Naive Set Theory: Solutions Manual

George Mplikas

September 2015

Contents

2 Misc. Exersizes 32.1 Logical Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Logical Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 Schroder-Bernstein Theorem 49

23 Countable Sets 51

24 Cardinal Arithmetic 52

1

25 Cardinal Numbers 54

26 Extra Exercises 5626.1 Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

2

2 Misc. Exersizes

2.1 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 q p ∨ q q ∨ pT T T TT F T TF T T TF F F F

Thus, p ∨ q ⇐⇒ q ∨ p

Proof. Consider the logical sentences p and q and the following truth table:

p q p ∧ q q ∧ pT T T TT F F FF T F FF F F F

Thus, p ∧ q ⇐⇒ q ∧ p.

3

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 ∨ pT TF F

Thus, p ∨ p ⇐⇒ p

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

p p ∧ pT TF 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 q p ∨ qT T TT F TF T TF F 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 suchas {∅, {∅}}, formed by two of them; consider the pairs formed by any such twopairs, or else mixed pairs formed by a singleton and any pair; and proceed adinfinitum. 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 theaxiom 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 Aand B are singletons, or they are pairs.

• A and B are singletons: Say, A = {a}, and B = {b}. Thus, by the Axiomof Extension, we have a = b. Implying that A and B were formed in thesame way.

• A and B are pairs: Let A = {a, a′}, B = {b, b′}. Without loss of general-ity, suppose that a = b, and a′ = b′. If this is the case, then by the Axiomof Extension, again, we have that a, b were obtained in the same fashion,as well as a′, b′. But in this case, A and B were constructed in the samemanner.

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, ofsets. For simplicity we will assume that C = {A,B,C} for arbitrary non-emptysets (unless specified) A, B, C and will make liberal use of the notation {x :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 (5)

Proof. (⇐⇒ ) Since ∅ has no elements, for all a ∈⋃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) (2)

Proof. (⇒)

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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 thatsimilar 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 theproof above, we have the following:

(A ∩B) ∪ C = (A ∪ C) ∩ (B ∪ C) = A ∩ (B ∪ C) (3)

(⇒)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

5 Complements and Powers

Throughout these exercises we assume that ”all sets are subsets of one and thesame set E , and that all complements (unless otherwise specified) are formedrelative 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 (1)

A ⊂ B ⇐⇒ A−B = ∅ (2)

A− (A−B) = A ∩B (3)

A ∩ (A− C) = (A ∩B)− (A ∩ C) (4)

A ∩B ⊂ (A ∩ C) ∪ (B ∩ Cc) (5)

(A ∪ C) ∩ (B ∪ Cc) ⊂ A ∪B (6)

Proof. (1) (⇐⇒ )

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

⇐⇒ x ∈ A ∩Bc

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, whichimplies x ∈ B: A contradiction.

(⇐) Suppose that A−B = ∅ and that there exists some x ∈ A which is not inB, 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) (⇐⇒ )

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By the proof above,

A− (A−B) = A− (A ∩Bc) = {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 ⊂⋃C, as stated above.

x ∈ A ∩B ⇒ x ∈ {x ∈ E : x ∈ A ∧ x ∈ B}⇒ x ∈ A ∧ x ∈ B⇒ (x ∈ A ∧ x ∈ B) ∧ (x ∈ C ∨ x ∈ Cc)

(1)

x ∈ C ∧ (1)⇒ x ∈ A ∩ C ∧ x ∈ B ∩ C (2)

x ∈ Cc ∧ (1)⇒ x ∈ A ∩ Cc ∧ x ∈ B ∩ Cc (3)

Thus, either way, x ∈ (A ∩ C) or x ∈ (B ∩ Cc), i.e. x ∈ (A ∩ C) ∪ (A ∩ Cc)

Proof. (6) (⇒)

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x ∈ (A ∪ C) ∩ (B ∪ Cc)⇒ x ∈ A ∪ C ∧ x ∈ B ∪ Cc

⇒ (x ∈ A ∨ x ∈ C) ∧ (x ∈ B ∨ x ∈ Cc)(1)

We note that either x ∈ C, or x ∈ Cc:

x ∈ C ∧ (1)⇒ x /∈ Cc ∧ (x ∈ B ∨ x ∈ A)

⇒ x ∈ A ∪B(2)

x ∈ Cc ∧ (1)⇒ x /∈ C ∧ (x ∈ A ∨ x ∈ B)

⇒ x ∈ A ∪B(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}, andP(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 thatif A, B are sets, then (x ⊂ A ∨ x ⊂ B ⇒ x ⊂ A ∪ B). Indeed; if x ⊂ A, thenx ⊂ A ∪B and likewise, x ⊂ B implies x ⊂ B ∪A. With this in mind, we havethe 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 )

1Halmos, P. ”Naive Set Theory”. pg 19

12

Exercise 5.4. We show that⋂

X∈CP(X) = P(⋂

X∈C X):

Proof. Let C be a non-empty collection of sets.

(⇒)

z ∈⋂X∈C

P(X)⇒ (∀X ∈ C)(z ∈P(X))

⇒ (∀X ∈ C)({z} ⊂P(X))

⇒ (∀X ∈ C)(z ∈ X)

⇒ z ∈⋂X∈C

X

⇒ {z} ⊂⋂X∈C

X

⇒ z ∈P(⋂X∈C

X)

(⇐)

z ∈P(⋂X∈C

X)⇒ z ∈ {Y : Y ⊂⋂X∈C

X}

⇒ z ⊂⋂X∈C

X

⇒ (∀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 ∈⋂X∈C

P(X)

And this completes the proof.

Exercise 5.5. We show that⋃

X∈CP(X) ⊂P(⋃

X∈C X):

Proof. Let C be a collection of sets. A common fact used in the following proofis 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 ∈⋃X∈C

P(X)⇒ (∃Y ∈ C)(z ∈P(Y ))

⇒ z ⊂ Y ∈ C

⇒ z ⊂ {x : x ∈ X(∀X ∈ C)} =⋃X∈C

X

⇒ (∀s ∈ z)(s ∈⋃X∈C

X)

⇒ (∀s ∈ z)(s ∈P(⋃X∈C

X))

⇒ z ∈P(⋃X∈C

X)

Which completes the proof.

Exercise 5.6. We show that⋂

X∈P(E)X = ∅:

Proof. Let E be a set and consider P(E) = {Y : Y ⊂ E}. We note that thisis non-empty because ∅ ∈P(E), since ∅ is a subset of every set. We also notethat the statement (x ∈ ∅ ⇒ x ∈

⋂X∈P(E)X) is vacuously satisfied since ∅ has

no elements. Thus, we only show the forward inclusion:

(⇒)

x ∈⋂

X∈P(E)

X ⇒ (∀X ∈P(E))(x ∈ X)

⇒ 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 =⋃

P(E) =⋃

X∈P(E)X:

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

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(⇐)

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.

Exercise 5.9. We show that P´⋃

E¯

is a set that includes E, typically, as a

proper subset.

Proof. Indeed, P´⋃

E¯

= P(E) = {X ∈ E : X ⊂ E}. If E is non-empty,

then it is clear that E Ĺ P(E). However, if E = ∅, then P(E) = ∅ andE = P(E).

15

6 Ordered Pairs

Throughout this section, we will consider A, B, C, X, Y to be any sets, andwill use the notation {x : S(x)} in place of the more accurate, yet cumbersomenotation {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 = ∅) (4)

(A ⊂ X) ∧ (B ⊂ Y ) ⇐⇒ (A×X ⊂ X × Y ) (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 )

(by Commutativity) ⇐⇒ (c ∈ A ∧ d ∈ X) ∧ (c ∈ B ∧ d ∈ Y )

⇒ (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 noelements. Consider A× B = {(a, b) : a ∈ A ∧ b ∈ B}. Then, since there are noa ∈ A, {(a, b) : x ∈ A ∧ x ∈ B} = ∅.

(⇐) Suppose to the contrary that A 6= ∅ and B 6= ∅, but A×X = ∅. SinceA 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 contradictionsince 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 thatA ⊂ 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 transitiveproperties, there exists a relation that does not have that property but does havethe other two.

Proof. Let X be a set. We first show that if X = ∅, then any relation in X isan 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 anequivalence relation, since all statements are vacuously satisfied.

Likewise, suppose that X = {a}; i.e. X is a singleton. We claim, also that anyrelation in X is an equivalence relation:

Lemma. Let X = {a}. Then, by the comments of the previoussection, 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 Itis easy to verify that the transitive condition is vacuously satisfiedfor R. And, it is symmetric, since (a, a) = (a, a). And, clearly, Ris reflexive. Thus, R is an equivalence relation.

We also show that for X = {a, b} each of which is distinct, there does not exista relation, which is symmetric, reflexive, but not transitive:

Lemma. Let X = {a, b}. Then, the relation R in question mustcontain {(a, a), (b, b)} = I. However, it is easy to verify that this isan equivalence relation.3 Thus, we must add an element of X ×Xwhich maintains its symmetry, yet is not transitive. However, thisis not possible since either way (a, b), (b, a) must be in R andso we would have R = {(a, a), (b, b), (a, b), (b, a)}.And, this is anequivalence relation since (x, x) ∈ R for all x ∈ X and, by thecomments above, R is symmetric, as well as transitive.4

Thus, let X = {a, b, c....}, for distinct elements. We first show the relationalproperties which include reflexivity:

2see Halmos pg. 273see the above lemma4this is left to the reader

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 whichis transitive, but not symmetric, and symmetric but not transitive:

• ConsiderR = I∪{(a, b), (b, a), (b, c), (c, b)}. Then, this relation is symmet-ric, 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 tocheck (a, b), (a, a), (b, b) for transitivity, since they are the only orderedpairs 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 asabove:

• ConsiderR = {(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, reflexivesince (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 indeeda 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.

5the 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 rangeprojection.

• We first note that if X, Y have a pair or more of distinct elements, theng 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, theng((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 andy = g((x, y)) 6= x = g((y, x). We note that this covers the case in whichx = y.

• Next, we show that if either X or Y is a singleton and the other is anarbitrary 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 previousexercise; which is trivially 1-1. Else, we assume Y has more than a pairof 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 gis 1-1.

Exercise 8.2. We show that Y ∅ has exactly one element, namely ∅, whether Yis empty or not; i.e. Y ∅ = {∅}.

Proof. Suppose that Y is a set. Consider f = Y ∅. From page thirty, and by aprevious theorem, we have f ⊂ P(∅ × Y ) = P(∅) = {∅}, where f is a set ofelements which are ordered pairs. From the comments in section seven, ∅ is aset 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, thereexists some b ∈ ∅ such that (x, b) ∈ f . However, this is not true, since ∅ containsno elements. Thus, ∅X = ∅.

22

9 Families

Exercise 9.1. After the comments on page thirty-five, we formulate and provea generalized version of the commutative law for unions.

Proposition 9.1. Let K be an arbitrary, non-empty set, and let p ∈ Aut(KK)be an element of the set of one-to-one and onto functions from K to K. We notethat in the current notation this means that p is the family and {pi} denotesthe values of p evaluating some i ∈ K. Let {Ak} be a family of functions withdomain K. We claim that

⋃k∈K Ak =

⋃i∈K Api .

Proof.(⇒)

x ∈⋃k∈K

Ak ⇒ (∃k0 ∈ K)(x ∈ Ak0)

(since p is onto) ⇒ (x ∈ Ak0) ∧ (∃s ∈ K)(k0 = ps)

⇒ x ∈ Ak0= Aps

⇒ x ∈⋃i∈K

Api

(⇐)

x ∈⋃i∈K

Api⇒ (∃s ∈ K)(x ∈ Aps

)

(since ps ∈ K and onto) ⇒ (x ∈ Aps) ∧ (∃k0 ∈ K)(ps = k0)

⇒ x ∈ Aps= Ak0

⇒ x ∈⋃k∈K

Ak

Which completes the proof.

To illustrate the previous proof, consider K = {0, 1} and pi =

{1 i = 0

0 i = 1; it

is easy to show that p is 1-1 and onto. From the statement above, we have⋃k∈K

Ak = A0 ∪A1 (1)⋃i∈K

Api = Ap0 ∪Ap1 = A1 ∪A0 (2)

23

Exercise 9.2. We show that if both {Ai} and {Bi} are families of sets, withdomains I, J , then

´⋃i

Ai

¯

∩´⋃

j

Bj

¯

=⋃i,j

(Ai ∩Bj) (1)

´⋂i

Ai

¯

∪´⋂

j

Bj

¯

=⋂i,j

(Ai ∪Bj) (2)

Where the symbols such as⋂

i,j mean⋂

(i,j)∈I×J

Proof. (1)

(⇐⇒ )

x ∈´⋃

i

Ai

¯

∩´⋃

j

Bj

¯

⇐⇒ x ∈´⋃

i

Ai

¯

∧ x ∈´⋃

j

Bj

¯

⇐⇒ (∃i0 ∈ I)(x ∈ Ai0) ∧ (∃j0 ∈ J)(x ∈ Bj0)

⇐⇒ (∃(i0, j0) ∈ I × J)(x ∈ (Ai0 ∩Bj0))

⇐⇒ x ∈⋃i,j

(Ai ∩Bj)

Proof. (2)

(⇐⇒ )

x ∈´⋂

i

Ai

¯

∪´⋂

j

Bj

¯

⇐⇒ x ∈´⋂

i

Ai

¯

∨ x ∈´⋂

j

Bj

¯

⇐⇒ (∀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 ∈⋂

(i,j)∈I×J

(Ai ∪Bj)

24

Exercise 9.3. Let {Ai}, {Bj} be families of sets with domains I 6= ∅, J 6= ∅,respectively. We show that

´⋃i

Ai

¯

×´⋃

j

Bj

¯

=⋃i,j

(Ai ×Bj) (1)

´⋂i

Ai

¯

×´⋂

j

Bj

¯

=⋂i,j

(Ai ×Bj) (2)

Proof. (1)

(⇐⇒ )

(a, b) ∈´⋃

i

Ai

¯

×´⋃

j

Bj

¯

⇐⇒ a ∈⋃i

Ai ∧ b ∈⋃j

Bj

⇐⇒ (∃i0 ∈ I)(∃j0 ∈ J)(a ∈ Ai0 ∧ b ∈ Bj0)

⇐⇒ (∃(i0, j0) ∈ I × J)((a, b) ∈ Ai0 ×Bj0)

⇐⇒ (a, b) ∈⋃

(i,j)∈I×J

(Ai ×Bj)

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

(⇐⇒ )

(a, b) ∈´⋂

i

Ai

¯

×´⋂

j

Bj

¯

⇐⇒ a ∈⋂i

Ai ∧ b ∈⋂j

Bj

⇐⇒ (∀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) ∈⋂i,j

Ai ×Bj

Exercise 9.4. Suppose that for each i ∈ I, Xi is non-empty. Consider thefamily of sets {Xi} with domain I. We show that

⋂iXi ⊂ Xj ⊂

⋃iXi for each

index j ∈ I.

25

Proof. Let {Xi} be stated as above, and let j ∈ I be arbitrary.

x ∈⋂i

Xi ⇒ (∀i ∈ I)(x ∈ Xi)

⇒ 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 ∈⋃j

Xj

⇒ x ∈´⋃

j

Xj ∪⋃

z∈I−JXz

¯

⇒ x ∈⋃i

Xi

Which prove the second inclusion.

Exercise 9.5. We show the minimality condition of intersection and unions:(1) That is, if Xj ⊂ Y for each index j and some set Y , then

⋃iXi ⊂ Y and

that it is unique; (2) Likewise, if Y ⊂⋂

j Xj for each index j and some set Y ,then Y ⊂

⋂iXi 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 )⇒⋃j

Xj ⊂ Y

⇒⋃i

Xi ⊂ Y

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 ⊂⋂j

Xj ⇒ Y ⊂⋂i

Xi

26

10 Inverses and Composites

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

f ′´⋃

i

Ai

¯

=⋃i

f ′(Ai)

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 ∈⋃i

f ′(Ai)⇒ (∃i0 ∈ I)(x ∈ f ′(Ai0))

(since Ai0 ⊂⋃i

Ai) ⇒ x ∈ f ′´⋃

i

Ai

¯

(⇐)

x ∈ f ′´⋃

i

Ai

¯

⇒ (∃i0 ∈ I)(x ∈ f ′(Ai0))

⇒ x ∈´ ⋃

z∈I−{i0}

f ′(Az)¯

∪ f ′(Ai0) =⋃i

f ′(Ai)

And the result follows.

Exercise 10.2. We show that, in general,

f ′´⋂

i

Ai

¯

=⋂i

f ′(Ai)

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 ′ : P(X)→P(Y ) as the following:

f ′(∅) = ∅f ′({a}) = {d}f ′({b}) = {e}f ′({c}) = {d}

f ′({a, b}) = {d, e}f ′({b, c}) = {d}f ′({a, c}) = {e}

f ′({a, b, c}) = ∅

Then, let {Ai} be a family of subsets of X with domain I = 2, such thatA0 = {a} and A1 = {a, c}. Then, we have the following:

f ′´⋂

i

Ai

¯

= f ′({a}) = {d}⋂i

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

Which completes the proof.

Exercise 10.3. Let f : X → Y . We show that f maps X onto Y iff for everynon-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 thatf(x) = y. Since this is the case, if B is a non-empty subset of Y , then for anyb ∈ 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 followsimmediately.

Exercise 10.4. In the context of page forty-one, R := ”son”, S := ”brother”,what do R−1, S−1, RS and R−1S−1 mean?

Proof. (R−1): From the comments on page forty-one, xRy implies that that xis son-related to y. That is, y is the father of x. Thus, yR−1x means that y hasthe 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−1x implies y is the brother of x.

28

(T = R ◦ S): Also, by the previous comments, if xTz, then there exists ysuch that xSy and yRz. Thus, x is the brother of y and z is the son to y. Thatis, x is the uncle of y.

(T = R−1 ◦ S−1): Likewise, if xTz, then there exists a y such that xS−1yand yR−1z. Thus, x is the brother of y and y is the father to z. Thus, x is theuncle of z.

Exercise 10.5. Assume that for each of the following f : X → Y . We showthe following:

1. If g : Y → Z such that g ◦ f is the identity on X, then f is 1-1 and gmaps 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. Sinceg ◦ 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 ontoX 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 somex ∈ A ∩ B. Thus, x ∈ A and x ∈ B. This implies that f(x) = y ∈ f(A) andf(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 withf(a) = y. Also, y ∈ f(B), so there is some b ∈ B such that y = f(b). Nowf(a) = y = f(b), so f being 1-1 now implies a = b. Thus, a ∈ A ∩ B andf(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 thecomments on page thirty-nine, we have that

f−1f((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 subsetA ⊂ X. Then, there exists an x0 ∈ X − A such that f(x0) = y. Suppose tothe contrary that y /∈ Y − f(A). Then, f(x0) = y ∈ (Y − f(A))c = f(A). Thisimplies that there exists some a ∈ A such that f(x0) = y = f(a). However, thisis 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, eithery ∈ 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 thaty = 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 sox ∈ Ac = X − A. It follows from he definition that y = f(x) ∈ f(X − A) andthe result follows.

30

11 Numbers

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

i

Ai

is a successor set.

Proof. For all i ∈ I, Ai is a successor set, and so 0 ∈ Ai. This implies that0 ∈

⋂iAi. Likewise, suppose that n ∈

⋂iAi; then n ∈ Ai for all i ∈ I and so

n+ ∈ Ai for each i ∈ I; from the definition of intersection n+ ∈⋂

iAi. Thiscompletes 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 theaxiom of extension we have that n+ ⊂ n. However, then we would have thatn+ = 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 somem ∈ 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 forwhich 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 theinduction 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 suchthat 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 nat-ural number, then there exists an element k ∈ E such that k ∈ m whenever mis an element of E and distinct from k — This says that any finite set has alower bound, namely a greatest lower bound.

Proof. We proceed by mathematical induction. Let S be the set of z ∈ ω suchthat the above holds. We note that {} = 0 ∈ S, vacuously. Suppose that n ∈ Sfor some n ∈ ω; we show that n+ ∈ S. Suppose that E is a non-empty subsetof n+ = n ∪ {n}. If E is a singleton, then we are done; the above is vacuouslysatisfied. 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, sospecified, will work even if n ∈ E; Indeed from a previous proof, we know thatno 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 ofSubstitution, and the conversation I had with S. Bleiler, and later sections, itis still valid to start induction at some x ∈ ω; we have used this fact a coupletimes.

Exercise 13.1. Prove that if m < n for some natural numbers m, n, thenm+ k < n+ k for any k ∈ ω.

Proof. We proceed by mathematical induction. Suppose that for some naturalnumbers 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, itfollows that m = m + 0 < n + 0 = n, and so, 0 ∈ S. Now, suppose that k ∈ Sfor some k ∈ ω. We show that k+ ∈ S.

Note, sn(k+) = (sn(k))+ = sn(k)∪ {sn(k)}, and by the definition of <, andthe induction hypothesis, we have that {sn(k)} ⊂ sm(k) ⊂ (sm(k)∪{sm(k)}) =sm(k+). This implies that sn(k+) = sn(k) ∪ {sn(k)} ⊂ sm(k) ∪ {sm(k)} =sm(k+). That is, S = ω

Exercise 13.2. Prove that if m < n for some m,n ∈ ω, and k 6= 0, thenm · k < n · k.

Proof. We proceed by mathematical induction. Suppose that m < n for somem,n ∈ ω. Let S be the set of z ∈ ω, which satisfy the following statement: ifz 6= 0, then m · z < n · z. We note that 1 ∈ S. Suppose that k ∈ S for somek ∈ ω; we show that k+ ∈ S. Indeed: by definition of product, we have thefollowing:

pm(k+) = pm(k) +m

= sm(pm(k)) (1)

< sm(pn(k)) (2)

= pn(k) +m

< pn(k) + n (3)

= pn(k+)

We only note that (1),(2), and (3) follow from the inductive hypothesis and theprevious exercise. Therefore, S = ω.

Lemma 13.1. For any n ∈ ω, n is transitive.

Proof. Suppose that E = n ∈ ω and y ∈ E. Since y ∈ E, y is a proper subsetof E. Now, if x ∈ y, then x is a proper subset of y and it follows directly thatx ⊂ y ⊂ E. That is, x is a proper subset of E. That is x ∈ E.

Lemma 13.2. If m,n ∈ ω such that m 6= n, then m ∈ n iff m ⊂ n

Proof. Let m, n be stated as above.

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(⇒)Suppose that m ∈ n. We show that m ⊂ n: let x ∈ m, then by the lemma

above, x ∈ n. Thus, m ⊂ n.

(⇐)Suppose that m ⊂ n. If m = n, then we are done. Otherwise, m 6= n. If

this is the case, then n ∈ m cannot happen (for them m would be a subset ofone of its elements), and therefor m ∈ n; this is by the comparability of andx, y ∈ ω.

Exercise 13.3. We prove that if E is a non-empty set of natural numbers, thenthere exists an element k ∈ E such that k ≤ m for all m ∈ E. — This statementsays that any non-empty subset in ω has a minimum.

Proof. Let E ⊂ ω such that E 6= ∅. Since E is non-empty, we can form theintersection of elements in E. Consider

z =⋂x∈E

x

It follows directly from the formulation of z that z ⊂ x, for all x ∈ E: z ≤ x.Thus, it is only necessary to prove that z ∈ E. From the previous lemma we seethat since z ∈ x for all x ∈ E, and x ∈ E, that z ∈ E. The result follows.

Exercise 13.4. A set E is called finite if it is equivalent to some natural number;otherwise E is infinite. We use this definition to prove that ω is infinite.

Proof. Suppose, to the contrary that ω is finite. Then, there exists a 1-1 andonto function, f , from ω to n for some n ∈ ω: f ⊂ ω × n. Since f is 1-1 andonto, for ever y ∈ n, there exists a unique x ∈ ω such that (x, y) ∈ f . Let g bethe union of all such x, which is an element of ω. Then, g+ = g ∪ {g} is in ωbut not in g. Since f is a function, we must have (g+, y1) ∈ f for some y1 ∈ n;likewise since f is onto, there exists some x ∈ g such that (x, y1) ∈ f . Fromhere we see that we must have (g+, y1) = (x, y1), implying g+ = x; which is acontradiction since g+ /∈ g, whereas x ∈ g.

Exercise 13.5. If A is a finite set, then A is not equivalent to a proper subsetof A. The contrastive of this statement is that if A is equivalent to a propersubset of A, then A is not finite (infinite). We use this definition to show thatω is infinite.

Proof. We show that ω ∼ A, where A = ω − {0}. First, it is clear that Ais a non-empty proper subset of ω. Consider the map f ⊂ ω × ω, given byf(n) = n+. First, we show that f is indeed a function, and then that it is 1-1and onto:

• Well-Defined : Suppose that x = y for some x, y ∈ ω. From a previoustheorem, we have x+ = y+. This shows that f is well-defined.

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• Everywhere-Defined : This follows directly from the definition of ω as beinga successor set. This shows that f is everywhere defined.

• 1-1 : Suppose that n+ = m+. Again, from a previous exercise (page 46),we have n = m. This shows that f is 1-1.

• Onto: Suppose that n ∈ A = ω − {0}. Then no n ∈ A is 0. Thus, by aprevious theorem, we have that n = m+ = f(m) for some m ∈ ω. Thisimplies that f is onto.

Exercise 13.6. We prove that union of a finite set of finite sets is finite.

Proof. We proceed by mathematical induction. Let S be a finite set, such that#(S) = n, for some n ∈ ω. Let B be the set of all N ∈ ω such that

⋃S is finite.

Note 0 ∈ B since, if #(S) = 0, then S = {} and⋃{} = {} (which is finite).

Next, suppose that k ∈ B: that is, if S is a finite set such that #(S) = k, then⋃S is finite. We show that k+ ∈ B. Suppose that #(S) = k+. Then, there

exists some 1-1 and onto function f i.e. f : k+ ↔ S. And we have the following:⋃S =

⋃i∈k+

f(i) =´ ⋃

i∈k

f(i)¯

∪ f(k)

From the comments on page 53, and the induction hypothesis, we see thatk+ ∈ B. Thus, B = ω.

Exercise 13.7. We prove the following: If E is finite, then P(E) is finite and,moreover, #(P(E)) = 2#(E).

Proof. It is only necessary to show that if #(E) = n for some n ∈ ω, then# P(E) = 2#(E); we conclude that 2#(E) ∈ ω by a previous exercise. Weproceed by mathematical induction.

Let S be the set of all N ∈ ω such that #(E) = N and # P(E) = 2#(E).Note that 0 ∈ S, since #(E) = 0 implies E = {} and #(P(E)) = #({∅, {∅}}) =2 = 20. Next, assuming that k ∈ S, we show that k+ ∈ S:

Suppose that #(E) = k+ = k+ 1, and consider the set E∗ = E−{0}. Note,#(E∗) = k. We note that either 0 is in a subset of E, or it is not. If it is not,then we are forming subsets of E∗. From the inductive hypothesis, there are2#(E∗) = 2k subsets of E∗. By similar logic, there are 2k subsets which include0. It follows that there are 2k + 2k = 2 · 2k = 2k+1 = 2#(E) subsets of E. Thus,k + 1 = k+ ∈ S. That is, S = ω.

Exercise 13.8. If E is a non-empty finite set of natural numbers, then thereexists and element k ∈ E such that m ≤ k for all m ∈ E — This statementsays that for any finite set E, non-empty, of natural numbers, there is an upperbound, actually a maximum.

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Proof. Let E be a finite, non-empty, subset of natural numbers. We proceed bymathematical induction. Let S be the set of all N ∈ ω such that #(E) = Nand there exists a k ∈ E such that the above holds.

Note that 1 ∈ S; 1 = #(E) implies E = n for some n ∈ ω, and clearly n ≤ n.Next, assuming that k ∈ S, for some k ≥ 1, we show that k+ ∈ S:

Suppose that #(E) = k+ = k + 1. Fix some x ∈ E, and consider E∗ =E − {x}. We note that #(E∗) = k, and so, it follows from the inductionhypothesis that there exists some g ∈ E∗ such that m ≤ g, for all m ∈ E∗. Bythe comparability of elements in ω, we must have that either x ∈ g, or g ∈ x; weonly note that x 6= g, since otherwise we would have #(E) = k. In the formercase, if x ∈ g, then x < g, and in particular x ≤ g. In the later case, supposingthat g ∈ x, we have g < x; and, again, in particular g ≤ x. But then m ≤ gand g ≤ x for all m ∈ E = E∗ ∪ {x}, implying m ≤ x for all m ∈ E by thetransitivity of ≤. This shows that k+ ∈ S, and so S = ω.

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14 Order

Exercise 14.1. We express the conditions of antisymmetry and totality for arelation R by means of equations involving R and its inverse — This exerciseillustrates that we can determine a partial, or total order, by looking the theco(domains), of a relation.

Proof. From the comments on page 54 and the notation in section 10, we adoptthe usual assumptions and notation. That is, assume that R is a relation in X.And, we have the following:Antisymmetry:

xRy ∧ yRx⇒ x = y

iff (R(x) = y ∧R−1(x) = y)⇒ R(x) = R−1(y)

Totality:

(∀x, y ∈ X)(xRy ∨ yRx) ⇐⇒ (∀x, y ∈ X)(R(x) = y ∨R−1(x) = y)

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15 Axiom of Choice

Exercise 15.1. We prove that every relation includes a function with the samedomain.

Proof. Suppose that X and Y are sets and let R be a relation from X into Y .By definition, R ⊂ X × Y , and so if X or Y is empty, the X × Y is empty andwe are done. Otherwise, there exists at least an element in each of X, Y . LetXR be the subset of X given by the first coordinates of R. Claim: there existssome function f : XR → Y .

Let y ∈ Y be arbitrary and consider Ty = {x ∈ XR : xRy}. Let I be theset of z ∈ Y such that Tz 6= ∅. Then, define T =

⋃y∈I{Ty} = {Ti}. Assuming

the axiom of choice, and by the comments on page sixty, {Ti} is a family ofnon-empty sets indexed by I and so, there exists a family {ti}, i ∈ I such thatti ∈ Ti for each i ∈ I. Since R is a relation on XR, so defined, it follows thatf = {(ti, b) : i ∈ I ∧ b = R(ti)} is a candidate function.

Indeed; we verify that f is well-defined and everywhere defined:

• Well-Defined : Suppose that ti = tj for some i, j ∈ I. By construction,and the previous supposition, we have that Tti = {x ∈ XR : xRti} = {x ∈XR : xRtj} = Ttj . Thus, (ti, R(ti)) = (tj , R(tj)) = f(ti) = f(tj).

• Every-Where Defined : This is clear by construction.

Thus, f is a function.

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16 Zorn’s Lemma

Exercise 16.1. We show that Zorn’s Lemma is equivalent to the axiom ofchoice.

Proof. Let X be a set and define F to be the following set of functions:

f ∈ F ⇐⇒

dom f ⊂P(X)

Im f ⊂ X(∀A ∈ dom f)(f(A) ∈ A)

Next, consider the relation ≤, in X, given by the following: for all f1, f2 ∈ F ,f1 ≤ f2 iff f2 is an extension of f1 i.e. f1 ⊂ f2. As the comments on page 65point out, ≤ is a partial ordering. We note that (F,≤) is a partially orderedset. Assuming Zorn’s Lemma, we show that there exists some maximal elementf ∈ F :F is already partially ordered. So, let X be a chain in F . Note that X is a

collection of ordered pairs, satisfying the above criterion. We claim that⋃

Xis the upper bound for X ; we only need to show that

⋃X is a function:

• Function: Suppose that (a, x), (a, y) ∈⋃

X , then (a, x) ∈ f for somef ∈ X and, likewise, (a, y) ∈ g for some g ∈ X . Then, since X is achain, either f ≤ g, f = g, g ≤ f . Without loss of generality, we have(a, x), (a, y) ∈ g. Since g is a function, we have x = y. The fact that

⋃X

is every-where defined is obvious.

• The rest of the properties follow by definition.

By Zorn’s Lemma, there exists a maximal element in F , call it f . To completethe proof, we show that dom f = P(X)− {∅}:

• Suppose that dom f 6= P(X) − {∅}. Then, there exists some a ∈ P (X)such that a /∈ dom f . Let g = f ∪ {(a, b)}, for any b ∈ X. It is clear thatg ∈ F and that g ≥ f . However, this contradicts the minimality of f .

We note that using the Axiom of Choice to prove Zorn’s Lemma was shown intext. This completes the proof.

Exercise 16.2. We prove that the following is equivalent to the axiom of choice:Every partially ordered set has a maximal chain — This is the Hausdorff Max-imality Principal (HMP).

Proof. We show that HMP is equivalent to Zorn’s Lemma and use the fact thatZorn’s Lemma is equivalent to the Axiom of Choice to conclude.

• (⇒) Suppose the HMP. Let (X,≤) be the partially ordered set, and A themaximal chain. By the assumptions of Zorn’s Lemma, A has an upper-bound, call it a. Now, a ∈ A, since otherwise, we could form B = A∪{a},

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which is again a totally ordered set, containing A; for all s ∈ B, theelements of A are comparably, likewise, s ≤ a. To complete the proof inthis direction, we note that if z ∈ X is any element such that z ≥ a, thenz is an upper-bound and must be in A, by the previous discussion. Thisimplies that a ≥ z for all z ∈ X. Specifically, since A is partially ordered,we have a ≤ z and z ≤ a implying a = z. Thus, a is maximal.

• (⇐) Suppose Zorn’s Lemma. Let (X,≤) be a partially ordered set. Con-sider F be the set of all totally ordered subsets of X. The inclusion orderon F is a partial order. Note that if S is a maximal element in F , thenit will be a maximal chain in X. We show that any chain A in F has anupper-bound:

– (The following was suggested by this example): Indeed, consider A′ =⋃E∈AE. By definition, A′ contains all elements of A. We only show

that A′ ∈ F : To show this take x and y, elements of A′. Then thereare sets Ex and Ey, members of A, with x ∈ Ex and y ∈ Ey. Since Ais partially ordered by inclusion, Ex ⊂ Ey or Ey ⊂ Ex, that is, eitherx, y ∈ Ey or x, y ∈ Ex. Since both Ex and Ey are linearly orderedin X then either x ≤ y or y ≤ x. This shows that A′ is a totallyordered set, and so a chain in X i.e. A′ ∈ F .

This completes the proof.

Exercise 16.3. We show that the following statements are equivalent to theaxiom of choice: (i) Every partially ordered set has a maximal chain — This isthe Hausdorff Maximality Principal (ii) Every chain in a partially ordered set isincluded in some maximal chain. (iii) Every partially ordered set in which eachchain has a least upper bound has a maximal element.

Proof. To do so, we show that i ⇒ ii ⇒ iii ⇒ ZL ⇒ i, noting that Zorn’sLemma is equivalent to the axiom of choice. Throughout these exercises, weassume that (X,≤) is the partially ordered set in discussion.

• (i ⇒ ii) Suppose that every partially ordered set has a maximal chain.Since (X,≤) is partially ordered, there exists some maximal chain A ⊂ X.We note that for chains, inclusion is the partial order. Thus, by definition,for all other chains A′ ⊂ X, we must have A′ ⊂ A, since otherwise, A Ĺ A′,contradicting the maximality of A.

• (ii ⇒ iii) Suppose that every chain in a partially ordered set is includedin some maximal chain. Let (X,≤) be such a partially ordered set. Then,there exists some maximal chain A such that A′ ⊂ A for all other chainsA′ ⊂ X. We show that each chain A′ of X has a least upper bound:

– Let G be the set of all upper-bounds of some chain A′. Then, wehave have A′ ≤ g for all g ∈ G. Likewise G ∩A′ = A′ implying A′ isthe least upper-bound of A′.

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From the antecedent in (iii), we see that A has a maximal element in A,call it a. We also note that a is maximal in X: If not, then we can defineT = A∪{a}, which is a totally ordered set in which A ⊂ T ; contradictingthe maximality of A.

• (iii ⇒ Zorn’s Lemma) Suppose that every partially ordered set in whicheach chain has a least upper bound has a maximal element. Let X be sucha set. Then for any chain A ⊂ X, A has a l.u.b. call it δA. In particular,δA is an upper-bound. By assumption X has a maximal element.

• (Zorn’s Lemma ⇒ i) This was proved in the previous exercise.

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17 Well Ordering

Exercise 17.1. A subset A of a partially ordered set X is cofinal in X in casefor each element x ∈ X, there exists an element a ∈ A such that x ≤ a. Weprove that every totally ordered set has a cofinal well ordered subset.

Proof. Let (X,≤) be a non-empty totally ordered set. If X has a maximumelement, then we are done. Thus, suppose that X has no maximum. Let F bethe set of all well ordered subsets of X with the following order: for all f, g ∈ F ,f � g iff f is an initial segment of g. We seek to apply Zorn’s Lemma.

• From the previous result, � is a partial order on F .

• Suppose that Z is chain in F . We show that Z has an upper bound inF . Consider

⋃Z. It is clear that Z �

⋃Z. We show that

⋃Z is well

ordered; Indeed, these are the comments on page 67.

We have satisfied the assumptions in Zorn’s Lemma, and so it follows that thereexists some maximal element M ∈ F . We claim that M is a well ordered cofinalsubset of X.

First, it is a subset. Second, it is well ordered by the definition of F . Next,if x ∈ X, then for any m ∈M , either m ≤ x, x ≤ m or x = m. But, since M ismaximal, if m ≤ x, then x ∈M ; otherwise, we could form N = M ∪ {x} whichis a contradiction to the maximality of M . Thus, M is the well ordered cofinalsubset of X.

Exercise 17.2. Does any such condition apply to partially ordered sets?

Proof.

Exercise 17.3. We prove that a totally ordered set is well ordered iff the set ofstrict predecessor of each element is well ordered.

Proof.(⇒) Suppose that we have a well ordered set (X,≤). Note that the set ofstrict predecessors of some x ∈ X forms a subset. The rest follows from thedefinition.(⇐) Suppose that (X,≤) is a partially ordered set and that for any x ∈ X thestrict initial segment s(x) is well ordered. Let J be a subset of X. Pick somej0 ∈ J and consider s(j0) = {x ∈ X : x < j0}. If s(j0) = ∅, j0 is the leastelement, and we are done. Otherwise, s(j0) 6= ∅. But since s(j0) is well ordered,we have J ∩ s(j0) ⊂ s(j0) and J ∩ s(j0) ⊂ J . Implying that J ∩ s(j0) has a leastelement in J , call it z. We note that z is the least element for J ; Otherwise,there is some x ∈ J , x 6= z such that x < z < j for all j ∈ J and in particularfor j0. This implies that x ∈ s(j0). However, then we must have x ∈ J ∩ s(j0)and so x = z; a contradiction.

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Exercise 17.4. We prove that the well ordering theorem implies the Axiom ofChoice.

Proof. The following proof was adapted from this website:Let C be a collection of nonempty sets. Then

⋃S∈C S is a set. By the well-

ordering principle,⋃

S∈C S is well-ordered under some relation <. Since each Sis a nonempty subset of

⋃S∈C S, each S has a least member mS with respect

to the relation <. Define f : C →⋃

S∈C S by f(S) = mS . We show that f is achoice function:

• Well-Defined : Suppose that (a, x), (a, y) ∈ f . Then, there exists someS1 ∈ C such that x ∈ S1. Likewise, there exists some S2 ∈ C suchthat y ∈ S2. Without loss of generality, it follows that x, y ∈ S2, and sox = y = mS2 .

• Everywhere Defined : This follows fromt he discussion above.

Therefore, f is a choice function. Hence, the axiom of choice holds.

Exercise 17.5. We prove that if R is a partial order in a set X, then thereexists a total order S in X such that R ⊂ S; in other words, every partial ordercan be extended to a total order without enlarging the domain.

Proof. Let (R,≤) be a partially ordered subset of X. Let C be the set of all Ain R of partial orders, with respect to ≤. Let (C,⊂) be the partial ordered set6.Let X be the collection of all chains in C. Note that for all y ∈X y ≤

⋃y∈X y.

Thus, every chain has an upper-bound. By Zorn’s Lemma, C has a maximalelement. Denote this by c. Claim: c is a total order in X:

• Since c ∈ C, c is a partial order in X. Since it is maximal, for all c′ ∈ C,we have c′ ⊂ c. Thus, for all x, y ∈ X, either (x, y), (y, x) ∈ c. Either way,we must have x ≤ y, y ≤ x or x = y.

6This has been shown other exercises.

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18 Transfinite Recursion

Exercise 18.1. We prove that a similarity preserves ≺ (in the same sense inwhich the definition demands the preservation of ≤).

Proof. Let (X,�), (Y,≤) be two partially ordered sets and that there existssome function f : X → Y , 1-1 and onto. f has the property that for all a, b ∈ Xwith a � b iff f(a) ≤ f(b), for all f(a), f(b) ∈ Y .

Suppose, to the contrary, that a ≺ b and f(a) ≥ f(b). However, this is acontradiction, since, by definition, we must have b � a. Thus f preserves strictorder.

Exercise 18.2. We prove the following statement: A 1-1 function, f that mapsone partially ordered set onto another is a similarity iff f preserves < (strictorder).

Proof. Throughout this exercise, we assume that (X,�) and (Y,≤) are partiallyordered sets and f : X → Y is 1-1 and onto.

• (⇒): Suppose that f is a similarity, then the previous exercise prove thisconclusion.

• (⇐): Suppose that f is as stated above, such that for any a, b ∈ X wehave a ≺ b iff f(a) < f(b). To show that f is a similarity, we show thata = b iff f(a) = f(b):

– (⇐): Since f is 1-1, then f(a) = f(b) implies a = b.

– (⇒): Suppose, to the contrary, that a = b and, (without loss general-ity) f(a) < f(b). Since f preserves strict order, we must have a ≺ b.However, this is a contradiction.

This completes the proof.

Exercise 18.3. We show that a subset Y of a well ordered set X is, again, wellordered.

Proof. Let (X,≤) be a well-ordered set, and Y ⊂ X. If Y = X or Y = ∅, thenY is clearly a well-order. Thus, suppose that Y is proper subset of X. Supposethat Y was not well-ordered. Then, some subset, Z of Y does not have a leastelement. However, Z ⊂ X, and so by definition of a well-ordered set, it has aleast element. A contradiction.

Exercise 18.4. We prove that each subset of a well-ordered set X is similareither to X, or an initial segment of X.

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Proof. Suppose that (X,�) is a well-ordered set. If X = ∅, then every subset ofX is equivalent to X. So, suppose X 6= ∅, and let S be a non-empty subset of X.Since X ⊂ X, X has a least element, call it x. Likewise, S has a least element,s. Consider the map f given by f(s) = x. And then, since X − {x} ⊂ X ithas a least element, call it x′. Likewise, S − {s} has a least element, call it s′

and define f(s′) = x′, and so on. We show that f is a sequence of type S in X.Now, either this process terminates, or it does not. If it does, then it terminatesat some x∗, s∗ i.e. f(s∗) = x∗. Claim: S ∼= s(x∗).

• The fact that f is a function is clear by previous definition. It only needto be shown that f is 1-1, onto and preserves ≺. The fact that f is 1-1 isclear. Thus, if x ≺ x∗, then by construction, s(x) = f(t) for some t ∈ S.

• f preserves ≺: This follows from the definition of initial segments give onpage 65, 78.

Otherwise, the process does not terminate. In this case, the claim is thatS ∼= X. The fact that f is a 1-1 and onto function is shown similarly as before.Likewise for ≺.

Exercise 18.5. We prove that if (X,≺), (Y,≤) are well ordered sets, and X ∼=Y , then the similarity maps the least upper bound (if any) of each subset of Xonto the least upper bound of the image of that subset.

Proof. Let X,Y be stated as above. And suppose that S ⊂ X has a least upperbound, s∗. This implies that for all s ∈ S, s ≤ s∗ and for any other upperbound x ∈ X, x ≤ s∗ implies x = s∗. Suppose, to the contrary, that f(s∗) wasnot the least upper-bound of f(S). Since f is order preserving, we have thatf(s∗) is an upper-bound for f(S). Let α 6= f(s∗) be the l.u.b. of f(S) (since fis onto there exists some x ∈ X such that α = f(x). Thus from here, α = f(x));for all f(s) ∈ f(S), we have f(s) ≤ f(x), and that for any other upper-boundz of f(S), z ≤ f(x) implies z = f(x), in particular, this implies that s∗ ≥ x.But then, since f is order-preserving, we have that f(s) ≤ f(x) implies thats ≤ x for all s ∈ S. But since s∗ is the l.u.b. for S, we must have s∗ ≤ x. SinceX, and consequently S is well-ordered, it is partially ordered, and so s∗ = x, acontradiciton.

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19 Ordinal Numbers

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20 Sets Of Ordinal Numbers

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21 Ordinal Arithmetic

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22 The Schroder-Bernstein Theorem

Exercise 22.1. Let X, and Y be sets. Suppose that f : X → Y , is a function,and g : Y → X, is a function. We prove that there exist subsets A ⊂ X, B ⊂ Ysuch that f(A) = B, and g(Y −B) = X − A. (This statement is used to provethe Schroder-Bernstein theorem.)

Proof. Let X, Y , f and g be stated as above. Throughout this exercise, we willuse f [A] instead of the more common, yet less accurate, f(A). We look at fourcases:

• f onto, and g is not: In this case, we have f [X] = Y , and g[Y − f(X)] =g[Y − Y ] = g[∅] = {} = X − X. Thus, choosing A = X ⊂ X andB = {} ⊂ Y , the result follows.

• g onto, and f is not: In this case, we have f [∅] = ∅, and g[Y −∅] = g(Y ) =X = X − {}. Thus, choosing A = {} ⊂ X, and B = {} ⊂ Y , the resultfollows.

• f onto, g onto: Again, in this case, letting A = ∅ ⊂ X we have f [∅] = ∅,and g[Y − ∅] = g[Y ] = X = X − ∅; the result follows.

• f not onto, g not onto: We point out that since neither f nor g is onto,there exists some x ∈ X such that the fiber of x under g is empty. Likewise,there exists some y ∈ Y such that the fiber of y under f is empty. Let F{}be the set of all such x ∈ X.

For any x ∈ X such that the fiber of x under g is not empty, eitherg−1[x] = {f(x)} or f(x) ∈ g−1[x], and g−1[x] 6= {f(x)}. In the first case,let F{f(x)} be the set of all x ∈ X such that g−1[x] = {f(x)} and then letFx be the set of all x ∈ X for which f(x) ∈ g−1[x], and g−1[x] 6= {f(x)}.From the section on relations, we have that X is partitioned by F{}, Fx

and F{f(x)}. That is, these sets are pairwise disjoint, and their union isX.

These sets have some interesting properties. Namely g[Y − f [F{}]] =g[f [Fx] ∪ f [F{f(x)}]]. This is from the fact that F{}, Fx, and F{f(x)}partition X and that if g is not mapping from to the set F{}, then it mustmap to the other two. Likewise, g[Y − f [F{f(x)}]] = g[f [Fx] ∪ f [F{}]].

The claim is that the subset A of X, so desired is, F{} ∪ F{f(x)}. We

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compute:

g”

Y − pf [F{}] ∪ f [F{f(x)}]qı

= g”

Y − f [F{}]ı

∪ g”

Y − f [F{f(x)}]ı

= g”

f [Fx] ∪ f [F{f(x)}]ı

∪ g”

f [Fx] ∪ f [F{}]ı

= g”

f [Fx]ı

∪ g”

f [F{}]ı

= g”

f [Fx]ı

∪ ∅ (1)

= g”

f [Fx]ı

= Fx

Where the second line follows form the comments above, and (1) is fromthe construction of F. Indeed, from the comments above, X − (F{} ∪F{f(x)}) = Fx.

Exercise 22.2. We prove the Schroder-Bernstein theorem using the above ex-ercise.

Proof. Suppose that X À Y , and Y À X. Then, there exist some f : X → Ysuch that f is 1-1 and onto a subset Y ′ of Y . Likewise, there exists someg : Y → X which is 1-1 and onto a subset X ′ of X. We note that if eitherX ′ = X or Y ′ = Y , then we are done. So, suppose that X ′ 6= X and Y ′ 6= Y .Likewise, if either X ′ or Y ′ was the empty set, then we are done, since it istrivially true that X ∼ Y .So, suppose that X ′ Ĺ X, such that X ′ 6= ∅, andY ′ Ĺ Y and Y ′ 6= Y .

So, consider f |X−X′ and g|Y−Y ′ . From the previous theorem, there exists asubset X ′′ ⊂ X −X ′ and some Y ′′ ⊂ Y − Y ′ such that f |X−X′ [X ′′] = Y ′′ andg|Y−Y ′ [(Y − Y ′) − Y ′′] = (X −X ′) −X ′′. We claim that ∅ = (X −X ′) −X ′′i.e. X ′′ = X −X ′; if not, then g|Y−Y ′ [(Y − Y ′)− Y ′′] 6= ∅ and so there is someg(y) ∈ X ′ such that y ∈ (Y − Y ′) − Y ′′. But then, we have f(g(y)) ∈ Y ′ 6⊂(Y − Y ′)− Y ′′; this implies that y ∈ Y ′, a contradiction. Thus, X ′′ = X −X ′.

Since this is the case, f |X−X′ [X ′′] ⊂ Y ′. But then, g|Y−Y ′ [(Y − Y ′) −f |X−X′ [X ′′]] = g|Y−Y ′ [Y − Y ′] = (X − X ′) − X ′′ = ∅. This implies thatY − Y ′ = ∅, or in other words, Y = Y ′. A contradiction.

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23 Countable Sets

Exercise 23.1. We prove that the set of all finite subsets of a countable set iscountable.

Proof. The following proof was suggested by this website:Let A(n) be the set of subsets of A with no more than n elements. Thus,

A(0) = {a}, A(1) = A(0) ∪ {{a} : a ∈ A} and for all n ≥ 0:

A(n+ 1) = {a(n) ∪ a(1) : a(n) ∈ A(n) ∧ a(1) ∈ A(1)}

We verify, by induction, that each A(n) is countable. Let A be countable. Then,A(1) is countable, by construction. Suppose A(n) is countable. Then by sincethe union of countable sets is countable, A(n+ 1) is also countable.

Denote with Af as the set of finite subsets of A. It is apparent that everyfinite subset is in some A(n), and so:

Af =⋃n∈N

A(n)

And since the union of countable subsets is countable, Af is countable.

Exercise 23.2. We prove that if every countable subset of a totally ordered setX is well ordered, then X itself is well ordered.

Proof. Let A ⊂ X be an arbitrary countable subset. Then, A is well-ordered,and every subset a of A is well-ordered, so they have least elements. Let

⋃nAn

be the union of all countable subsets of X. Since⋃

nAn is the union of well-ordered sets, it is well-ordered. The claim is that

⋃nAn = X:

Suppose otherwise; Since⋃

nAn ⊂ X, it follows that X =⋃

nAn ∪ (X −⋃nAn) =

⋃nAn ∪ Y . Since

⋃nAn is the union of all countable subsets of X,

Y must be finite; otherwise, we have missed a countable subset. But, since Yis finite, Y is included in some countable subset of X. Thus,

X =⋃n

An ∪ Y =⋃n

An

This contradiction concludes the proof.

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24 Cardinal Arithmetic

Exercise 24.1. We prove that if a, b, c and d are cardinal numbers such thata ≤ b and c ≤ d, then a+ c ≤ b+ d.

Proof. Let a, b, c, d be stated as given. By definition, there exists sets A,B,C,D(without loss of generality, pairwise disjoint) with cardinality a, b, c, d, respec-tively. Using the stated conditions, we have A À B, C À D. Thus, there existsB′ ⊂ B, and D′ ⊂ D such that A ∼ B′ and C ∼ D′ i.e. cardA = cardB′ andcardC = cardD′. Then, a+ b = b′ + d′ ≤ b+ d.

Exercise 24.2. We prove that if a, b, c and d are cardinal numbers such thata ≤ b and c ≤ d, then ac ≤ bd.

Proof. We proceed in a fashion similar to exercise 24.1. Let a, b, c, d be statedas given. By definition, there exists sets A,B,C,D (without loss of generality,pairwise disjoint) with cardinality a, b, c, d, respectively. Using the stated con-ditions, we have A À B, C À D. Thus, there exists B′ ⊂ B, and D′ ⊂ Dsuch that A ∼ B′ and C ∼ D′ i.e. cardA = cardB′ and cardC = cardD′.Thus, ab = card(A × B) = card(B′ × D′). From a previous exercise andthe comments on page 94, we have B′ × D′ À B × D, and so it follows thatcard(B′ ×D′) ≤ card(B ×D).

Exercise 24.3. We prove that if {ai} (i ∈ I) and {bi} (i ∈ I) are families ofcardinal numbers such that ai < bi for each i ∈ I, then

∑i ai ≤

∏i bi.

Proof. The following proof was suggested to me by Sophie C..Note, ∑

i

ai = a0 + a1 + . . . ≤ a0 × a1 × · · · =∏i

ai

We proceed by mathematical induction.

• Initial Case: a0 < b0 — By assumption.

• Hypothesis: Suppose that∏

0≤i≤n ai <∏

0≤i≤n bi

• Step:∏

0≤i≤n ai + an+1 <∏

0≤i≤n bi × bn+1 for an+1 < bn+1. Therefore,∏0≤i≤n+1 ai <

∏0≤i≤n+1 bi

From the above, we have∑

i ai ≤∏

i ai <∏

i bi. Which completes the proof.

Exercise 24.4. We prove that if a, b and c are cardinal numbers such thata ≤ b, then ac ≤ bc.

Proof. The following proof was suggested by Sophie C.. We take the definitionof ac as multiply a by itself c times. And likewise for bc. We proceed bymathematical induction:

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• Initial Case: a1 ≤ b1.

• Inductive Hypothesis: Suppose that∏

i∈I ai ≤∏

i∈I bi such that card I =C.

• Step: If∏

i∈I ai ≤∏

i∈I bi, then∏

i∈I ai × a ≤∏

i∈I bi × b, since a ≤ b.

Thus,∏

i∈I ai ≤∏

i∈I bi where card I = C + 1. And so, ac ≤ bc

Exercise 24.5. Prove that if a and b are finite, greater than 1, and if c isinfinite, then ac = bc.

Proof. The following proof was suggested by Sophie C.. We take the definitionof ac as multiply a by itself c times. And likewise for bc. If c is countable then,ac, bc are countable, and if c is uncountable, then ac, bc are uncountable. Thus,ac = bc

Exercise 24.6. We prove that if a and b are cardinal numbers at least one ofwhich is infinite, then a+ b = ab.

Proof. The following proof was suggest by Sophie C.. Suppose that a, b aregiven as stated. Without loss of generality, there exists sets A,B such thatcardA = a, cardB = b, and A ∼ B′, for some B′ ⊂ B. Thus, a+ b = cardA ∼cardB′ ∪ cardB = card(A ∪ B) ∼ card(B′ ∪ B) = cardB = b = ab. Therefore,a+ b = ab.

Suppose that a and b are both infinite, then as above, we have a+b = a+a =a and ab = aa = a. Therefore, a+ b = ab.

Suppose, without loss of generality, that a, and b are both infinite such thatcardA < cardB. Then, a+ b = b and ab = b, thus, a+ b = ab.

Exercise 24.7. We prove that if a and b are cardinal numbers such that a isinfinite and b is finite, then ab = a.

Proof. The following proof was suggested by Sophie C.. Suppose that a, b aregiven, as above. Then, ab =

∏i∈I ai where card I = b and where ai = a, for all

i ∈ I. Then, we have the following:

ab = card(A×A× . . .×A) where cardA = a

= a× a× . . .× a b times

= a× a× . . .× a b/2 times, since a× a = a

= a× a× . . .× a b/4 times, since a× a = a

...

= a

This completes the proof.

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25 Cardinal Numbers

Exercise 25.1. We prove that each infinite cardinal number is a limit number.

Proof. This proof was inspired from this website.Suppose to the contrary that k is an infinite cardinal which was not a limit

number. Then, k = a+ for some ordinal a+. If a is finite, we are done. Other-wise, we proceed to show that k = a.

Not that since k is infinite, ω À k, and in particular, ω À a. Thus, considerφ : a → k given by φ(0) = a, and φ(m+) = m (if m ∈ ω), and φ(x) = x (ifx ∈ a− ω). We show that φ is a 1-1 and onto function:

• Well-Defined : Since no ordinal is equivalent to its successor, it followsthat φ is well-defined.

• Everywhere Defined : By construction, this is clear.

• 1-1 : If φ(x) = φ(y), then in any of the cases, we have x = y.

• Onto: Let k1 ∈ k. Then, either k1 ∈ a, or k1 = a. In either case, thereexists a ` ∈ a for which φ(`) = a.

Thus k = a, a contradiction.

Exercise 25.2. We answer a few questions: If cardA = a, what is the cardinalnumber of the set of all 1-1 and onto in A? What is the cardinal number of theset of all countably infinite subsets of A?

Proof. If A is infinite, then ω À A and we can consider the infinitely manlymaps fi, such that i ∈ ω, given by fi(x) = x + 1 (if x ∈ ω), and fi(x) = x (ifx ∈ A−ω). However, if A is finite, then a simple counting argument show thatthe number is the multiplication of all predecessors of a, except zero.

This answer was suggested by this website: ”A has aω countably infinitesubsets, and there’s not much more that you can say unless you know somethingabout the cardinal a. For example, 2 ≤ a ≤ 2ω, then aω = 2ω. If cf a = ω, i.e.,if a has cofinality ω, then aω > a.”

Exercise 25.3. We prove that if α, β are ordinal numbers, then card(α+ β) =cardα+ cardβ and cardαβ = cardα cardβ. We use the ordinal interpretationof the operations on the left side and the cardinal interpretation on the right.

Proof. The following proof was suggested by Sophie C.. We compute:

• cardαβ = card(ordA ordB) = card(ord(A×B)) = card(A×B)

• (cardα)(cardβ) = (card(ordA))(card(ordB)) =card(A) card(B) = card(A×B)

Thus, cardαβ = (cardα)(cardβ)

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Proof. The following proof was suggest by Sophie C.. We compute:

• card(α + β) = card(ordA + ordB) = card(ord(A ∪ B)) = ord(A ∪ B) =ord(A) + ord(B) = card(ord(A)) + card(ord(B)) = cardα+ cardβ.

Thus, card(α+ β) = cardα+ cardβ.

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26 Extra Exercises

Many of these exercises were collaborations with other student. They arenot marked exercises in the text, however, they are useful.

26.1 Arithmetic

Exercise 26.1. We show that addition for elements of ω is associative. Thefollowing proof was suggested by Aidan H. and David H., and I attribute allcredit to them.

Proof. We proceed by mathematical induction: By the recursion theorem thereexists a function sm : ω → ω such that sm(0) = m and sm(n+) = (sm(n))+

where m,n ∈ ω and sm(n) = m + n.First note sk+m(n) = (k + m) + n, andsimilarly sk(m + n) = k + (m + n). Consider the set T = {n ∈ ω | sk+m(n) =sk(m+ n)}.

Initial case: See that 0 ∈ T since sk+m(0) = k + m = sk(m + 0) =k + (m+ 0) = k +m.

Inductive hypothesis: Let n ∈ T hold. Then sk+m(n) = sk(m+ n).This implies (k +m) + n = k + (m+ n).Now sk+m(n+) = (k +m) + n+ = (sk+m(n))+ = ((k +m) + n)+.Similarly, sk((m+ n)+) = k + (m+ n)+ = (sk(m+ n))+ = (k + (m+ n))+.

Since sk+m(n) = sk(m+ n) by the inductive hypothesis. We obtain

(k +m) + n+ = ((k +m) + n)+ = (k + (m+ n))+ = k + (m+ n)+

Note that sm(n) = m+ n, so (sm(n))+ = (m+ n)+ = m+ n+ = sm(n+).Thus, (k +m) + n+ = k + (m+ n+)

Conclusion: Therefore, n+ ∈ T , and thus T = ω, and thus addition isassociative.

Exercise 26.2. We prove that addition is commutative.

Proof. The following proof was suggested by Aidan H. and David H., and Iattribute all credit to them.

Let B = {m ∈ ω | sm(n) = sn(m)}.Initial case: 0 ∈ B, since s0(n) = sn(0) = n.Inductive hypothesis: Let b ∈ B hold. That is, sb(n) = b + n = n + b =sn(b). Then sn(b+) = (sn(b))+ = (n + b)+, by definition. Similarly, sb+(n) =(sb(n))+ = (b + n)+. Then, by the inductive hypothesis, sb+(n) = (sb(n))+ =(sn(b))+ = sn(b+).Inductive conclusion: b+ ∈ B. Thus, by the induction principle, B = ω. This

56

implies sm(n) = sn(m) ∀m,n ∈ ω, and thus m + n = n + m ∀m,n ∈ ω, andthus addition is commutative.

Exercise 26.3. We prove the properties of exponentiation.

Proof. The following proof was suggested by Aidan H. and David H., and Iattribute all credit to them.

First note em : ω → ω is a function defined as

em(0) = 1

em(n+) = em(n) ·m where em(n) = mn for m,n ∈ ω

I want to show m(n+l) = mn ·ml, that is em(n+l) = em(n)·em(l) for m,n, l ∈ ω.Define Pm : ω → ω such that Pm(0) = 0 and Pm(n+) = Pm(n) + m, andPm(n) = m · n for m,n ∈ ω. First I will show P1(n) = n ∀n ∈ ω by induction.Consider the set B = {n ∈ ω | P1(n) = n}. Initial case: Note 0 ∈ B sinceP1(0) = 0 by definition.Inductive hypothesis: Let n ∈ B hold, then P1(n) = n. Then P1(n+) = P1(n)+1 = n+ 1 = n+.Inductive conclusion: Thus n+ ∈ B and by principle of mathematical induction,B = ω. Thus, P1(n) = n ∀n ∈ ω.Now to show m(n+l) = mn ·ml. Let F = {n ∈ ω | em(n+ l) = em(n) · em(l)}.Initial case: Note 0 ∈ F since em(0 + l) = em(l) and em(0) · em(l) = 1 · em(l) =P1(em(l)) = em(l).Inductive hypothesis: Let n ∈ F hold, then em(n+ l) = em(n) · em(l). Then

em(n+ + l) = em((n+ l)+)

= em(n+ l) ·m= em(n) · em(l) ·m= em(n) ·m · em(l)

= em(n+) · em(l)

Inductive conclusion: Thus, n+ ∈ F , and by the principle of mathematicalinduction, F = ω. Thus, m(n+l) = mn ·ml ∀m,n, l ∈ ω.

Exercise 26.4. We prove the left distributive law.

Proof. The following proof was suggested by Aidan H. and David H., and Iattribute all credit to them.

We proceed by mathematical induction:Initial case:

k · (m+ 0) = k ·m+ k · 0

k ·m = k ·m+ 0 = k ·m

Inductive hypothesis: k · (mn) = k ·m+ k · n Then

k · (m+ n+) = (k · (m+ n))+ by definition of addition

57

= (k ·m+ k · n)+ = k ·m+ (k · n)+ by hypothesis and addition

= k ·m+ k · n+ by definition of addition

Conclusion: k · (m+ n+) = k ·m+ k · n+

58