ (c ∈ A ∧ c ∈ / B) ∧ d ∈ X ⇐⇒ c ∈ (A − B) ∧ d ∈ X ⇒ {{c}, {c, d}} ⊂ P((A − B) ∪ X) ⇐⇒ {{c}, {c, d}} ∈ P(P((A − B) ∪ X)) ⇐⇒ (c, d) ∈ (A − B) × X

Which completes the proof. Proof. (4) (⇒) Suppose, without loss of generality, that A = ∅. Then, A contains no elements. Consider A × B = {(a, b) : a ∈ A ∧ b ∈ B}. Then, since there are no a ∈ A, {(a, b) : x ∈ A ∧ x ∈ B} = ∅. (⇐) Suppose to the contrary that A 6= ∅ and B 6= ∅, but A × X = ∅. Since A 6= ∅ and B 6= ∅, there exists some a ∈ A and b ∈ B. And so, {{a}, {a, b}} ⊂ P(A ∪ B) and further we have (a, b) ∈ A × B. However, this is a contradiction since we assumed A × B = ∅. Proof. (5) (⇒) We show that if A ⊂ X and B ⊂ Y , then A×B ⊂ X ×Y . Since A ⊂ X, for all a ∈ A, a ∈ X and likewise for B. (c, d) ∈ A × B ⇐⇒ c ∈ A ∧ d ∈ B (by Assumption)

⇒c∈X ∧d∈Y ⇒c∈X ∪Y ∧d∈X ∪Y ⇒ {{c}, {c, d}} ⊂ P(X ∪ Y ) ⇒ (c, d) ∈ P(P(X ∪ Y )) ⇐⇒ (c, d) ∈ X × Y

18

(⇐) Suppose that A × X 6= ∅. And thatA × B ⊂ X × Y . We show that A ⊂ X and B ⊂ Y . a ∈ A ∧ b ∈ B ⇒ {{a}, {a, b}} ∈ P(P(A ∪ B)) ⇐⇒ (a, b) ∈ A × B ⇒ (a, b) ∈ X × Y ⇒ (a, b) ∈ {(x, y) : x ∈ X ∧ y ∈ Y } ⇒a∈X ∧b∈Y

Which completes the proof.

19

7

Relations

Exercise 7.1. We show that for each of symmetric, reflexive, and transitive properties, there exists a relation that does not have that property but does have the other two. Proof. Let X be a set. We first show that if X = ∅, then any relation in X is an equivalence relation: Lemma. Let X = ∅. Then, by a previous theorem, we have X × X = ∅. And so, if R is a relation in X, R = ∅. However, R is an equivalence relation, since all statements are vacuously satisfied. Likewise, suppose that X = {a}; i.e. X is a singleton. We claim, also that any relation in X is an equivalence relation: Lemma. Let X = {a}. Then, by the comments of the previous section, we have X ×X = {(a, a)}. Thus, since R is the set {(a, b) : a ∈ X ∧ b ∈ X} it follows that R = {(a, a)} for any relation.2 It is easy to verify that the transitive condition is vacuously satisfied for R. And, it is symmetric, since (a, a) = (a, a). And, clearly, R is reflexive. Thus, R is an equivalence relation. We also show that for X = {a, b} each of which is distinct, there does not exist a relation, which is symmetric, reflexive, but not transitive: Lemma. Let X = {a, b}. Then, the relation R in question must contain {(a, a), (b, b)} = I. However, it is easy to verify that this is an equivalence relation.3 Thus, we must add an element of X × X which maintains its symmetry, yet is not transitive. However, this is not possible since either way (a, b), (b, a) must be in R and so we would have R = {(a, a), (b, b), (a, b), (b, a)}.And, this is an equivalence relation since (x, x) ∈ R for all x ∈ X and, by the comments above, R is symmetric, as well as transitive.4 Thus, let X = {a, b, c....}, for distinct elements. We first show the relational properties which include reflexivity: 2 see

Halmos pg. 27 the above lemma 4 this is left to the reader 3 see

20

Let R be a relation in X which is reflexive. Then, we must have I = {(x, x) : x ∈ X} ⊂ R, by definition. Thus, we show a representation of R which is transitive, but not symmetric, and symmetric but not transitive: • Consider R = I ∪{(a, b), (b, a), (b, c), (c, b)}. Then, this relation is symmetric, since (x, x) = (x, x) for all x ∈ R and {(a, b), (b, a), (b, c), (c, b)} ⊂ R. However, R is not transitive, since (a, b) ∈ R and (b, c) ∈ R and yet (a, c) ∈ / R. • Consider R = I ∪ {(a, b)}. Then, R is transitive; note, we only need to check (a, b), (a, a), (b, b) for transitivity, since they are the only ordered pairs which share a first, or second coordinate, where only (a, a), or (a, b) come first. Correctly, (a, b) ∈ R and (b, b) ∈ R implies (a, b) ∈ R; (a, a) ∈ R and (a, b) ∈ R implies (a, b) ∈ R. However, R is not symmetric: (a, b) ∈ R, however, (b, a) ∈ / R. Finally, we show the relation in X which is not reflexive. Let X be given as above: • Consider R = {(a, b), (b, a), (a, a), (b, b)} ⊂ X ×X. This is symmetric since (x, x) = (x, x) for all x ∈ X and {(a, b), (b, a)} ⊂ R. It is, also, reflexive since (a, b) ∈ R and (b, a) ∈ R implies (a, a) ∈ R; likewise, (b, a) ∈ R and (a, b) ∈ R implies (b, b) ∈ R.5 However, R is not reflexive since c ∈ X, yet (c, c) ∈ / R.

Exercise 7.2. We show that the set of all equivalence classes, X/R, is indeed a set. Proof. Let R be an equivalence relation on a set X. We show that X/R ⊂ P(X): z ∈ X/R ⇒ (∃x ∈ X)(z = {x/R} = {y ∈ X : (x, y) ∈ R} ⊂ X) ⇒ z ∈ P(X) ⇒ {z} ⊂ X

Which completes the proof.

5 the

others follow

21

8

Functions

Exercise 8.1. We show the special cases in which the projections are 1-1. Proof. Without loss of generality, we only consider g : X × Y → Y , the range projection. • We first note that if X, Y have a pair or more of distinct elements, then g is not 1-1: Lemma 8.1. Suppose X, Y , have more than a pair of distinct elements, then g is not 1-1: To see this, let {a, b} ⊂ X, and {c, d} ⊂ Y , then {(a, c), (a, d), (b, c), (b, d)} ⊂ X × Y . If we suppose g is one-to-one, then g((a, c)) = g((b, d)) implies (a, c) = (b, d). But this is a contradiction, since a = b and yet a, b are distinct. • Next, if X = {x} and Y = {y}, then we claim g is 1-1. Lemma 8.2. Let X = {x} and Y = {y}; then, X × Y = {(x, y), (y, x)}. Upon inspection, we see that g is 1-1: if (x, y) 6= (y, x), then y 6= x and y = g((x, y)) 6= x = g((y, x). We note that this covers the case in which x = y. • Next, we show that if either X or Y is a singleton and the other is an arbitrary set, g is one-to-one. Lemma 8.3. Suppose, without loss of generality, that X is the singleton. Then either Y = ∅ or Y 6= ∅. If Y = ∅, then X × Y = ∅, by previous exercise; which is trivially 1-1. Else, we assume Y has more than a pair of distinct elements. Since X = {x} for some x, it follows that {(a, b) : a ∈ X ∧ b ∈ Y } = {(x, b) : b ∈ Y }, which means g is 1-1. • Lastly, if X = Y = ∅, then all conditions are vacuously satisfied, and so g is 1-1.

Exercise 8.2. We show that Y ∅ has exactly one element, namely ∅, whether Y is empty or not; i.e. Y ∅ = {∅}. Proof. Suppose that Y is a set. Consider f = Y ∅ . From page thirty, and by a previous theorem, we have f ⊂ P(∅ × Y ) = P(∅) = {∅}, where f is a set of elements which are ordered pairs. From the comments in section seven, ∅ is a set of ordered pairs. It follows that f = {∅}. Exercise 8.3. We show that if X 6= ∅, then ∅X is empty; i.e. ∅X = ∅. Proof. Suppose that there is some function f ∈ ∅X . Then, for all x ∈ X, there exists some b ∈ ∅ such that (x, b) ∈ f . However, this is not true, since ∅ contains no elements. Thus, ∅X = ∅.

22

9

Families

Exercise 9.1. After the comments on page thirty-five, we formulate and prove a generalized version of the commutative law for unions. Proposition 9.1. Let K be an arbitrary, non-empty set, and let p ∈ Aut(K K ) be an element of the set of one-to-one and onto functions from K to K. We note that in the current notation this means that p is the family and {pi } denotes the values of p evaluating some S i ∈ K. SLet {Ak } be a family of functions with domain K. We claim that k∈K Ak = i∈K Api . Proof. (⇒) x∈

[

Ak ⇒ (∃k0 ∈ K)(x ∈ Ak0 )

k∈K

(since p is onto)

⇒ (x ∈ Ak0 ) ∧ (∃s ∈ K)(k0 = ps ) ⇒ x ∈ Ak 0 = Ap s [ ⇒x∈ Ap i i∈K

(⇐) [

x∈

Api ⇒ (∃s ∈ K)(x ∈ Aps )

i∈K

(since ps ∈ K and onto)

⇒ (x ∈ Aps ) ∧ (∃k0 ∈ K)(ps = k0 ) ⇒ x ∈ Aps = Ak0 [ ⇒x∈ Ak k∈K

Which completes the proof. (

1 i=0 ; it 0 i=1 is easy to show that p is 1-1 and onto. From the statement above, we have [ Ak = A0 ∪ A1 (1) To illustrate the previous proof, consider K = {0, 1} and pi =

k∈K

[

Api = Ap0 ∪ Ap1 = A1 ∪ A0

i∈K

23

(2)

Exercise 9.2. We show that if both {Ai } and {Bi } are families of sets, with domains I, J, then ´[ ¯ ´[ ¯ [ Ai ∩ Bj = (Ai ∩ Bj ) (1) i

´\

j

¯ Ai ∪

i

Where the symbols such as

´\

i,j

¯ Bj

\ = (Ai ∪ Bj )

j

T

i,j

mean

(2)

i,j

T

(i,j)∈I×J

Proof. (1)

( ⇐⇒ ) x∈

´[

¯ ´[ ¯ ´[ ¯ ´[ ¯ Ai ∩ Bj ⇐⇒ x ∈ Ai ∧ x ∈ Bj

i

j

i

j

⇐⇒ (∃i0 ∈ I)(x ∈ Ai0 ) ∧ (∃j0 ∈ J)(x ∈ Bj0 ) ⇐⇒ (∃(i0 , j0 ) ∈ I × J)(x ∈ (Ai0 ∩ Bj0 )) [ ⇐⇒ x ∈ (Ai ∩ Bj ) i,j

Proof. (2)

( ⇐⇒ ) x∈

´\ i

¯ ´\ ¯ ´\ ¯ ´\ ¯ Ai ∪ Bj ⇐⇒ x ∈ Ai ∨ x ∈ Bj j

i

j

⇐⇒ (∀i ∈ I)(x ∈ Ai ) ∨ (∀j ∈ J)(x ∈ Bj ) ⇐⇒ (∀i ∈ I)(∀j ∈ J)(x ∈ Ai ∨ x ∈ Bj ) ⇐⇒ (∀i ∈ I)(∀j ∈ J)(x ∈ Ai ∪ x ∈ Bj ) ⇐⇒ (∀(i, j) ∈ I × J)(x ∈ (Ai ∪ Bj )) \ ⇐⇒ x ∈ (Ai ∪ Bj ) (i,j)∈I×J

24

Exercise 9.3. Let {Ai }, {Bj } be families of sets with domains I 6= ∅, J = 6 ∅, respectively. We show that ´[ ¯ ´[ ¯ [ Ai × Bj = (Ai × Bj ) (1) i

j

´\

¯ Ai ×

i

´\

i,j

¯ Bj =

j

\

(Ai × Bj )

(2)

i,j

Proof. (1)

( ⇐⇒ ) (a, b) ∈

´[

¯ ´[ ¯ [ [ Ai × Bj ⇐⇒ a ∈ Ai ∧ b ∈ Bj

i

j

i

j

⇐⇒ (∃i0 ∈ I)(∃j0 ∈ J)(a ∈ Ai0 ∧ b ∈ Bj0 ) ⇐⇒ (∃(i0 , j0 ) ∈ I × J)((a, b) ∈ Ai0 × Bj0 ) [ ⇐⇒ (a, b) ∈ (Ai × Bj ) (i,j)∈I×J

Proof. (2) Suppose that for each i ∈ I, Ai 6= ∅ and for each j ∈ J, Bj 6= ∅.

( ⇐⇒ ) (a, b) ∈

´\ i

¯ ´\ ¯ \ \ Ai × Bj ⇐⇒ a ∈ Ai ∧ b ∈ Bj j

i

j

⇐⇒ (∀i ∈ I)(a ∈ Ai ) ∧ (∀j ∈ J)(b ∈ Bj ) ⇐⇒ (∀i ∈ I)(∀j ∈ J)(a ∈ Ai ∧ b ∈ Bj ) ⇐⇒ (∀i ∈ I)(∀j ∈ J)((a, b) ∈ Ai × Bj ) ⇐⇒ (∀(i, j) ∈ I × J)((a, b) ∈ Ai × Bj ) \ ⇐⇒ (a, b) ∈ Ai × B j i,j

Exercise 9.4. Suppose that for each i ∈ I, Xi T is non-empty. SConsider the family of sets {Xi } with domain I. We show that i Xi ⊂ Xj ⊂ i Xi for each index j ∈ I.

25

Proof. Let {Xi } be stated as above, and let j ∈ I be arbitrary. \ x∈ Xi ⇒ (∀i ∈ I)(x ∈ Xi ) i

⇒ x ∈ Xj

(since j ∈ I)

Which proves the first inclusion. For the second, suppose j ∈ I is arbitrary, then we have the following: [ x ∈ Xj ⇒ x ∈ Xj j

⇒x∈

´[

⇒x∈

[

[

Xj ∪

j

¯ Xz

z∈I−J

Xi

i

Which prove the second inclusion. Exercise 9.5. We show the minimality condition of intersection S and unions: (1) That is, if Xj ⊂ Y for each index T j and some set Y , then i Xi ⊂ Y and that it is unique; (2) Likewise, if Y ⊂ j Xj for each index j and some set Y , T then Y ⊂ i Xi and it is unique. Proof. (1) Suppose that {Xi } is a family of sets with domain I, such that j ∈ I, and let Y be a set. Then, we have the following: [ (∀j)(Xj ⊂ Y ) ⇒ Xj ⊂ Y j

⇒

[

Xi ⊂ Y

i

Proof. (2) Suppose that {Xi } is a family of sets with domain I, such that j ∈ I. And let Y be a set. Then, we have the following: \ \ Y ⊂ Xj ⇒ Y ⊂ Xi j

i

26

10

Inverses and Composites

Exercise 10.1. Suppose that f : X → Y . Let {Ai } be a family of subsets of X. Define f 0 : P(X) → P(Y ), as the mapping from each Ai to the image subset, f 0 (Ai ), as stated in the text. We claim that ´[ ¯ [ f0 Ai = f 0 (Ai ) i

i

Proof. After the comments on page thirty-four, we note that A : I → P(X), for some index set I 6= ∅. Then, we have the following: (⇒) x∈

[

(since Ai0 ⊂

[

f 0 (Ai ) ⇒ (∃i0 ∈ I)(x ∈ f 0 (Ai0 ))

i

Ai )

⇒ x ∈ f0

´[

i

¯ Ai

i

(⇐) x ∈ f0

´[

¯ Ai ⇒ (∃i0 ∈ I)(x ∈ f 0 (Ai0 ))

i

´

¯ [ f 0 (Az ) ∪ f 0 (Ai0 ) = f 0 (Ai )

[

⇒x∈

i

z∈I−{i0 }

And the result follows. Exercise 10.2. We show that, in general, ´\ ¯ \ f0 Ai = f 0 (Ai ) i

i

does not hold. The following proof was suggested by David. H.. Proof. Consider X = {a, b, c} and Y = {d, e} and suppose all the elements are

27

distinct. Then define f 0 : P(X) → P(Y ) as the following: f 0 (∅) = ∅ f 0 ({a}) = {d} f 0 ({b}) = {e} f 0 ({c}) = {d} f 0 ({a, b}) = {d, e} f 0 ({b, c}) = {d} f 0 ({a, c}) = {e} 0

f ({a, b, c}) = ∅

Then, let {Ai } be a family of subsets of X with domain I = 2, such that A0 = {a} and A1 = {a, c}. Then, we have the following: ´\ ¯ f0 Ai = f 0 ({a}) = {d} i

\

f 0 (Ai ) = f 0 (A0 ) ∩ f 0 (A1 ) = {d} ∩ {e} = ∅

i

Which completes the proof. Exercise 10.3. Let f : X → Y . We show that f maps X onto Y iff for every non-empty subset B of Y , f −1 (B) 6= ∅. Proof. (⇒) If f maps X onto Y , then for every y ∈ Y , there exists an x ∈ X such that f (x) = y. Since this is the case, if B is a non-empty subset of Y , then for any b ∈ B, b ∈ Y . And we have that there exists an x0 ∈ X such that f (x0 ) = b; therefore, f −1 (B) = {x ∈ X : f (x) ∈ B} = 6 ∅. (⇐) Suppose that f −1 (B) 6= ∅ for any B ⊂ Y . In particular, for each y ∈ Y , {y} ⊂ Y and f −1 ({y}) = {x ∈ X : f (x) = y} = 6 ∅. The result follows immediately. Exercise 10.4. In the context of page forty-one, R := ”son”, S := ”brother”, what do R−1 , S −1 , RS and R−1 S −1 mean? Proof. (R−1 ): From the comments on page forty-one, xRy implies that that x is son-related to y. That is, y is the father of x. Thus, yR−1 x means that y has the father relation to x. (S −1 ): Likewise, xSy means that x bears the brother-relation to y. That is, x is the brother of y. Thus, yS −1 x implies y is the brother of x. 28

(T = R ◦ S): Also, by the previous comments, if xT z, then there exists y such that xSy and yRz. Thus, x is the brother of y and z is the son to y. That is, x is the uncle of y. (T = R−1 ◦ S −1 ): Likewise, if xT z, then there exists a y such that xS −1 y and yR−1 z. Thus, x is the brother of y and y is the father to z. Thus, x is the uncle of z. Exercise 10.5. Assume that for each of the following f : X → Y . We show the following: 1. If g : Y → Z such that g ◦ f is the identity on X, then f is 1-1 and g maps Y onto X. 2. f (A ∩ B) = f (A) ∩ f (B) for all A, B ⊂ X iff f is 1-1. 3. f (X − A) ⊂ Y − f (A) for all subsets A of X iff f is 1-1. 4. Y − f (A) ⊂ f (X − A) for all subsets A of X iff f maps X onto Y . Proof. (1): Suppose that g : Y → Z such that g ◦ f is the identity on X. Since g ◦ f is the identity on X, for all x ∈ X, g(f (x)) = x. Thus, if f (x0 ) = f (x1 ), then x0 = g(f (x0 )) = g(f (x1 )) = x1 and f is 1-1. To show that g maps Y onto X let x ∈ X, then x = g(f (x)) where f (x) ∈ Y . Thus, g maps Y onto X. Proof. (2): (⇒) By assumption, we have that f (A ∩ B) = f (A) ∩ f (B) for all A, B ⊂ X. Suppose that f (x) = f (y) for some x, y ∈ X. If x = y, then we are done. So, suppose that x 6= y. Letting A = {x}, B = {y}, we have that f (A ∩ B) = ∅ 6= {f (x) = f (y)} = f (A) ∩ f (B), a contradiction. (⇐) Suppose that f is 1-1. If y ∈ f (A ∩ B), then y = f (x) for some x ∈ A ∩ B. Thus, x ∈ A and x ∈ B. This implies that f (x) = y ∈ f (A) and f (x) = y ∈ f (B). Thus, y ∈ f (A) ∩ f (B). Now take y ∈ f (A) ∩ f (B). Then y ∈ f (A) so there is some a ∈ A with f (a) = y. Also, y ∈ f (B), so there is some b ∈ B such that y = f (b). Now f (a) = y = f (b), so f being 1-1 now implies a = b. Thus, a ∈ A ∩ B and f (a) = y ∈ f (A ∩ B). Proof. (3):

29

(⇒) Suppose that f (x) = f (y) for some x, y ∈ X. Let A = {x, y}. Then, f (X − {x, y}) ⊂ Y − f ({x, y}) = Y − {f (x) = f (y)} = Y − {f (x)}.By the comments on page thirty-nine, we have that f −1 f ((X − A)) ⊂ f −1 (Y − {f (x)}) ⇒ X − A = X − f −1 ({f (x)}) ⇒ X − {x, y} = X − {x}

It follows that {x, y} = {x} i.e. y = x. (⇐) Suppose that f is 1-1. And let y ∈ f (X − A) for any non-empty subset A ⊂ X. Then, there exists an x0 ∈ X − A such that f (x0 ) = y. Suppose to the contrary that y ∈ / Y − f (A). Then, f (x0 ) = y ∈ (Y − f (A))c = f (A). This implies that there exists some a ∈ A such that f (x0 ) = y = f (a). However, this is a contradiction since f being 1-1 implies a = x0 ∈ A. Proof. (4): (⇒) Let y ∈ Y and consider an arbitrary subset A of X. Now, either y ∈ f (A) or it is not. If it is, we are done. So suppose that it is not. Then, y ∈ Y − f (A) ⊂ f (X − A) and so, there exists some x0 ∈ X − A such that y = f (x0 ); and the result follows. (⇐) Let A ⊂ X and suppose that y ∈ Y − f (A). Since f maps X onto Y , there exists some x ∈ X such that y = f (x) ∈ Y − f (A). Thus, x ∈ / A and so x ∈ Ac = X − A. It follows from he definition that y = f (x) ∈ f (X − A) and the result follows.

30

11

Numbers

Exercise 11.1. Let {Ai } be a family of non-empty successor sets for some non-empty domain I. We show that \ Ai i

is a successor set. Proof. set, and so 0 ∈ Ai . This implies that T For all i ∈ I, Ai is a successor T 0 ∈ i Ai . Likewise, suppose that n ∈ i Ai ; then n ∈ Ai for all iT∈ I and so n+ ∈ Ai for each i ∈ I; from the definition of intersection n+ ∈ i Ai . This completes the proof.

31

12

Peano Axioms

Exercise 12.1. We prove that if n ∈ ω then n 6= n+ . Proof. Suppose, to the contarary, that n = n+ for some n ∈ ω. Then, from the axiom of extension we have that n+ ⊂ n. However, then we would have that n+ = n ∪ {n} ⊂ n implying that n ∈ n which is a contradiction. Exercise 12.2. Let n ∈ ω. We show that if n 6= 0, then n = m+ for some m ∈ ω — We note that this show any element n ∈ ω has a predecessor. Proof. We proceed by mathematical induction: Let S be that subset of ω for which if n 6= 0, then n = m+ for some m ∈ ω. We note that 1 ∈ S, since 1 = 0+ and 0 ∈ ω. Suppose that for some n ∈ ω, n = m+ for some m ∈ ω; indeed, if n+ = m for some m ∈ ω we have (n+ )+ = m+ = z for some z ∈ ω by the induction hypothesis. This completes the proof. Exercise 12.3. We prove that ω is transitive. That is, if x ∈ y and y ∈ ω, then x ∈ ω. Proof. We proceed by mathematical induction. Let S be the set of all z ∈ ω such that if x ∈ z, then x ∈ ω. We note that {} = 0 ∈ S, vacuously. Thus, let y ∈ S. We show that y + ∈ S. That is, let x be any element in y + : x ∈ y + = y ∪ {y} . Then, either x ∈ y or x = y. In the former cases, we that x ∈ ω, by assumption; in the later case we have that y = x ∈ S. Thus, ω = S and so ω is transitive. Exercise 12.4. We show the following: if E is non-empty subset of some natural number, then there exists an element k ∈ E such that k ∈ m whenever m is an element of E and distinct from k — This says that any finite set has a lower bound, namely a greatest lower bound. Proof. We proceed by mathematical induction. Let S be the set of z ∈ ω such that the above holds. We note that {} = 0 ∈ S, vacuously. Suppose that n ∈ S for some n ∈ ω; we show that n+ ∈ S. Suppose that E is a non-empty subset of n+ = n ∪ {n}. If E is a singleton, then we are done; the above is vacuously satisfied. We note that if E = n, then we are done, by the induction hypothesis. Thus, there exists some k ∈ E, as specified above. We claim that this k, so specified, will work even if n ∈ E; Indeed from a previous proof, we know that no element is equal to its successor and so it follows that for any a ∈ E, a ∈ n. In particular, this is true for k. Therefore, S = ω.

32

13

Arithmetic

In many of these exercises, we use induction. As pointed out in the Axiom of Substitution, and the conversation I had with S. Bleiler, and later sections, it is still valid to start induction at some x ∈ ω; we have used this fact a couple times. Exercise 13.1. Prove that if m < n for some natural numbers m, n, then m + k < n + k for any k ∈ ω. Proof. We proceed by mathematical induction. Suppose that for some natural numbers m, n, m < n. Let S be the set of all z ∈ ω such that m + z < n + z. We show that S = ω. Indeed; we note that since m + 0 = m and n + 0 = n, it follows that m = m + 0 < n + 0 = n, and so, 0 ∈ S. Now, suppose that k ∈ S for some k ∈ ω. We show that k + ∈ S. Note, sn (k + ) = (sn (k))+ = sn (k) ∪ {sn (k)}, and by the definition of