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Discrete Mathematics ICS127

Lecturer: Dr. Alex Tiskinhttp://www.dcs.warwick.ac.uk/˜tiskin

Department of Computer Science

University of Warwick

Discrete Mathematics I – p. 1/292

Discrete Mathematics IMathematics relevant to computer scienceUsed in other CS courses

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Discrete Mathematics IClass test in week 7 (new from 2002/03!)

Exam in week 1 of Summer TermResults at the end of Summer Term

Discrete Mathematics ILecture notes available at lectures

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Discrete Mathematics IDiscrete Maths II — a Summer Term optionLecturer: Dr. Mike Joy

Discrete maths in depth, highly recommended!

Discrete Mathematics IRecommended books:

Discrete MathematicsRoss and Wright (Prentice Hall, 2003)

Discrete Mathematics and its ApplicationsRosen (McGraw-Hill, 2003)

Discrete Mathematics for Computer ScientistsTruss (Addison-Wesley, 1999)

Discrete Mathematics IWhich is the best?

• Ross and Wright: the most helpful. . .

• Rosen: the most interesting. . .

• Truss: the most advanced. . .

Hundreds more, the choice is yours!

Discrete Mathematics IAlso:

Proofs and Fundamentals: a First Course in AbstractMathematicsBloch (Birkhäuser, 2002)

Does not cover whole course, but helps with proofs

Discrete Mathematics IAlso:

Proofs and Fundamentals: a First Course in AbstractMathematicsBloch (Birkhäuser, 2002)

Does not cover whole course, but helps with proofs

Introduction

IntroductionMathematics:

the science of abstraction

Natural numbers: 0, 1, 2, 3, 4, 5, 6, 7, . . .What does “5” mean?

Five apples, five hats, five lottery numbers. . .Abstraction of all five-element sets

What set does 0 represent? The empty set

IntroductionMathematics: the science of abstraction

What set does 0 represent?

The empty set

IntroductionA set is any collection of elements

{Peter Pan, Gingerbread Man, Wrestling Fan}

{♠,♥,♣,♦}{0, 1, 2, 3, 4, 5, 6, 7, . . . } = N natural numbers

{0, 2, 4, 6, 8, 10, 12, 14, . . . } = Neven even naturals

{. . . ,−3,−2,−1, 0, 1, 2, 3, . . . } = Z integers

First two sets finite, last three infinite

{. . . ,−3,−2,−1, 0, 1, 2, 3, . . . } = Z integers

{♠,♥,♣,♦}

{0, 1, 2, 3, 4, 5, 6, 7, . . . } = N natural numbers

{. . . ,−3,−2,−1, 0, 1, 2, 3, . . . } = Z integers

IntroductionThe empty set: {} = ∅Plays a role for sets similar to 0 for numbers

IntroductionSome “tough” questions:

Do infinite sets have “sizes”?Yes, but these are beyond natural numbers

Can infinite sets have different sizes?Yes, a huge variety of possible sizes

Is there a set of all sets?No! Not even a set of all (infinite) set sizes

Why? It can be proved!

Do infinite sets have “sizes”?

Yes, but these are beyond natural numbers

Can infinite sets have different sizes?

Yes, a huge variety of possible sizes

Is there a set of all sets?

No! Not even a set of all (infinite) set sizes

It can be proved!

IntroductionNatural sciences are based on evidence

Mathematics is based on proof(but evidence helps to understand proofs)

IntroductionNatural sciences are based on evidence

Mathematics is based on proof(but evidence helps to understand proofs)

IntroductionTo write proofs, we need a special language:

• precise (unambiguous)

• concise (clear and relatively brief)

The “grammar” of this language is logic

IntroductionAll eagles can flySome pigs cannot fly

Therefore, some pigs are not eagles

Proof.

Consider all creatures. If it is an eagle, it can fly.Hence, if it cannot fly, it is not an eagle.

There is a creature that is a pig and cannot fly.By the above statement, it is not an eagle.

Proof.

IntroductionThe same in mathematical notation:

If for all x, eagle(x) ⇒ canfly(x),and for some y, pig(y) ∧ ¬canfly(y),

then for some z, pig(z) ∧ ¬eagle(z)

Proof. Consider all x ∈ Creatures .

By first condition, ¬canfly(x) ⇒ ¬eagle(x).

Take any y such that pig(y) ∧ ¬canfly(y).By the above, we have pig(y) ∧ ¬eagle(y).Take z = y.

IntroductionSome concepts are basic, i.e. need no definitionExamples: set, natural number

All other concepts must be definedExamples: finite set, even number

Some facts are axioms, i.e. need no proofExample: equal sets have the same elements

All other facts must be provedExample: the set of even numbers is infinite

This is the axiomatic method

IntroductionCourse structure:

• Logic

• Sets

• More fun: relations, functions, graphs

Any questions?

IntroductionCourse structure:

• Logic

• Sets

• More fun: relations, functions, graphs

Any questions?

In everyday life, we use all sorts of sentences:

Five is less than ten. Welcome to Tweedy’s farm!Pigs can fly. What’s in the pies?There is life on Mars. It’s not as bad as it looks. . .

A statement is a sentence that is either true or false(but not both!).

Five is less than ten.

Welcome to Tweedy’s farm!

Pigs can fly.

What’s in the pies?

There is life on Mars.

It’s not as bad as it looks. . .

False and true are Boolean values: B = {F, T}

(After G. Boole, 1815–1864)

value(5 < 10) = T

value(“Pigs can fly”) = F

value(“It’s not as bad as it looks”) — ?

value(“The pie is not as bad as it looks”) = F

False and true are Boolean values: B = {F, T}(After G. Boole, 1815–1864)

value(5 < 10) = T

Often need compound statements:

(5 < 10) AND (Pigs can fly)

. . . i.e. T AND F = F — similar e.g. to 3 + 4 = 7

Often need compound statements:

(5 < 10) AND (Pigs can fly)

. . . i.e. T AND F = F — similar e.g. to 3 + 4 = 7

Boolean operators on B:

¬A NOT A negationA ∧B A AND B conjunctionA ∨B A OR B disjunctionA ⇒ B IF A THEN B implicationA ⇔ B A ⇒ B AND B ⇒ A equivalence

Definition: by truth tables

¬A NOT A negation

A ∧B A AND B conjunctionA ∨B A OR B disjunctionA ⇒ B IF A THEN B implicationA ⇔ B A ⇒ B AND B ⇒ A equivalence

¬A NOT A negationA ∧B A AND B conjunction

A ∨B A OR B disjunctionA ⇒ B IF A THEN B implicationA ⇔ B A ⇒ B AND B ⇒ A equivalence

¬A NOT A negationA ∧B A AND B conjunctionA ∨B A OR B disjunction

A ⇒ B IF A THEN B implicationA ⇔ B A ⇒ B AND B ⇒ A equivalence

¬A NOT A negationA ∧B A AND B conjunctionA ∨B A OR B disjunctionA ⇒ B IF A THEN B implication

A ⇔ B A ⇒ B AND B ⇒ A equivalence

Negation (NOT A): ¬A

True if A false, false if A true

Negation (NOT A): ¬A

True if A false, false if A true

Examples:

¬[5 < 10]

⇐⇒ ¬T ⇐⇒ F

¬[Pigs can fly] ⇐⇒ ¬F ⇐⇒ T

Examples:

¬[5 < 10] ⇐⇒ ¬T

⇐⇒ F

Examples:

¬[5 < 10] ⇐⇒ ¬T ⇐⇒ F

Examples:

¬[5 < 10] ⇐⇒ ¬T ⇐⇒ F

¬[Pigs can fly]

⇐⇒ ¬F ⇐⇒ T

Examples:

¬[5 < 10] ⇐⇒ ¬T ⇐⇒ F

¬[Pigs can fly] ⇐⇒ ¬F

⇐⇒ T

Examples:

¬[5 < 10] ⇐⇒ ¬T ⇐⇒ F

Conjunction (A AND B): A ∧B

True if both A and B true

A B A ∧B

Conjunction (A AND B): A ∧B

True if both A and B true

A B A ∧B

Example:

[5 < 10] ∧ [Pigs can fly]

⇐⇒ T ∧ F ⇐⇒ F

Example:

[5 < 10] ∧ [Pigs can fly] ⇐⇒ T ∧ F

⇐⇒ F

Example:

[5 < 10] ∧ [Pigs can fly] ⇐⇒ T ∧ F ⇐⇒ F

Disjunction (A OR B): A ∨B

True if either A or B true (or both)

A B A ∨B

Disjunction (A OR B): A ∨B

True if either A or B true (or both)

A B A ∨B

Example:

[5 < 10] ∨ [Pigs can fly]

⇐⇒ T ∨ F ⇐⇒ T

Example:

[5 < 10] ∨ [Pigs can fly] ⇐⇒ T ∨ F

⇐⇒ T

Example:

[5 < 10] ∨ [Pigs can fly] ⇐⇒ T ∨ F ⇐⇒ T

Implication (IF A THEN B)

In everyday life, often ambiguous:

If the bird is happy, then it sings loud

Happy — definitely singsUnhappy — may or may not sing

Happy — definitely singsUnhappy — ?

may or may not sing

Implication (IF A THEN B): A ⇒ B

True if A false; true if B true; false otherwise

A B A ⇒ B

Everything implies truth; false implies anything

A B A ⇒ B

Examples:[

[5 < 10] ⇒ [Pigs fly]]

⇐⇒ [T ⇒ F ] ⇐⇒ F[

[Pigs fly] ⇒ [5 < 10]]

⇐⇒ [F ⇒ T ] ⇐⇒ T[

[Pigs fly] ⇒ [5 > 10]]

⇐⇒ [F ⇒ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇒ [Pigs fly]]

⇐⇒ [T ⇒ F ]

⇐⇒ F[

[Pigs fly] ⇒ [5 < 10]]

⇐⇒ [F ⇒ T ] ⇐⇒ T[

[Pigs fly] ⇒ [5 > 10]]

⇐⇒ [F ⇒ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇒ [Pigs fly]]

⇐⇒ [T ⇒ F ] ⇐⇒ F

[Pigs fly] ⇒ [5 < 10]]

⇐⇒ [F ⇒ T ] ⇐⇒ T[

[Pigs fly] ⇒ [5 > 10]]

⇐⇒ [F ⇒ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇒ [Pigs fly]]

⇐⇒ [T ⇒ F ] ⇐⇒ F[

[Pigs fly] ⇒ [5 < 10]]

⇐⇒ [F ⇒ T ] ⇐⇒ T[

[Pigs fly] ⇒ [5 > 10]]

⇐⇒ [F ⇒ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇒ [Pigs fly]]

⇐⇒ [T ⇒ F ] ⇐⇒ F[

[Pigs fly] ⇒ [5 < 10]]

⇐⇒ [F ⇒ T ]

⇐⇒ T[

[Pigs fly] ⇒ [5 > 10]]

⇐⇒ [F ⇒ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇒ [Pigs fly]]

⇐⇒ [T ⇒ F ] ⇐⇒ F[

[Pigs fly] ⇒ [5 < 10]]

⇐⇒ [F ⇒ T ] ⇐⇒ T

[Pigs fly] ⇒ [5 > 10]]

⇐⇒ [F ⇒ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇒ [Pigs fly]]

⇐⇒ [T ⇒ F ] ⇐⇒ F[

[Pigs fly] ⇒ [5 < 10]]

⇐⇒ [F ⇒ T ] ⇐⇒ T[

[Pigs fly] ⇒ [5 > 10]]

⇐⇒ [F ⇒ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇒ [Pigs fly]]

⇐⇒ [T ⇒ F ] ⇐⇒ F[

[Pigs fly] ⇒ [5 < 10]]

⇐⇒ [F ⇒ T ] ⇐⇒ T[

[Pigs fly] ⇒ [5 > 10]]

⇐⇒ [F ⇒ F ]

⇐⇒ T

Examples:[

[5 < 10] ⇒ [Pigs fly]]

⇐⇒ [T ⇒ F ] ⇐⇒ F[

[Pigs fly] ⇒ [5 < 10]]

⇐⇒ [F ⇒ T ] ⇐⇒ T[

[Pigs fly] ⇒ [5 > 10]]

⇐⇒ [F ⇒ F ] ⇐⇒ T

Example (by G. Hardy):2 + 2 = 5 =⇒ I am Count Dracula

“Proof”:2 + 2 = 5 =⇒ 4 = 5 =⇒ 5 = 4 =⇒ 2 = 1

Dracula and I are two =⇒Dracula and I are one

“Proof”:2 + 2 = 5 =⇒ 4 = 5 =⇒ 5 = 4 =⇒ 2 = 1

Example: 2 + 2 = 5 =⇒ Grass is green

Proof:2 + 2 = 5 =⇒ 4 = 5 =⇒ 5 = 4 =⇒

4 + 5 = 5 + 4 =⇒ T

T =⇒ Grass is green

Proof:2 + 2 = 5 =⇒ 4 = 5 =⇒ 5 = 4 =⇒

4 + 5 = 5 + 4 =⇒ T

Proof:2 + 2 = 5 =⇒ 4 = 5 =⇒ 5 = 4 =⇒

4 + 5 = 5 + 4 =⇒ T

Implication A ⇒ B can have many disguises:

A implies B B is implied by A

A leads to B B follows from A

A is stronger than B B is weaker than A

A is sufficient for B B is necessary for A

A implies B

B is implied by A

A leads to B

B follows from A

A is stronger than B

B is weaker than A

A is sufficient for B

B is necessary for A

Examples:

For a number to be divisible by 4, it is

necessary

thatit is even

For a triangle to be isosceles, it is sufficient that it isequilateral

Examples:

For a number to be divisible by 4, it is necessary thatit is even

Examples:

For a triangle to be isosceles, it is

sufficient

that it isequilateral

Examples:

Equivalence (A IF AND ONLY IF B): A ⇔ B

True if A and B agree; false otherwise

A B A ⇔ B

IF AND ONLY IF often contracted to IFF

A B A ⇔ B

Examples:[

[5 < 10] ⇔ [Pigs fly]]

⇐⇒ [T ⇔ F ] ⇐⇒ F[

[Pigs fly] ⇔ [5 < 10]]

⇐⇒ [F ⇔ T ] ⇐⇒ F[

[Pigs fly] ⇔ [5 > 10]]

⇐⇒ [F ⇔ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇔ [Pigs fly]]

⇐⇒ [T ⇔ F ]

⇐⇒ F[

[Pigs fly] ⇔ [5 < 10]]

⇐⇒ [F ⇔ T ] ⇐⇒ F[

[Pigs fly] ⇔ [5 > 10]]

⇐⇒ [F ⇔ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇔ [Pigs fly]]

⇐⇒ [T ⇔ F ] ⇐⇒ F

[Pigs fly] ⇔ [5 < 10]]

⇐⇒ [F ⇔ T ] ⇐⇒ F[

[Pigs fly] ⇔ [5 > 10]]

⇐⇒ [F ⇔ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇔ [Pigs fly]]

⇐⇒ [T ⇔ F ] ⇐⇒ F[

[Pigs fly] ⇔ [5 < 10]]

⇐⇒ [F ⇔ T ] ⇐⇒ F[

[Pigs fly] ⇔ [5 > 10]]

⇐⇒ [F ⇔ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇔ [Pigs fly]]

⇐⇒ [T ⇔ F ] ⇐⇒ F[

[Pigs fly] ⇔ [5 < 10]]

⇐⇒ [F ⇔ T ]

⇐⇒ F[

[Pigs fly] ⇔ [5 > 10]]

⇐⇒ [F ⇔ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇔ [Pigs fly]]

⇐⇒ [T ⇔ F ] ⇐⇒ F[

[Pigs fly] ⇔ [5 < 10]]

⇐⇒ [F ⇔ T ] ⇐⇒ F

[Pigs fly] ⇔ [5 > 10]]

⇐⇒ [F ⇔ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇔ [Pigs fly]]

⇐⇒ [T ⇔ F ] ⇐⇒ F[

[Pigs fly] ⇔ [5 < 10]]

⇐⇒ [F ⇔ T ] ⇐⇒ F[

[Pigs fly] ⇔ [5 > 10]]

⇐⇒ [F ⇔ F ] ⇐⇒ T

Examples:[

[5 < 10] ⇔ [Pigs fly]]

⇐⇒ [T ⇔ F ] ⇐⇒ F[

[Pigs fly] ⇔ [5 < 10]]

⇐⇒ [F ⇔ T ] ⇐⇒ F[

[Pigs fly] ⇔ [5 > 10]]

⇐⇒ [F ⇔ F ]

⇐⇒ T

Examples:[

[5 < 10] ⇔ [Pigs fly]]

⇐⇒ [T ⇔ F ] ⇐⇒ F[

[Pigs fly] ⇔ [5 < 10]]

⇐⇒ [F ⇔ T ] ⇐⇒ F[

[Pigs fly] ⇔ [5 > 10]]

⇐⇒ [F ⇔ F ] ⇐⇒ T

Implication and equivalence are often used to statetheorems

Examples:

Axiom. If n is in N, then n + 1 is in N.That is, for all n, [n in N] =⇒ [n + 1 in N]

Theorem. Number n is even iff n + 1 is odd.That is, for all n in N, [n even] ⇐⇒ [n + 1 odd]

Examples:

Axiom. If n is in N, then n + 1 is in N.

That is, for all n, [n in N] =⇒ [n + 1 in N]

Examples:

Theorem. Number n is even iff n + 1 is odd.

That is, for all n in N, [n even] ⇐⇒ [n + 1 odd]

Examples:

More examples (from geometry):

Axiom. If two points are distinct, then there is exactlyone line connecting them.

Theorem. A triangle has two equal sides, if and onlyif it has two equal angles.

More examples (from geometry):

Axiom. If two points are distinct, then there is exactlyone line connecting them.

Theorem. A triangle has two equal sides, if and onlyif it has two equal angles.

Implication and equivalence are used in proofs

Example:

Theorem. If number n is even, then n + 2 is even.

Proof.[n even] ⇒ [n + 1 odd] ⇒ [(n + 1) + 1 even]

Example:

When proving “⇔”, must prove both “⇒” and “⇐”!

Example (a stronger theorem):

Theorem. Number n is even, iff n + 2 is even.

Proof.

“⇒” as before

“⇐”: [n + 2 = (n + 1) + 1 even] ⇒ [n + 1 odd] ⇒[n even]

Proof.

“⇒” as before

Proof.

“⇒” as before

To cut down on brackets, we use priorities

Highest priority: ¬, then ∧, ∨, then ⇒, ⇔

Example:

¬(A ∧B) ⇔ ¬A ∨ ¬B means¬(A ∧B) ⇔ ((¬A) ∨ (¬B))

To cut down on brackets, we use priorities

Highest priority: ¬, then ∧, ∨, then ⇒, ⇔Example:

¬(A ∧B) ⇔ ¬A ∨ ¬B means¬(A ∧B) ⇔ ((¬A) ∨ (¬B))

Truth table completely define logical operators.

Not always convenient:

(¬(T ∧ F ) ∨ ¬(F ⇒ T )) ⇔ (¬¬(F ∨ T ) ∨ F )

— true or false?

((A ∨B) ∧ C) ∨ ((¬A ∧ ¬B) ∨ ¬C)

— true for all A, B, C?

To simplify expressions, will use laws of logic

(¬(T ∧ F ) ∨ ¬(F ⇒ T )) ⇔ (¬¬(F ∨ T ) ∨ F )

— true or false?

((A ∨B) ∧ C) ∨ ((¬A ∧ ¬B) ∨ ¬C)

(¬(T ∧ F ) ∨ ¬(F ⇒ T )) ⇔ (¬¬(F ∨ T ) ∨ F )

— true or false?

((A ∨B) ∧ C) ∨ ((¬A ∧ ¬B) ∨ ¬C)

(¬(T ∧ F ) ∨ ¬(F ⇒ T )) ⇔ (¬¬(F ∨ T ) ∨ F )

— true or false?

((A ∨B) ∧ C) ∨ ((¬A ∧ ¬B) ∨ ¬C)

Laws of Boolean logic (hold for any A, B, C):

¬¬A ⇐⇒ A double negation

A ∧ A ⇐⇒ A ∧ idempotentA ∨ A ⇐⇒ A ∨ idempotent

A ∧B ⇐⇒ B ∧ A ∧ commutativeA ∨B ⇐⇒ B ∨ A ∨ commutative

More laws of Boolean logic:

(A ∧B) ∧ C ⇐⇒ A ∧ (B ∧ C) ∧ associative(A ∨B) ∨ C ⇐⇒ A ∨ (B ∨ C) ∨ associative

A ∧ (B ∨ C) ⇐⇒ (A ∧B) ∨ (A ∧ C)∧ distributes over ∨

A ∨ (B ∧ C) ⇐⇒ (A ∨B) ∧ (A ∨ C)∨ distributes over ∧

Compare a · (b + c) = a · b + a · c,but a + b · c 6= (a + b) · (a + c)

De Morgan’s laws:

¬(A ∧B) ⇐⇒ ¬A ∨ ¬B¬(A ∨B) ⇐⇒ ¬A ∧ ¬B

Thus, A ∧B ⇐⇒ ¬(¬A ∨ ¬B),so ∧ can be expressed via ¬, ∨Alternatively, A ∨B ⇐⇒ ¬(¬A ∧ ¬B),so ∨ can be expressed via ¬, ∧(Cannot remove both ∧, ∨ at the same time!)

De Morgan’s laws:

¬(A ∧B) ⇐⇒ ¬A ∨ ¬B¬(A ∨B) ⇐⇒ ¬A ∧ ¬B

Thus, A ∧B ⇐⇒ ¬(¬A ∨ ¬B),so ∧ can be expressed via ¬, ∨

Alternatively, A ∨B ⇐⇒ ¬(¬A ∧ ¬B),so ∨ can be expressed via ¬, ∧(Cannot remove both ∧, ∨ at the same time!)

De Morgan’s laws:

¬(A ∧B) ⇐⇒ ¬A ∨ ¬B¬(A ∨B) ⇐⇒ ¬A ∧ ¬B

Thus, A ∧B ⇐⇒ ¬(¬A ∨ ¬B),so ∧ can be expressed via ¬, ∨Alternatively, A ∨B ⇐⇒ ¬(¬A ∧ ¬B),so ∨ can be expressed via ¬, ∧

(Cannot remove both ∧, ∨ at the same time!)

De Morgan’s laws:

¬(A ∧B) ⇐⇒ ¬A ∨ ¬B¬(A ∨B) ⇐⇒ ¬A ∧ ¬B

Thus, A ∧B ⇐⇒ ¬(¬A ∨ ¬B),so ∧ can be expressed via ¬, ∨Alternatively, A ∨B ⇐⇒ ¬(¬A ∧ ¬B),so ∨ can be expressed via ¬, ∧(Cannot remove both ∧, ∨ at the same time!)

Still more laws of Boolean logic:

A ∧ T ⇐⇒ A A ∨ F ⇐⇒ A identity laws

A ∧ F ⇐⇒ F A ∨ T ⇐⇒ Tannihilation laws

A ∧ ¬A ⇐⇒ F A ∨ ¬A ⇐⇒ Tlaws of excluded middle

A ∧ (A ∨B) ⇐⇒ A ⇐⇒ A ∨ (A ∧B)absorption laws

Finally,

(A ⇒ B) ⇐⇒ (¬A ∨B) ⇐⇒ ¬(A ∧ ¬B)

(A ⇔ B) ⇐⇒ (A ⇒ B) ∧ (B ⇒ A) ⇐⇒(A ∧B) ∨ (¬A ∧ ¬B)

So, ⇒ and ⇔ are redundant (but convenient)

Finally,

(A ⇒ B) ⇐⇒ (¬A ∨B) ⇐⇒ ¬(A ∧ ¬B)

(A ⇔ B) ⇐⇒ (A ⇒ B) ∧ (B ⇒ A) ⇐⇒(A ∧B) ∨ (¬A ∧ ¬B)

Finally,

(A ⇒ B) ⇐⇒ (¬A ∨B) ⇐⇒ ¬(A ∧ ¬B)

(A ⇔ B) ⇐⇒ (A ⇒ B) ∧ (B ⇒ A) ⇐⇒(A ∧B) ∨ (¬A ∧ ¬B)

All these laws can be verified by truth tables

Example: ¬(A ∧B) ⇐⇒ (¬A ∨ ¬B)

A B A ∧B ¬(A ∧B) ¬A ¬B (¬A ∨ ¬B)

T T T F F F F

T F F T F T T

F T F T T F T

F F F T T T T

Example: ¬(A ∧B) ⇐⇒ (¬A ∨ ¬B)

A B A ∧B ¬(A ∧B) ¬A ¬B (¬A ∨ ¬B)

T T T F F F F

T F F T F T T

F T F T T F T

F F F T T T T

Example: ¬(A ∧B) ⇐⇒ (¬A ∨ ¬B)

A B A ∧B ¬(A ∧B) ¬A ¬B (¬A ∨ ¬B)

T F F F F

F T F T T

F T T F T

F T T T T

Example: ¬(A ∧B) ⇐⇒ (¬A ∨ ¬B)

A B A ∧B ¬(A ∧B) ¬A ¬B (¬A ∨ ¬B)

F F F F

T F T T

T T F T

T T T T

Example: ¬(A ∧B) ⇐⇒ (¬A ∨ ¬B)

A B A ∧B ¬(A ∧B) ¬A ¬B (¬A ∨ ¬B)

T T T F

T F F T

F T F T

F F F T

Example: ¬(A ∧B) ⇐⇒ (¬A ∨ ¬B)

A B A ∧B ¬(A ∧B) ¬A ¬B (¬A ∨ ¬B)

T T T F F F

T F F T F T

F T F T T F

F F F T T T

Example: ¬(A ∧B) ⇐⇒ (¬A ∨ ¬B)

A B A ∧B ¬(A ∧B) ¬A ¬B (¬A ∨ ¬B)

T T T F F F F

T F F T F T T

F T F T T F T

F F F T T T T

Example: ¬(A ∧B) ⇐⇒ (¬A ∨ ¬B)

A B A ∧B ¬(A ∧B) ¬A ¬B (¬A ∨ ¬B)

T T T F F F F

T F F T F T T

F T F T T F T

F F F T T T T

Using laws of logic

Prove: (A ⇒ B) ⇐⇒ (¬B ⇒ ¬A).

Proof.

(¬B ⇒ ¬A) ⇐⇒ (¬¬B ∨ ¬A) ⇐⇒(B ∨ ¬A) ⇐⇒ (¬A ∨B) ⇐⇒ (A ⇒ B)

¬B ⇒ ¬A is the contapositive of A ⇒ B

B ⇒ A is the converse of A ⇒ B

Statement A ⇒ B is equivalent to its contrapositive,but not to its converse

Equivalence with contrapositive allows proof bycontradiction

Using laws of logic

Prove: (A ⇒ B) ⇐⇒ (¬B ⇒ ¬A).

Proof.

(¬B ⇒ ¬A)

⇐⇒ (¬¬B ∨ ¬A) ⇐⇒(B ∨ ¬A) ⇐⇒ (¬A ∨B) ⇐⇒ (A ⇒ B)

Using laws of logic

Prove: (A ⇒ B) ⇐⇒ (¬B ⇒ ¬A).

Proof.

(¬B ⇒ ¬A) ⇐⇒ (¬¬B ∨ ¬A)

⇐⇒(B ∨ ¬A) ⇐⇒ (¬A ∨B) ⇐⇒ (A ⇒ B)

Using laws of logic

Prove: (A ⇒ B) ⇐⇒ (¬B ⇒ ¬A).

Proof.

(¬B ⇒ ¬A) ⇐⇒ (¬¬B ∨ ¬A) ⇐⇒(B ∨ ¬A)

⇐⇒ (¬A ∨B) ⇐⇒ (A ⇒ B)

Using laws of logic

Prove: (A ⇒ B) ⇐⇒ (¬B ⇒ ¬A).

Proof.

(¬B ⇒ ¬A) ⇐⇒ (¬¬B ∨ ¬A) ⇐⇒(B ∨ ¬A) ⇐⇒ (¬A ∨B)

⇐⇒ (A ⇒ B)

Using laws of logic

Prove: (A ⇒ B) ⇐⇒ (¬B ⇒ ¬A).

Proof.

(¬B ⇒ ¬A) ⇐⇒ (¬¬B ∨ ¬A) ⇐⇒(B ∨ ¬A) ⇐⇒ (¬A ∨B) ⇐⇒ (A ⇒ B)

Using laws of logic

Prove: (A ⇒ B) ⇐⇒ (¬B ⇒ ¬A).

Proof.

(¬B ⇒ ¬A) ⇐⇒ (¬¬B ∨ ¬A) ⇐⇒(B ∨ ¬A) ⇐⇒ (¬A ∨B) ⇐⇒ (A ⇒ B)

Using laws of logic

Prove: (A ⇒ B) ⇐⇒ (¬B ⇒ ¬A).

Proof.

(¬B ⇒ ¬A) ⇐⇒ (¬¬B ∨ ¬A) ⇐⇒(B ∨ ¬A) ⇐⇒ (¬A ∨B) ⇐⇒ (A ⇒ B)

Holmes: I see our visitor was absent-minded. . .

Watson: But why!??

Holmes: Elementary, my dear Watson! Alert peoplenever leave things behind. But he left his walkingstick. Therefore, he must be absent-minded.

A = “person is alert”B = “person does not leave things behind”

Holmes’ argument: (A ⇒ B) =⇒ (¬B ⇒ ¬A)

Watson: But why!??

Sign in a restaurant:

Good food is not cheap.

Cheap food is not good.Is it repeating the same thing twice?

Yes! The statements are contrapositive to each other.

Sign in a restaurant:

Good food is not cheap.

Cheap food is not good.Is it repeating the same thing twice?

Yes! The statements are contrapositive to each other.

Somebody walks into a pub and says:

If I drink, everybody drinks!

Can this statement be true?

Answer: yes, it can!

Proof. We know the world is nonempty.

Case 1: Suppose everybody in the world drinks. Thenevery person can say that.

Case 2: Suppose Joe does not drink. Then Joe can saythat.

Somebody walks into a pub and says:

If anybody drinks, I drink!

Can this statement be true?

Case 1: Suppose nobody in the world drinks. Thenevery person can say that.

Case 2: Suppose Jack drinks. Then Jack can saythat.

So far — statements about individual objects:

Five is less than tenThe pie is not as bad as it looks

Often need to say more:

Some natural numbers are less than tenAll pies are not as bad as they look

So far — statements about individual objects:

Five is less than tenThe pie is not as bad as it looks

Often need to say more:

Some natural numbers are less than tenAll pies are not as bad as they look

Some natural numbers are less than ten

Could try to specify an instance:

Five is less than ten

What if we do not have an instance?

Could try to use Boolean operators:

(0 < 10) ∨ (1 < 10) ∨ (2 < 10) ∨ (3 < 10) ∨ · · ·All pies are not as bad as they look

(Chicken pie. . . ) ∧ (Mushroom pie. . . ) ∧(Cabbage pie. . . ) ∧ · · ·Cannot have infinite conjunction/disjunction!

(0 < 10) ∨ (1 < 10) ∨ (2 < 10) ∨ (3 < 10) ∨ · · ·

All pies are not as bad as they look

(Chicken pie. . . ) ∧ (Mushroom pie. . . ) ∧(Cabbage pie. . . ) ∧ · · ·

Cannot have infinite conjunction/disjunction!

A predicate is a sentence with variables

Becomes true or false when values are substituted forvariables

Values are taken from a particular set (the range)

Always assume range is nonempty

A predicate is a sentence with variables

Becomes true or false when values are substituted forvariables

Values are taken from a particular set (the range)

Always assume range is nonempty

Examples:

x < 10 (x in N)

“Pie p is not as bad as it looks” (p in Pies)

Can be true or false, depending on x, p

Examples:

x < 10 (x in N)

Examples:

x < 10 (x in N)

A predicate can have more than one variable

Examples:

x < y (x, y in N)

“Pie p is better than pie q” (p, q in Pies)

Examples:

x < y (x, y in N)

Examples:

x < y (x, y in N)

Predicate with no variables: ordinary statement

Examples:

5 < 10

“My pie is better than your pie”

Examples:

5 < 10

Examples:

5 < 10

Let P (x) be a predicate with variable x

Can make statements by quantifiers

Existential (FOR SOME x, P (x)): ∃x : P (x)

Universal (FOR ALL x, P (x)): ∀x : P (x)

A particular range of x always assumed

Range often made explicit

Examples:

∃x ∈ N : x < 10

∀p ∈ Pies : “p is not as bad as it looks”

∀x ∈ N : ∃y ∈ N : x < y

∃y ∈ N : ∀x ∈ N : x < y — note the difference!

Examples:

∃x ∈ N : x < 10

∀x ∈ N : ∃y ∈ N : x < y

Examples:

∃x ∈ N : x < 10

∀x ∈ N : ∃y ∈ N : x < y

Examples:

∃x ∈ N : x < 10

∀x ∈ N : ∃y ∈ N : x < y

Examples:

∃x ∈ N : x < 10

∀x ∈ N : ∃y ∈ N : x < y

Quantifier variable can be changed

∀p ∈ Pies : “p is not as bad as it looks” ⇐⇒∀π ∈ Pies : “π is not as bad as it looks”

Variable under a quantifier bound, otherwise free

Example:

∃y ∈ N : x > y x free, y bound

Truth value depends on x, but not on y

∀p ∈ Pies : “p is not as bad as it looks” ⇐⇒

∀π ∈ Pies : “π is not as bad as it looks”

Example:

P (x, y) ⇐⇒ x > y x, y free

P (u, v) ⇐⇒ u > v u, v free

Q1(x) ⇐⇒ P (x, 5) ⇐⇒ x > 5 x free

Q2(z) ⇐⇒ P (3, z) ⇐⇒ 3 > z z free

Q3(y) ⇐⇒ ∀x : P (x, y) ⇐⇒ ∀u : P (u, y)⇐⇒ ∀u : u > y y free x, u bound

Q4(v) ⇐⇒ ∃y : P (v, y) ⇐⇒ ∃w : P (v, w)⇐⇒ ∃w : v > w v free y, w bound

Q1(x) ⇐⇒ P (x, 5) ⇐⇒ x > 5 x free

Q2(z) ⇐⇒ P (3, z) ⇐⇒ 3 > z z free

Q1(x) ⇐⇒ P (x, 5) ⇐⇒ x > 5 x free

Q2(z) ⇐⇒ P (3, z) ⇐⇒ 3 > z z free

Q1(x) ⇐⇒ P (x, 5) ⇐⇒ x > 5 x free

Q2(z) ⇐⇒ P (3, z) ⇐⇒ 3 > z z free

Q1(x) ⇐⇒ P (x, 5) ⇐⇒ x > 5 x free

Q2(z) ⇐⇒ P (3, z) ⇐⇒ 3 > z z free

Q1(x) ⇐⇒ P (x, 5) ⇐⇒ x > 5 x free

Q2(z) ⇐⇒ P (3, z) ⇐⇒ 3 > z z free

Q5 ⇐⇒ ∃z : P (0, z)⇐⇒ ∃z : 0 > z ⇐⇒ F z bound

Q6 ⇐⇒ ∀y : ∃x : P (x, y)⇐⇒ ∀y : ∃x : x > y ⇐⇒ T x, y bound

Q7 ⇐⇒ P (3, 5) ⇐⇒ 3 > 5 ⇐⇒ F

Q5 ⇐⇒ ∃z : P (0, z)⇐⇒ ∃z : 0 > z ⇐⇒ F z bound

Q7 ⇐⇒ P (3, 5) ⇐⇒ 3 > 5 ⇐⇒ F

Q5 ⇐⇒ ∃z : P (0, z)⇐⇒ ∃z : 0 > z ⇐⇒ F z bound

Q7 ⇐⇒ P (3, 5) ⇐⇒ 3 > 5 ⇐⇒ F

Q5 ⇐⇒ ∃z : P (0, z)⇐⇒ ∃z : 0 > z ⇐⇒ F z bound

Q7 ⇐⇒ P (3, 5) ⇐⇒ 3 > 5 ⇐⇒ F

Suppose set S finite: S = {a1, . . . , an}

∀x ∈ S : P (x) ⇐⇒ P (a1) ∧ · · · ∧ P (an)

∃x ∈ S : P (x) ⇐⇒ P (a1) ∨ · · · ∨ P (an)

On a finite range, quantifiers can be expressed byBoolean operators

Not so on an infinite range

Suppose set S finite: S = {a1, . . . , an}∀x ∈ S : P (x) ⇐⇒ P (a1) ∧ · · · ∧ P (an)

∃x ∈ S : P (x) ⇐⇒ P (a1) ∨ · · · ∨ P (an)

Laws of predicate logic:

(∀x : T ) ⇐⇒ T (∀x : F ) ⇐⇒ F

(∃x : T ) ⇐⇒ T (∃x : F ) ⇐⇒ F

∀x : (P (x) ∧Q) ⇐⇒ (∀x : P (x)) ∧Q

(if Q does not contain free x)

Holds for ∀, ∃, and for each of ¬, ∧, ∨, ⇒, ⇔

(∀x : T ) ⇐⇒ T (∀x : F ) ⇐⇒ F

(∃x : T ) ⇐⇒ T (∃x : F ) ⇐⇒ F

∀x : (P (x) ∧Q) ⇐⇒ (∀x : P (x)) ∧Q

(∀x : T ) ⇐⇒ T (∀x : F ) ⇐⇒ F

(∃x : T ) ⇐⇒ T (∃x : F ) ⇐⇒ F

∀x : (P (x) ∧Q) ⇐⇒ (∀x : P (x)) ∧Q

(∀x : T ) ⇐⇒ T (∀x : F ) ⇐⇒ F

(∃x : T ) ⇐⇒ T (∃x : F ) ⇐⇒ F

∀x : (P (x) ∧Q) ⇐⇒ (∀x : P (x)) ∧Q

(∀x : T ) ⇐⇒ T (∀x : F ) ⇐⇒ F

(∃x : T ) ⇐⇒ T (∃x : F ) ⇐⇒ F

∀x : (P (x) ∧Q) ⇐⇒ (∀x : P (x)) ∧Q

(∀x : T ) ⇐⇒ T (∀x : F ) ⇐⇒ F

(∃x : T ) ⇐⇒ T (∃x : F ) ⇐⇒ F

∀x : (P (x) ∧Q) ⇐⇒ (∀x : P (x)) ∧Q

De Morgan’s laws for predicates:

¬∀x : P (x) ⇐⇒ ∃x : ¬P (x)

¬∃x : P (x) ⇐⇒ ∀x : ¬P (x)

Quantifiers — handle with care!

∀x : (P (x) ∧Q(x)) ⇐⇒ (∀x : P (x)) ∧ (∀x : Q(x))

∃x : (P (x) ∨Q(x)) ⇐⇒ (∃x : P (x)) ∨ (∃x : Q(x))

∀x : (P (x) ∨Q(x)) 6⇒ (∀x : P (x)) ∨ (∀x : Q(x))(“⇐” still holds)

∃x : (P (x) ∧Q(x)) 6⇐ (∃x : P (x)) ∧ (∃x : Q(x))(“⇒” still holds)

∀x : (P (x) ∧Q(x)) ⇐⇒ (∀x : P (x)) ∧ (∀x : Q(x))

∃x : (P (x) ∨Q(x)) ⇐⇒ (∃x : P (x)) ∨ (∃x : Q(x))

∀x : (P (x) ∨Q(x)) 6⇒ (∀x : P (x)) ∨ (∀x : Q(x))

(“⇐” still holds)

∀x : (P (x) ∧Q(x)) ⇐⇒ (∀x : P (x)) ∧ (∀x : Q(x))

∃x : (P (x) ∨Q(x)) ⇐⇒ (∃x : P (x)) ∨ (∃x : Q(x))

∀x : (P (x) ∧Q(x)) ⇐⇒ (∀x : P (x)) ∧ (∀x : Q(x))

∃x : (P (x) ∨Q(x)) ⇐⇒ (∃x : P (x)) ∨ (∃x : Q(x))

∃x : (P (x) ∧Q(x)) 6⇐ (∃x : P (x)) ∧ (∃x : Q(x))

(“⇒” still holds)

∀x : (P (x) ∧Q(x)) ⇐⇒ (∀x : P (x)) ∧ (∀x : Q(x))

∃x : (P (x) ∨Q(x)) ⇐⇒ (∃x : P (x)) ∨ (∃x : Q(x))

Using quantifiers — an example

P (x) true for at least one x in S: ∃x ∈ S : P (x)

P (x) true for exactly one x in S:

∃x ∈ S : (P (x) ∧ ∀y ∈ S : P (y) ⇒ (x = y)) ⇐⇒∃x ∈ S : (P (x)∧∀y ∈ S : (x 6= y) ⇒ ¬P (y)) ⇐⇒∃x ∈ S : (P (x) ∧ ∀y ∈ S : (x = y) ∨ ¬P (y)) ⇐⇒∃x ∈ S : (P (x) ∧ ¬∃y ∈ S : (x 6= y) ∧ P (y))

Notation: ∃!x ∈ S : P (x)

Exercise: P (x) true for all but one x in S

∃x ∈ S : (P (x) ∧

∀y ∈ S : P (y) ⇒ (x = y)) ⇐⇒∃x ∈ S : (P (x)∧∀y ∈ S : (x 6= y) ⇒ ¬P (y)) ⇐⇒∃x ∈ S : (P (x) ∧ ∀y ∈ S : (x = y) ∨ ¬P (y)) ⇐⇒∃x ∈ S : (P (x) ∧ ¬∃y ∈ S : (x 6= y) ∧ P (y))

∃x ∈ S : (P (x) ∧ ∀y ∈ S : P (y) ⇒ (x = y))

⇐⇒∃x ∈ S : (P (x)∧∀y ∈ S : (x 6= y) ⇒ ¬P (y)) ⇐⇒∃x ∈ S : (P (x) ∧ ∀y ∈ S : (x = y) ∨ ¬P (y)) ⇐⇒∃x ∈ S : (P (x) ∧ ¬∃y ∈ S : (x 6= y) ∧ P (y))

∃x ∈ S : (P (x) ∧ ∀y ∈ S : P (y) ⇒ (x = y)) ⇐⇒∃x ∈ S : (P (x)∧∀y ∈ S : (x 6= y) ⇒ ¬P (y))

⇐⇒∃x ∈ S : (P (x) ∧ ∀y ∈ S : (x = y) ∨ ¬P (y)) ⇐⇒∃x ∈ S : (P (x) ∧ ¬∃y ∈ S : (x 6= y) ∧ P (y))

∃x ∈ S : (P (x) ∧ ∀y ∈ S : P (y) ⇒ (x = y)) ⇐⇒∃x ∈ S : (P (x)∧∀y ∈ S : (x 6= y) ⇒ ¬P (y)) ⇐⇒∃x ∈ S : (P (x) ∧ ∀y ∈ S : (x = y) ∨ ¬P (y))

⇐⇒∃x ∈ S : (P (x) ∧ ¬∃y ∈ S : (x 6= y) ∧ P (y))

SetsSet: a basic (undefined) concept

By a set we shall understand any collectioninto a whole M of definite, distinct objects ofour intuition or of our thought. These objectsare called the elements of M .

G. Cantor (1845–1918)

SetsSet: a basic (undefined) concept

By a set we shall understand any collectioninto a whole M of definite, distinct objects ofour intuition or of our thought. These objectsare called the elements of M .

G. Cantor (1845–1918)

SetsAnything can be an element of a set

Planets = {Mercury, Venus, . . . , Pluto}Neven = {0, 2, 4, 6, 8, 10, . . .}Junk = {239, banana, ace of spades}A set can be an element of another set

SuperJunk = {239, Junk , ∅} ={239, {banana, ace of spades, 239}, ∅}

Planets = {Mercury, Venus, . . . , Pluto}

Neven = {0, 2, 4, 6, 8, 10, . . .}Junk = {239, banana, ace of spades}A set can be an element of another set

Planets = {Mercury, Venus, . . . , Pluto}Neven = {0, 2, 4, 6, 8, 10, . . .}

Junk = {239, banana, ace of spades}A set can be an element of another set

Planets = {Mercury, Venus, . . . , Pluto}Neven = {0, 2, 4, 6, 8, 10, . . .}Junk = {239, banana, ace of spades}

A set can be an element of another set

SetsOrder of elements does not matter

Junk = {banana, ace of spades, 239}

Repetition of elements does not matter

Junk ={banana, banana, ace of spades, 239, 239, 239}

SetsOrder of elements does not matter

Junk = {banana, ace of spades, 239}Repetition of elements does not matter

Junk ={banana, banana, ace of spades, 239, 239, 239}

SetsThe empty set: ∅ = {}

A singleton: any one-element set

MorningStars = {Venus}NonpositiveNaturals = {0}EmptySets = {∅} 6= ∅

SetsThe empty set: ∅ = {}A singleton: any one-element set

MorningStars = {Venus}

NonpositiveNaturals = {0}EmptySets = {∅} 6= ∅

MorningStars = {Venus}NonpositiveNaturals = {0}

EmptySets = {∅} 6= ∅

SetsElement x is in set A: x ∈ A

Jupiter ∈ Planets , orange 6∈ Junk

Set A is a subset of set B, if all elements of A are alsoelements of B (but not necessarily the other wayround)

A ⊆ B ⇐⇒ ∀x : x ∈ A ⇒ x ∈ B

SetsIn particular, for any set A: A ⊆ A ∅ ⊆ A

∅ ⊆ ∅Neven ⊆ N

∅ ⊆ {banana} ⊆ {banana, 239} ⊆ Junk

∅ ⊆ ∅

Neven ⊆ N

SetsWhat are the axioms of set theory?

The Law of Extensionality:

Two sets with all the same elements are equal

For any sets A, B: (A ⊆ B) ∧ (B ⊆ A) ⇒ A = B

In particular, there is only one empty set

SetsLet P (x) be a predicate

{x | P (x)}: the set of all x, such that P (x) is true

Range often made explicit: {x ∈ S | P (x)}In particular: {x ∈ S | T} = S {x ∈ S | F} = ∅

Range often made explicit: {x ∈ S | P (x)}

In particular: {x ∈ S | T} = S {x ∈ S | F} = ∅

Range often made explicit: {x ∈ S | P (x)}In particular: {x ∈ S | T} =

S {x ∈ S | F} = ∅

Range often made explicit: {x ∈ S | P (x)}In particular: {x ∈ S | T} = S

{x ∈ S | F} = ∅

Range often made explicit: {x ∈ S | P (x)}In particular: {x ∈ S | T} = S {x ∈ S | F} =

SetsExamples:

{x ∈ N | x > 0} = {1, 2, 3, 4, 5, 6, . . .}

{x ∈ Planets | x is red} = {Mars}{x ∈ N | x ≥ 0} = N

{x ∈ Planets | x is a banana} = ∅

SetsExamples:

{x ∈ N | x > 0} = {1, 2, 3, 4, 5, 6, . . .}{x ∈ Planets | x is red} = {Mars}

{x ∈ N | x ≥ 0} = N

SetsExamples:

{x ∈ N | x > 0} = {1, 2, 3, 4, 5, 6, . . .}{x ∈ Planets | x is red} = {Mars}{x ∈ N | x ≥ 0} = N

SetsExamples:

{x ∈ N | x > 0} = {1, 2, 3, 4, 5, 6, . . .}{x ∈ Planets | x is red} = {Mars}{x ∈ N | x ≥ 0} = N

SetsAnother axiom of set theory

The Law of Abstraction:

For any predicate P (x), there is a set {x | P (x)}

For any predicate P (x), there is a set {x | P (x)}Extensionality + Abstraction = Set Theory

For any predicate P (x), there is a set {x | P (x)}Extensionality + Abstraction = CONTRADICTION

SetsRussell’s paradox

A barber shaves everyone who does not shavehimself. Who shaves the barber?

Let P (x) = x 6∈ x

Let B = {x | P (x)} = {x | x 6∈ x}B ∈ B — true or false?

B ∈ B =⇒ B 6∈ B B 6∈ B =⇒ B ∈ B

Contradiction!

Let P (x) = x 6∈ x

B ∈ B =⇒ B 6∈ B B 6∈ B =⇒ B ∈ B

Contradiction!

Let P (x) = x 6∈ x

B ∈ B =⇒ B 6∈ B B 6∈ B =⇒ B ∈ B

Contradiction!

Let P (x) = x 6∈ x

Let B = {x | P (x)} = {x | x 6∈ x}

B ∈ B — true or false?

B ∈ B =⇒ B 6∈ B B 6∈ B =⇒ B ∈ B

Contradiction!

Let P (x) = x 6∈ x

B ∈ B =⇒ B 6∈ B B 6∈ B =⇒ B ∈ B

Contradiction!

Let P (x) = x 6∈ x

B ∈ B =⇒ B 6∈ B

B 6∈ B =⇒ B ∈ B

Contradiction!

Let P (x) = x 6∈ x

B ∈ B =⇒ B 6∈ B B 6∈ B =⇒ B ∈ B

Contradiction!

Let P (x) = x 6∈ x

B ∈ B =⇒ B 6∈ B B 6∈ B =⇒ B ∈ B

Contradiction!

SetsRussell’s paradox can only be explained byinconsistency of axioms

Set theory can be fixed — no details here

Extensionality + Abstraction =Naive set theory

Extensionality + Abstraction =

Naive set theory

Extensionality + Abstraction = Naive set theory

SetsOperations on sets:

Intersection: A ∩B = {x | (x ∈ A) ∧ (x ∈ B)}Union: A ∪B = {x | (x ∈ A) ∨ (x ∈ B)}Venn diagrams (illustration only!):

A B A ∩B A ∪B

Sets A, B are called disjoint, if A ∩B = ∅

Intersection: A ∩B = {x | (x ∈ A) ∧ (x ∈ B)}

Union: A ∪B = {x | (x ∈ A) ∨ (x ∈ B)}Venn diagrams (illustration only!):

A B A ∩B A ∪B

Intersection: A ∩B = {x | (x ∈ A) ∧ (x ∈ B)}Union: A ∪B = {x | (x ∈ A) ∨ (x ∈ B)}

Venn diagrams (illustration only!):

A B A ∩B A ∪B

A ∩B A ∪B

A B A ∩B

A ∪B

A B A ∩B A ∪B

SetsMore operations on sets:

Difference: A \B = {x | (x ∈ A) ∧ (x 6∈ B)}

A B A \B B \ A

If A, B disjoint, then A \B = A, B \ A = B

A \B B \ A

A B A \B

A B A \B B \ A

SetsLet S be a fixed (universal) set, A ⊆ S

Complement of A (with respect to S): A = S \ A

SetsExamples:

A = {a, b, c}, B = {c, a, f, g}

A ∪B = {a, b, c, f, g} A ∩B = {a, c}A \B = {b} B \ A = {f, g}Note: A ∪B = (A ∩B) ∪ (A \B) ∪ (B \ A)

Also: (A ∪B) \ (A ∩B) = (A \B) ∪ (B \ A)

SetsExamples:

A = {a, b, c}, B = {c, a, f, g}A ∪B =

{a, b, c, f, g} A ∩B = {a, c}A \B = {b} B \ A = {f, g}Note: A ∪B = (A ∩B) ∪ (A \B) ∪ (B \ A)

Also: (A ∪B) \ (A ∩B) = (A \B) ∪ (B \ A)

SetsExamples:

A = {a, b, c}, B = {c, a, f, g}A ∪B = {a, b, c, f, g}

A ∩B = {a, c}A \B = {b} B \ A = {f, g}Note: A ∪B = (A ∩B) ∪ (A \B) ∪ (B \ A)

Also: (A ∪B) \ (A ∩B) = (A \B) ∪ (B \ A)

SetsExamples:

A = {a, b, c}, B = {c, a, f, g}A ∪B = {a, b, c, f, g} A ∩B =

{a, c}A \B = {b} B \ A = {f, g}Note: A ∪B = (A ∩B) ∪ (A \B) ∪ (B \ A)

Also: (A ∪B) \ (A ∩B) = (A \B) ∪ (B \ A)

SetsExamples:

A = {a, b, c}, B = {c, a, f, g}A ∪B = {a, b, c, f, g} A ∩B = {a, c}

A \B = {b} B \ A = {f, g}Note: A ∪B = (A ∩B) ∪ (A \B) ∪ (B \ A)

Also: (A ∪B) \ (A ∩B) = (A \B) ∪ (B \ A)

SetsExamples:

A = {a, b, c}, B = {c, a, f, g}A ∪B = {a, b, c, f, g} A ∩B = {a, c}A \B =

{b} B \ A = {f, g}Note: A ∪B = (A ∩B) ∪ (A \B) ∪ (B \ A)

Also: (A ∪B) \ (A ∩B) = (A \B) ∪ (B \ A)

SetsExamples:

A = {a, b, c}, B = {c, a, f, g}A ∪B = {a, b, c, f, g} A ∩B = {a, c}A \B = {b}

B \ A = {f, g}Note: A ∪B = (A ∩B) ∪ (A \B) ∪ (B \ A)

Also: (A ∪B) \ (A ∩B) = (A \B) ∪ (B \ A)

SetsExamples:

A = {a, b, c}, B = {c, a, f, g}A ∪B = {a, b, c, f, g} A ∩B = {a, c}A \B = {b} B \ A =

{f, g}Note: A ∪B = (A ∩B) ∪ (A \B) ∪ (B \ A)

Also: (A ∪B) \ (A ∩B) = (A \B) ∪ (B \ A)

SetsExamples:

A = {a, b, c}, B = {c, a, f, g}A ∪B = {a, b, c, f, g} A ∩B = {a, c}A \B = {b} B \ A = {f, g}

Note: A ∪B = (A ∩B) ∪ (A \B) ∪ (B \ A)

Also: (A ∪B) \ (A ∩B) = (A \B) ∪ (B \ A)

SetsExamples:

A = {a, b, c}, B = {c, a, f, g}A ∪B = {a, b, c, f, g} A ∩B = {a, c}A \B = {b} B \ A = {f, g}Note: A ∪B = (A ∩B) ∪ (A \B) ∪ (B \ A)

Also: (A ∪B) \ (A ∩B) = (A \B) ∪ (B \ A)

SetsExamples:

A = {a, b, c}, B = {c, a, f, g}A ∪B = {a, b, c, f, g} A ∩B = {a, c}A \B = {b} B \ A = {f, g}Note: A ∪B = (A ∩B) ∪ (A \B) ∪ (B \ A)

Also: (A ∪B) \ (A ∩B) = (A \B) ∪ (B \ A)

SetsLaws of set operations (hold for any A, B, C):

¯A = A double complement

A ∩ A = A ∩ idempotentA ∪ A = A ∪ idempotent

A ∩B = B ∩ A ∩ commutativeA ∪B = B ∪ A ∪ commutative

SetsMore laws of set operations:

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

A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)∩ distributes over ∪

A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C)∪ distributes over ∩

Compare with arithmetic and Boolean logic

SetsA ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

C B ∪ C A ∩ (B ∪ C)

A ∩B A ∩ C (A ∩B) ∪ (A ∩ C)

SetsA ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

B ∪ C A ∩ (B ∪ C)

A ∩B A ∩ C (A ∩B) ∪ (A ∩ C)

SetsA ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

C B ∪ C

A ∩ (B ∪ C)

A ∩B A ∩ C (A ∩B) ∪ (A ∩ C)

SetsA ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

C B ∪ C A ∩ (B ∪ C)

A ∩B A ∩ C (A ∩B) ∪ (A ∩ C)

SetsA ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

C B ∪ C A ∩ (B ∪ C)

A ∩B

A ∩ C (A ∩B) ∪ (A ∩ C)

SetsA ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

C B ∪ C A ∩ (B ∪ C)

A ∩B A ∩ C

(A ∩B) ∪ (A ∩ C)

SetsA ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

C B ∪ C A ∩ (B ∪ C)

A ∩B A ∩ C (A ∩B) ∪ (A ∩ C)

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

Proof. Consider any x.

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

Hence A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

x ∈ A∩(B∪C) ⇐⇒

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

Hence A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

x ∈ A∩(B∪C) ⇐⇒ (x ∈ A)∧(x ∈ (B∪C)) ⇐⇒

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

Hence A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

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

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

Hence A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

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

(x ∈ A∩B)∨(x ∈ A∩C) ⇐⇒ x ∈ (A∩B)∪(A∩C)

Hence A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

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

x ∈ (A∩B)∪(A∩C)

Hence A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

SetsDe Morgan’s laws:

A ∩B = A ∪ B A ∪B = A ∩ B

Thus, A ∩B = A ∪ B,so ∩ can be expressed via , ∪Alternatively, A ∪B = A ∩ B,so ∪ can be expressed via , ∩(Cannot remove both ∩, ∪ at the same time!)

A ∩B = A ∪ B A ∪B = A ∩ B

Thus, A ∩B = A ∪ B,so ∩ can be expressed via , ∪

Alternatively, A ∪B = A ∩ B,so ∪ can be expressed via , ∩(Cannot remove both ∩, ∪ at the same time!)

A ∩B = A ∪ B A ∪B = A ∩ B

Thus, A ∩B = A ∪ B,so ∩ can be expressed via , ∪Alternatively, A ∪B = A ∩ B,so ∪ can be expressed via , ∩

(Cannot remove both ∩, ∪ at the same time!)

A ∩B = A ∪ B A ∪B = A ∩ B

Thus, A ∩B = A ∪ B,so ∩ can be expressed via , ∪Alternatively, A ∪B = A ∩ B,so ∪ can be expressed via , ∩(Cannot remove both ∩, ∪ at the same time!)

A ∩B = A ∪ B

A ∩B A ∩B

A B A ∪ B

A ∩B = A ∪ B

A ∩B A ∩B

A B A ∪ B

A ∩B = A ∪ B

A ∩B

A B A ∪ B

A ∩B = A ∪ B

A ∩B A ∩B

A B A ∪ B

A ∩B = A ∪ B

A ∩B A ∩B

B A ∪ B

A ∩B = A ∪ B

A ∩B A ∩B

A ∪ B

A ∩B = A ∪ B

A ∩B A ∩B

A B A ∪ B

Prove A ∩B = A ∪ B

Proof. Consider any x ∈ S.

x ∈ A ∩B ⇐⇒ x 6∈ A ∩B ⇐⇒¬(x ∈ A ∧ x ∈ B) ⇐⇒ (x 6∈ A) ∨ (x 6∈ B) ⇐⇒(x ∈ A) ∨ (x ∈ B) ⇐⇒ x ∈ (A ∪ B)

Hence A ∩B = A ∪ B

x ∈ A ∩B ⇐⇒

x 6∈ A ∩B ⇐⇒¬(x ∈ A ∧ x ∈ B) ⇐⇒ (x 6∈ A) ∨ (x 6∈ B) ⇐⇒(x ∈ A) ∨ (x ∈ B) ⇐⇒ x ∈ (A ∪ B)

x ∈ A ∩B ⇐⇒ x 6∈ A ∩B ⇐⇒

¬(x ∈ A ∧ x ∈ B) ⇐⇒ (x 6∈ A) ∨ (x 6∈ B) ⇐⇒(x ∈ A) ∨ (x ∈ B) ⇐⇒ x ∈ (A ∪ B)

x ∈ A ∩B ⇐⇒ x 6∈ A ∩B ⇐⇒¬(x ∈ A ∧ x ∈ B) ⇐⇒

(x 6∈ A) ∨ (x 6∈ B) ⇐⇒(x ∈ A) ∨ (x ∈ B) ⇐⇒ x ∈ (A ∪ B)

x ∈ A ∩B ⇐⇒ x 6∈ A ∩B ⇐⇒¬(x ∈ A ∧ x ∈ B) ⇐⇒ (x 6∈ A) ∨ (x 6∈ B) ⇐⇒

(x ∈ A) ∨ (x ∈ B) ⇐⇒ x ∈ (A ∪ B)

x ∈ A ∩B ⇐⇒ x 6∈ A ∩B ⇐⇒¬(x ∈ A ∧ x ∈ B) ⇐⇒ (x 6∈ A) ∨ (x 6∈ B) ⇐⇒(x ∈ A) ∨ (x ∈ B) ⇐⇒

x ∈ (A ∪ B)

SetsStill more laws:

Let A ⊆ S

A ∩ S = A A ∪ ∅ = A identity laws

A ∩ ∅ = ∅ A ∪ S = S annihilation laws

A ∩ A = ∅ A ∪ A = Slaws of excluded middle

A ∩ (A ∪B) = A = A ∪ (A ∩B)absorption laws

Let A ⊆ S

SetsA structure with such properties is called a Booleanalgebra

Examples:

B = {F, T}operations ∧, ∨, ¬ identities F , T

Set of all subsets of fixed S

operations ∩, ∪,¯ identities ∅, S

Examples:

SetsThe powerset of S is the set of all subsets of S

P(S) = {A | A ⊆ S}A ∈ P(S) ⇐⇒ A ⊆ S

SetsExamples:

P(∅) =

{∅} (note: P(∅) 6= ∅!)

P({Bunty}) = {∅, {Bunty}}P({a, b, c}) =

{∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

SetsExamples:

P(∅) = {∅}

(note: P(∅) 6= ∅!)

{∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

SetsExamples:

P(∅) = {∅} (note: P(∅) 6= ∅!)

{∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

SetsExamples:

P(∅) = {∅} (note: P(∅) 6= ∅!)

P({Bunty}) =

{∅, {Bunty}}P({a, b, c}) =

{∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

SetsExamples:

P(∅) = {∅} (note: P(∅) 6= ∅!)

P({Bunty}) = {∅, {Bunty}}

P({a, b, c}) ={∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

SetsExamples:

P(∅) = {∅} (note: P(∅) 6= ∅!)

{∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

SetsExamples:

P(∅) = {∅} (note: P(∅) 6= ∅!)

{∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

SetsIf S finite, P(S) finite

If S has n elements, P(S) has 2n elements

If S infinite, P(S) infinite

(Sometimes P(S) denoted 2S , even if S infinite)

SetsProperties of P:

P(A ∩B) = P(A) ∩ P(B)

In general, P(A ∪B) 6= P(A) ∪ P(B)(but ⊇ holds)

In general, P(A \B) 6= P(A) \ P(B)(but ⊆ holds)

P(A ∩B) = P(A) ∩ P(B)

SetsProve or disprove: For all A, B,P(A) ∪ P(B) ⊆ P(A ∪B).

Proof. Consider any X .

X ∈ P(A) ∪ P(B) ⇐⇒(X ∈ P(A)) ∨ (X ∈ P(B)) ⇐⇒(X ⊆ A) ∨ (X ⊆ B) ⇐⇒(∀x ∈ X : x ∈ A) ∨ (∀x ∈ X : x ∈ B) =⇒∀x ∈ X : (x ∈ A) ∨ (x ∈ B) ⇐⇒∀x ∈ X : (x ∈ A ∪B) ⇐⇒X ⊆ A ∪B ⇐⇒ X ∈ P(A ∪B)

SetsProve or disprove: For all A, B,P(A) ∪ P(B) ⊆ P(A ∪B). True

X ∈ P(A) ∪ P(B) ⇐⇒

(X ∈ P(A)) ∨ (X ∈ P(B)) ⇐⇒(X ⊆ A) ∨ (X ⊆ B) ⇐⇒(∀x ∈ X : x ∈ A) ∨ (∀x ∈ X : x ∈ B) =⇒∀x ∈ X : (x ∈ A) ∨ (x ∈ B) ⇐⇒∀x ∈ X : (x ∈ A ∪B) ⇐⇒X ⊆ A ∪B ⇐⇒ X ∈ P(A ∪B)

X ∈ P(A) ∪ P(B) ⇐⇒(X ∈ P(A)) ∨ (X ∈ P(B)) ⇐⇒

(X ⊆ A) ∨ (X ⊆ B) ⇐⇒(∀x ∈ X : x ∈ A) ∨ (∀x ∈ X : x ∈ B) =⇒∀x ∈ X : (x ∈ A) ∨ (x ∈ B) ⇐⇒∀x ∈ X : (x ∈ A ∪B) ⇐⇒X ⊆ A ∪B ⇐⇒ X ∈ P(A ∪B)

X ∈ P(A) ∪ P(B) ⇐⇒(X ∈ P(A)) ∨ (X ∈ P(B)) ⇐⇒(X ⊆ A) ∨ (X ⊆ B) ⇐⇒

(∀x ∈ X : x ∈ A) ∨ (∀x ∈ X : x ∈ B) =⇒∀x ∈ X : (x ∈ A) ∨ (x ∈ B) ⇐⇒∀x ∈ X : (x ∈ A ∪B) ⇐⇒X ⊆ A ∪B ⇐⇒ X ∈ P(A ∪B)

X ∈ P(A) ∪ P(B) ⇐⇒(X ∈ P(A)) ∨ (X ∈ P(B)) ⇐⇒(X ⊆ A) ∨ (X ⊆ B) ⇐⇒(∀x ∈ X : x ∈ A) ∨ (∀x ∈ X : x ∈ B) =⇒

∀x ∈ X : (x ∈ A) ∨ (x ∈ B) ⇐⇒∀x ∈ X : (x ∈ A ∪B) ⇐⇒X ⊆ A ∪B ⇐⇒ X ∈ P(A ∪B)

X ∈ P(A) ∪ P(B) ⇐⇒(X ∈ P(A)) ∨ (X ∈ P(B)) ⇐⇒(X ⊆ A) ∨ (X ⊆ B) ⇐⇒(∀x ∈ X : x ∈ A) ∨ (∀x ∈ X : x ∈ B) =⇒∀x ∈ X : (x ∈ A) ∨ (x ∈ B) ⇐⇒

∀x ∈ X : (x ∈ A ∪B) ⇐⇒X ⊆ A ∪B ⇐⇒ X ∈ P(A ∪B)

X ∈ P(A) ∪ P(B) ⇐⇒(X ∈ P(A)) ∨ (X ∈ P(B)) ⇐⇒(X ⊆ A) ∨ (X ⊆ B) ⇐⇒(∀x ∈ X : x ∈ A) ∨ (∀x ∈ X : x ∈ B) =⇒∀x ∈ X : (x ∈ A) ∨ (x ∈ B) ⇐⇒∀x ∈ X : (x ∈ A ∪B) ⇐⇒

X ⊆ A ∪B ⇐⇒ X ∈ P(A ∪B)

X ∈ P(A) ∪ P(B) ⇐⇒(X ∈ P(A)) ∨ (X ∈ P(B)) ⇐⇒(X ⊆ A) ∨ (X ⊆ B) ⇐⇒(∀x ∈ X : x ∈ A) ∨ (∀x ∈ X : x ∈ B) =⇒∀x ∈ X : (x ∈ A) ∨ (x ∈ B) ⇐⇒∀x ∈ X : (x ∈ A ∪B) ⇐⇒X ⊆ A ∪B ⇐⇒

X ∈ P(A ∪B)

SetsProve or disprove: For all A, B,P(A ∪B) ⊆ P(A) ∪ P(B).

Proof. Need to find A, B, X such thatX ∈ P(A ∪B), X 6∈ P(A) ∪ P(B).

Let A = {0}, B = {1}. Let X = {0, 1}.

X ⊆ A ∪B = {0, 1} =⇒ X ∈ P(A ∪B)

X 6∈ P(A) = {∅, {0}} X 6∈ P(B) = {∅, {1}}=⇒ X 6∈ P(A) ∪ P(B)

SetsProve or disprove: For all A, B,P(A ∪B) ⊆ P(A) ∪ P(B). False

Let A = {0}, B = {1}. Let X = {0, 1}.

X ⊆ A ∪B = {0, 1} =⇒ X ∈ P(A ∪B)

X 6∈ P(A) = {∅, {0}} X 6∈ P(B) = {∅, {1}}=⇒ X 6∈ P(A) ∪ P(B)

Let A = {0}, B = {1}. Let X = {0, 1}.

X ⊆ A ∪B = {0, 1} =⇒ X ∈ P(A ∪B)

X 6∈ P(A) = {∅, {0}} X 6∈ P(B) = {∅, {1}}=⇒ X 6∈ P(A) ∪ P(B)

Let A = {0}, B = {1}.

Let X = {0, 1}.

X ⊆ A ∪B = {0, 1} =⇒ X ∈ P(A ∪B)

X 6∈ P(A) = {∅, {0}} X 6∈ P(B) = {∅, {1}}=⇒ X 6∈ P(A) ∪ P(B)

Let A = {0}, B = {1}. Let X = {0, 1}.

X ⊆ A ∪B = {0, 1} =⇒ X ∈ P(A ∪B)

X 6∈ P(A) = {∅, {0}} X 6∈ P(B) = {∅, {1}}=⇒ X 6∈ P(A) ∪ P(B)

Let A = {0}, B = {1}. Let X = {0, 1}.

X ⊆ A ∪B = {0, 1} =⇒ X ∈ P(A ∪B)

X 6∈ P(A) = {∅, {0}} X 6∈ P(B) = {∅, {1}}=⇒ X 6∈ P(A) ∪ P(B)

Let A = {0}, B = {1}. Let X = {0, 1}.

X ⊆ A ∪B = {0, 1} =⇒ X ∈ P(A ∪B)

X 6∈ P(A) = {∅, {0}}

X 6∈ P(B) = {∅, {1}}=⇒ X 6∈ P(A) ∪ P(B)

Let A = {0}, B = {1}. Let X = {0, 1}.

X ⊆ A ∪B = {0, 1} =⇒ X ∈ P(A ∪B)

X 6∈ P(A) = {∅, {0}} X 6∈ P(B) = {∅, {1}}

=⇒ X 6∈ P(A) ∪ P(B)

Let A = {0}, B = {1}. Let X = {0, 1}.

X ⊆ A ∪B = {0, 1} =⇒ X ∈ P(A ∪B)

X 6∈ P(A) = {∅, {0}} X 6∈ P(B) = {∅, {1}}=⇒ X 6∈ P(A) ∪ P(B)

SetsLet x1, x2, . . . , xn be any elements (n ∈ N)

A (finite) sequence: (x1, x2, . . . , xn)

JunkSeq1 = (239, banana, ace of spades)

JunkSeq2 = (banana, 239, ace of spades, 239)

JunkSeq1 6= JunkSeq2

A sequence is not a set!

(. . . and not a basic concept, will be defined later)

SetsFor sequences, repetitions and order matter

(x, y) 6= (y, x) 6= (y, x, x)

Number n is sequence length

length(JunkSeq1 ) = 3 length((x, y)) = 2

Sequence of length 2 is called an ordered pair

A direct definition: (x, y) means {{x, y}, x}

(x, y) 6= (y, x) 6= (y, x, x)

SetsThe Cartesian product of sets A, B is the set of allordered pairs (a, b), where a ∈ A, b ∈ B

(After R. Descartes, 1596–1650)

A×B = {(a, b) | (a ∈ A) ∧ (b ∈ B)}A2 = A× A the Cartesian square of A

A×B = {(a, b) | (a ∈ A) ∧ (b ∈ B)}

A2 = A× A the Cartesian square of A

SetsExamples:

∅ × A =

A× ∅ = ∅ for any set A

{Bunty} × {Fowler} = {(Bunty, Fowler)}{Fowler} × {Bunty} = {(Fowler, Bunty)}

SetsExamples:

∅ × A = A× ∅ =

∅ for any set A

SetsExamples:

∅ × A = A× ∅ = ∅ for any set A

SetsExamples:

{Bunty} × {Fowler} =

{(Bunty, Fowler)}{Fowler} × {Bunty} = {(Fowler, Bunty)}

SetsExamples:

{Bunty} × {Fowler} = {(Bunty, Fowler)}

{Fowler} × {Bunty} = {(Fowler, Bunty)}

SetsExamples:

{Bunty} × {Fowler} = {(Bunty, Fowler)}{Fowler} × {Bunty} =

{(Fowler, Bunty)}

SetsExamples:

SetsMore examples:

{a, b, c} × {d, e} =

{(a, d), (a, e), (b, d), (b, e), (c, d), (c, e)}N×Planets = {(n, x) | (n ∈ N)∧ (x ∈ Planets)} ={(5, Saturn), (239, Earth), . . .}N2 = N× N = {(m, n) | m, n ∈ N}

SetsMore examples:

{a, b, c} × {d, e} ={(a, d), (a, e), (b, d), (b, e), (c, d), (c, e)}

N×Planets = {(n, x) | (n ∈ N)∧ (x ∈ Planets)} ={(5, Saturn), (239, Earth), . . .}N2 = N× N = {(m, n) | m, n ∈ N}

SetsMore examples:

{a, b, c} × {d, e} ={(a, d), (a, e), (b, d), (b, e), (c, d), (c, e)}N×Planets =

{(n, x) | (n ∈ N)∧ (x ∈ Planets)} ={(5, Saturn), (239, Earth), . . .}N2 = N× N = {(m, n) | m, n ∈ N}

SetsMore examples:

{a, b, c} × {d, e} ={(a, d), (a, e), (b, d), (b, e), (c, d), (c, e)}N×Planets = {(n, x) | (n ∈ N)∧ (x ∈ Planets)} =

{(5, Saturn), (239, Earth), . . .}N2 = N× N = {(m, n) | m, n ∈ N}

SetsMore examples:

{a, b, c} × {d, e} ={(a, d), (a, e), (b, d), (b, e), (c, d), (c, e)}N×Planets = {(n, x) | (n ∈ N)∧ (x ∈ Planets)} ={(5, Saturn), (239, Earth), . . .}

N2 = N× N = {(m, n) | m, n ∈ N}

SetsMore examples:

{a, b, c} × {d, e} ={(a, d), (a, e), (b, d), (b, e), (c, d), (c, e)}N×Planets = {(n, x) | (n ∈ N)∧ (x ∈ Planets)} ={(5, Saturn), (239, Earth), . . .}N2 = N× N = {(m, n) | m, n ∈ N}

SetsIf A, B finite, A×B finite

If A has m elements, B has n elements, then A×Bhas m · n elements

(. . . hence the “×” sign)

If A infinite, B nonempty, then A×B, B ×A infinite

SetsProperties of ×:

In general, A×B 6= B × A

In general, (A×B)× C 6= A× (B × C)

. . . but = holds if we identify ((a, b), c) and (a, (b, c))

SetsA× (B ∩ C) = (A×B) ∩ (A× C)(A ∩B)× C = (A× C) ∩ (B × C)

A× (B ∪ C) = (A×B) ∪ (A× C)(A ∪B)× C = (A× C) ∪ (B × C)

A× (B \ C) = (A×B) \ (A× C)(A \B)× C = (A× C) \ (B × C)

× distributes over ∩,∪, \

A× (B ∪ C) = (A×B) ∪ (A× C)(A ∪B)× C = (A× C) ∪ (B × C)

A× (B \ C) = (A×B) \ (A× C)(A \B)× C = (A× C) \ (B × C)

A× (B ∪ C) = (A×B) ∪ (A× C)(A ∪B)× C = (A× C) ∪ (B × C)

A× (B \ C) = (A×B) \ (A× C)(A \B)× C = (A× C) \ (B × C)

A× (B ∪ C) = (A×B) ∪ (A× C)(A ∪B)× C = (A× C) ∪ (B × C)

A× (B \ C) = (A×B) \ (A× C)(A \B)× C = (A× C) \ (B × C)

SetsProve A× (B ∪ C) = (A×B) ∪ (A× C)

Proof. Consider any (x, y).(x, y) ∈ A× (B ∪ C) ⇐⇒(x ∈ A) ∧ (y ∈ (B ∪ C)) ⇐⇒(x ∈ A) ∧ (y ∈ B ∨ y ∈ C) ⇐⇒(x ∈ A ∧ y ∈ B) ∨ (x ∈ A ∧ y ∈ C) ⇐⇒((x, y) ∈ A×B) ∨ ((x, y) ∈ A× C) ⇐⇒(x, y) ∈ (A×B) ∪ (A× C)

Hence A× (B ∪ C) = (A×B) ∪ (A× C).

Proof. Consider any (x, y).

(x, y) ∈ A× (B ∪ C) ⇐⇒(x ∈ A) ∧ (y ∈ (B ∪ C)) ⇐⇒(x ∈ A) ∧ (y ∈ B ∨ y ∈ C) ⇐⇒(x ∈ A ∧ y ∈ B) ∨ (x ∈ A ∧ y ∈ C) ⇐⇒((x, y) ∈ A×B) ∨ ((x, y) ∈ A× C) ⇐⇒(x, y) ∈ (A×B) ∪ (A× C)

Hence A× (B ∪ C) = (A×B) ∪ (A× C).

Proof. Consider any (x, y).(x, y) ∈ A× (B ∪ C) ⇐⇒

(x ∈ A) ∧ (y ∈ (B ∪ C)) ⇐⇒(x ∈ A) ∧ (y ∈ B ∨ y ∈ C) ⇐⇒(x ∈ A ∧ y ∈ B) ∨ (x ∈ A ∧ y ∈ C) ⇐⇒((x, y) ∈ A×B) ∨ ((x, y) ∈ A× C) ⇐⇒(x, y) ∈ (A×B) ∪ (A× C)

Hence A× (B ∪ C) = (A×B) ∪ (A× C).

Proof. Consider any (x, y).(x, y) ∈ A× (B ∪ C) ⇐⇒(x ∈ A) ∧ (y ∈ (B ∪ C)) ⇐⇒

(x ∈ A) ∧ (y ∈ B ∨ y ∈ C) ⇐⇒(x ∈ A ∧ y ∈ B) ∨ (x ∈ A ∧ y ∈ C) ⇐⇒((x, y) ∈ A×B) ∨ ((x, y) ∈ A× C) ⇐⇒(x, y) ∈ (A×B) ∪ (A× C)

Hence A× (B ∪ C) = (A×B) ∪ (A× C).

Proof. Consider any (x, y).(x, y) ∈ A× (B ∪ C) ⇐⇒(x ∈ A) ∧ (y ∈ (B ∪ C)) ⇐⇒(x ∈ A) ∧ (y ∈ B ∨ y ∈ C) ⇐⇒

(x ∈ A ∧ y ∈ B) ∨ (x ∈ A ∧ y ∈ C) ⇐⇒((x, y) ∈ A×B) ∨ ((x, y) ∈ A× C) ⇐⇒(x, y) ∈ (A×B) ∪ (A× C)

Hence A× (B ∪ C) = (A×B) ∪ (A× C).

Proof. Consider any (x, y).(x, y) ∈ A× (B ∪ C) ⇐⇒(x ∈ A) ∧ (y ∈ (B ∪ C)) ⇐⇒(x ∈ A) ∧ (y ∈ B ∨ y ∈ C) ⇐⇒(x ∈ A ∧ y ∈ B) ∨ (x ∈ A ∧ y ∈ C) ⇐⇒

((x, y) ∈ A×B) ∨ ((x, y) ∈ A× C) ⇐⇒(x, y) ∈ (A×B) ∪ (A× C)

Hence A× (B ∪ C) = (A×B) ∪ (A× C).

Proof. Consider any (x, y).(x, y) ∈ A× (B ∪ C) ⇐⇒(x ∈ A) ∧ (y ∈ (B ∪ C)) ⇐⇒(x ∈ A) ∧ (y ∈ B ∨ y ∈ C) ⇐⇒(x ∈ A ∧ y ∈ B) ∨ (x ∈ A ∧ y ∈ C) ⇐⇒((x, y) ∈ A×B) ∨ ((x, y) ∈ A× C) ⇐⇒

(x, y) ∈ (A×B) ∪ (A× C)

Hence A× (B ∪ C) = (A×B) ∪ (A× C).

SetsThe Cartesian product of sets A1, A2, . . . , An is theset of all ordered sequences (a1, a2, . . . , an), whereai ∈ Ai for all i ∈ {1, . . . , n}

A1 × · · · × An ={(a1, . . . , an) | ∀i ∈ {1, . . . , n} : ai ∈ Ai}

An = A× A× · · · × A (n times)the n-th Cartesian power of A

SetsThe Cartesian product of sets A1, A2, . . . , An is theset of all ordered sequences (a1, a2, . . . , an), whereai ∈ Ai for all i ∈ {1, . . . , n}A1 × · · · × An =

{(a1, . . . , an) | ∀i ∈ {1, . . . , n} : ai ∈ Ai}

An = A× A× · · · × A (n times)the n-th Cartesian power of A

SetsThe Cartesian product of sets A1, A2, . . . , An is theset of all ordered sequences (a1, a2, . . . , an), whereai ∈ Ai for all i ∈ {1, . . . , n}A1 × · · · × An =

{(a1, . . . , an) | ∀i ∈ {1, . . . , n} : ai ∈ Ai}An = A× A× · · · × A (n times)

the n-th Cartesian power of A

SetsIf A1, A2, . . . , An finite, A1 × A2 × · · · × An finite

If for all i, Ai has ni elements, A1 × · · · × Ak hasn1 · . . . · nk elements

If one of A1, A2, . . . , An infinite, A1 × A2 × · · · × An

infinite (unless one of them is empty)

SetsTherefore:

If A finite, An finite

If A has k elements, An has kn elements

If A infinite, An infinite

Relations

RelationsConsider P (x, y) = x ≤ y x, y ∈ N

{(x, y) | x ≤ y} ⊆ N× N = N2

RelationsConsider P (x, y) = x ≤ y x, y ∈ N

{(x, y) | x ≤ y} ⊆ N× N = N2

RelationsA relation between sets A, B is a subset of A×B

Rp : A ↔ B ⇐⇒ Rp ⊆ A×B

Example:

R≤ = {(a, b) ∈ N× N | a ≤ b} ⊆ N× N = N2

Write a p b for (a, b) ∈ Rp

For example a ≤ b instead of (a, b) ∈ R≤

Rp : A ↔ B ⇐⇒ Rp ⊆ A×B

Example:

R≤ = {(a, b) ∈ N× N | a ≤ b} ⊆ N× N = N2

Rp : A ↔ B ⇐⇒ Rp ⊆ A×B

Example:

R≤ = {(a, b) ∈ N× N | a ≤ b} ⊆ N× N = N2

Rp : A ↔ B ⇐⇒ Rp ⊆ A×B

Example:

R≤ = {(a, b) ∈ N× N | a ≤ b} ⊆ N× N = N2

RelationsExamples of relations:

Equality relation R=A: A ↔ A

R=A= {(a, a) | a ∈ A}

Usually drop A: a = a

Empty relation ∅ : A ↔ A

Complete relation A2 : A ↔ A

R=A= {(a, a) | a ∈ A}

RelationsMore examples of relations:

R<, R≤, R>, R≥ N ↔ N

R| : N ↔ N m | n ⇐⇒ m divides n

m | n ⇐⇒ ∃k ∈ N : k ·m = n

Rq : People ↔ People x q y ⇐⇒ x is a child of y

Rt : People ↔ Animalsx t y ⇐⇒ x has y as a pet

R<, R≤, R>, R≥ N ↔ N

m | n ⇐⇒ ∃k ∈ N : k ·m = n

R<, R≤, R>, R≥ N ↔ N

m | n ⇐⇒ ∃k ∈ N : k ·m = n

R<, R≤, R>, R≥ N ↔ N

m | n ⇐⇒ ∃k ∈ N : k ·m = n

R<, R≤, R>, R≥ N ↔ N

m | n ⇐⇒ ∃k ∈ N : k ·m = n

Rs : N ↔ N m s n ⇐⇒ m2 = n

Rm : People ↔ Peoplex m y ⇐⇒ the mother of x is y

In these relations, for every x ∈ A, there is a uniquey ∈ B, such that x is related to y

Such relations are called functions

Rs : N ↔ N m s n ⇐⇒ m2 = n

RelationsRp, Rq : A ↔ B

Rp ∩Rq, Rp ∪Rq, Rp \Rq : A ↔ B

RelationsRp, Rq : A ↔ B

Rp ∩Rq, Rp ∪Rq, Rp \Rq : A ↔ B

RelationsRp : A ↔ B Rq : B ↔ C

The composition of p and q Rp ◦ q : A ↔ C

∀(a, c) ∈ A×C : a(p ◦ q)c ⇔ ∃b ∈ B : (a p b)∧(b q c)

RelationsExamples:

Rq ◦ t : People ↔ Animals

x(q ◦ t)z ⇐⇒ x has a parent with pet z

Rq ◦ q : People ↔ People

x(q ◦ q)z ⇐⇒ x is a grandchild of z

RelationsExamples:

Prove: if Rp, Rq functions, then Rp ◦ q a function

Proof. Consider any x ∈ A.

Rp function =⇒ ∃!y ∈ B : x p y

Rq function =⇒ ∃!z ∈ C : y q z

Hence ∃!z ∈ C : x(p ◦ q)z

Therefore, Rp ◦ q is a function.

RelationsRp : A ↔ B

The inverse of p Rp−1 : B ↔ A

∀(b, a) ∈ B × A : b(p−1)a ⇐⇒ a p b

Example:

Rq−1 : People ↔ People

x q−1 y ⇐⇒ x is a parent of y

∀(b, a) ∈ B × A : b(p−1)a ⇐⇒ a p b

Example:

∀(b, a) ∈ B × A : b(p−1)a ⇐⇒ a p b

Example:

∀(b, a) ∈ B × A : b(p−1)a ⇐⇒ a p b

Example:

∀(b, a) ∈ B × A : b(p−1)a ⇐⇒ a p b

Example:

RelationsRelation Rp : A ↔ A is reflexive, if ∀a ∈ A : a p a

Examples: R= A2 R≤ R|

Rp reflexive iff R= ⊆ Rp

Examples: R=

A2 R≤ R|

Examples: R= A2

R≤ R|

Examples: R= A2 R≤

RelationsRelation Rp : A ↔ A is symmetric, if

∀a, b ∈ A : a p b ⇒ b p a

Examples: R= A2 ∅R∗ : N ↔ N x ∗ y ⇔ (x + y = 10)

Rp symmetric iff Rp−1 = Rp

∀a, b ∈ A : a p b ⇒ b p a

Examples: R=

A2 ∅R∗ : N ↔ N x ∗ y ⇔ (x + y = 10)

∀a, b ∈ A : a p b ⇒ b p a

Examples: R= A2

∅R∗ : N ↔ N x ∗ y ⇔ (x + y = 10)

∀a, b ∈ A : a p b ⇒ b p a

Examples: R= A2 ∅

R∗ : N ↔ N x ∗ y ⇔ (x + y = 10)

∀a, b ∈ A : a p b ⇒ b p a

RelationsRelation Rp : A ↔ A is antisymmetric, if

∀a, b ∈ A : (a p b ∧ b p a) ⇒ a = b

Examples: R= ∅ R≤ R<

Rp antisymmetric iff Rp ∩Rp−1 ⊆ R=

Note non-symmetric 6⇔ antisymmetric (e.g. R=)

∀a, b ∈ A : (a p b ∧ b p a) ⇒ a = b

Examples: R=

∅ R≤ R<

∀a, b ∈ A : (a p b ∧ b p a) ⇒ a = b

Examples: R= ∅

R≤ R<

∀a, b ∈ A : (a p b ∧ b p a) ⇒ a = b

Examples: R= ∅ R≤

∀a, b ∈ A : (a p b ∧ b p a) ⇒ a = b

RelationsRelation Rp : A ↔ A is transitive, if

∀a, b, c ∈ A : (a p b ∧ b p c) ⇒ a p c

Examples: R= A2 ∅ R≤ R<

Rp transitive iff Rp ◦ p ⊆ Rp

∀a, b, c ∈ A : (a p b ∧ b p c) ⇒ a p c

Examples: R=

A2 ∅ R≤ R<

∀a, b, c ∈ A : (a p b ∧ b p c) ⇒ a p c

Examples: R= A2

∅ R≤ R<

∀a, b, c ∈ A : (a p b ∧ b p c) ⇒ a p c

Examples: R= A2 ∅

R≤ R<

∀a, b, c ∈ A : (a p b ∧ b p c) ⇒ a p c

Examples: R= A2 ∅ R≤

∀a, b, c ∈ A : (a p b ∧ b p c) ⇒ a p c

RelationsRelation R∼ : A ↔ A is an equivalence relation,if it is reflexive, symmetric and transitive

Relation R� : A ↔ A is a partial order,if it is reflexive, antisymmetric and transitive

RelationsExamples of equivalence relations:

Rp : People ↔ People

a p b ⇐⇒ a and b share a birthday

RelationsExamples of equivalence relations:

a p b ⇐⇒ a and b share a birthday

Relations{. . . ,−3,−2,−1, 0, 1, 2, 3, . . . } = Z integers

R| : N ↔ Z a divides b

a | b ⇐⇒ ∃k ∈ Z : k · a = b

R≡n: Z ↔ Z a ≡n b ⇐⇒ n|(a− b)

R≡nis called congruence modulo n

a | b ⇐⇒ ∃k ∈ Z : k · a = b

R≡n: Z ↔ Z a ≡n b ⇐⇒ n|(a− b)

a | b ⇐⇒ ∃k ∈ Z : k · a = b

R≡n: Z ↔ Z a ≡n b ⇐⇒ n|(a− b)

a | b ⇐⇒ ∃k ∈ Z : k · a = b

R≡n: Z ↔ Z a ≡n b ⇐⇒ n|(a− b)

RelationsProve: R≡n

is an equivalence for all n ∈ N, n ≥ 1.

Proof. Let n ∈ N, n ≥ 1.

Let x ∈ Z.

x ≡n x ⇐⇒ n | (x− x) ⇐⇒ n | 0 ⇐⇒∃k : k · n = 0 ⇐⇒ T (since 0 · n = 0)

Hence R≡nreflexive.

Let x ∈ Z.

RelationsLet x, y ∈ Z.

x ≡n y =⇒ n | (x− y) =⇒ ∃k : k · n = x− y

(−k) · n = −(x− y) = y − x =⇒n | (y − x) =⇒ y ≡n x

Hence R≡nsymmetric.

x ≡n y =⇒ n | (x− y) =⇒ ∃k : k · n = x− y

(−k) · n = −(x− y) = y − x =⇒n | (y − x) =⇒ y ≡n x

x ≡n y =⇒ n | (x− y) =⇒ ∃k : k · n = x− y

(−k) · n = −(x− y) = y − x =⇒n | (y − x) =⇒ y ≡n x

x ≡n y =⇒ n | (x− y) =⇒ ∃k : k · n = x− y

(−k) · n = −(x− y) = y − x =⇒n | (y − x) =⇒ y ≡n x

RelationsLet x, y, z ∈ Z.

x ≡n y =⇒ n | (x− y) =⇒ ∃k : k · n = x− y

y ≡n z =⇒ n | (y − z) =⇒ ∃l : l · n = y − z

(k + l) · n = (x− y) + (y − z) = x− z =⇒n | (x− z) =⇒ x ≡n z

Hence R≡ntransitive.

R≡nreflexive, symmetric and transitive, therefore

R≡nis an equivalence relation.

x ≡n y =⇒ n | (x− y) =⇒ ∃k : k · n = x− y

y ≡n z =⇒ n | (y − z) =⇒ ∃l : l · n = y − z

x ≡n y =⇒ n | (x− y) =⇒ ∃k : k · n = x− y

y ≡n z =⇒ n | (y − z) =⇒ ∃l : l · n = y − z

x ≡n y =⇒ n | (x− y) =⇒ ∃k : k · n = x− y

y ≡n z =⇒ n | (y − z) =⇒ ∃l : l · n = y − z

x ≡n y =⇒ n | (x− y) =⇒ ∃k : k · n = x− y

y ≡n z =⇒ n | (y − z) =⇒ ∃l : l · n = y − z

x ≡n y =⇒ n | (x− y) =⇒ ∃k : k · n = x− y

y ≡n z =⇒ n | (y − z) =⇒ ∃l : l · n = y − z

RelationsR∼ : A ↔ A — an equivalence relation

For any a ∈ A, the equivalence class of a is the set ofall elements related to a

[a]∼ = {x ∈ A | x ∼ a}By reflexivity, a ∈ [a]∼

Every a is a representative of [a]∼

The set of all equivalence classes of R∼ is the quotientset of A with respect to R∼

A/R∼ = {[a]∼ | a ∈ A}

[a]∼ = {x ∈ A | x ∼ a}

By reflexivity, a ∈ [a]∼

A/R∼ = {[a]∼ | a ∈ A}

RelationsExample:

x p y ⇐⇒ x and y share a birthday

Size ofPeople/Rp = 366

∀x, y ∈ People : ([x]p = [y]p) ∨ ([x]p ∩ [y]p = ∅)

RelationsExample:

Size ofPeople/Rp =

RelationsExample:

RelationsLet Fred, George ∈ People

Suppose Fred was born on 1 November

[Fred]p = {x ∈ People | x born on 1 November}Size of [Fred]p ≈ 6bln/365.25 ≈ 16mln

Suppose George was born on 29 February

[George]p = {x ∈ People | x born on 29 February}Size of [George]p ≈ 6bln/(4 ∗ 365.25) ≈ 4mln

[Fred]p = {x ∈ People | x born on 1 November}

Size of [Fred]p ≈ 6bln/365.25 ≈ 16mln

[George]p = {x ∈ People | x born on 29 February}

Size of [George]p ≈ 6bln/(4 ∗ 365.25) ≈ 4mln

RelationsAnother example:

R≡5: Z ↔ Z a ≡5 b ⇐⇒ 5 | (a− b)

[0]≡5= {. . . ,−10,−5, 0, 5, 10, . . . }

[1]≡5= {. . . ,−9,−4, 1, 6, 11, . . . }

[2]≡5= {. . . ,−8,−3, 2, 7, 12, . . . }

[3]≡5= {. . . ,−7,−2, 3, 8, 13, . . . }

[4]≡5= {. . . ,−6,−1, 4, 9, 14, . . . }

[a]≡5called residue classes modulo 5 (can be any n)

R≡5: Z ↔ Z a ≡5 b ⇐⇒ 5 | (a− b)

[0]≡5=

{. . . ,−10,−5, 0, 5, 10, . . . }[1]≡5

= {. . . ,−9,−4, 1, 6, 11, . . . }[2]≡5

= {. . . ,−8,−3, 2, 7, 12, . . . }[3]≡5

= {. . . ,−7,−2, 3, 8, 13, . . . }[4]≡5

= {. . . ,−6,−1, 4, 9, 14, . . . }[a]≡5

called residue classes modulo 5 (can be any n)

R≡5: Z ↔ Z a ≡5 b ⇐⇒ 5 | (a− b)

[0]≡5= {. . . ,−10,−5, 0, 5, 10, . . . }

[1]≡5= {. . . ,−9,−4, 1, 6, 11, . . . }

[2]≡5= {. . . ,−8,−3, 2, 7, 12, . . . }

[3]≡5= {. . . ,−7,−2, 3, 8, 13, . . . }

[4]≡5= {. . . ,−6,−1, 4, 9, 14, . . . }

R≡5: Z ↔ Z a ≡5 b ⇐⇒ 5 | (a− b)

[0]≡5= {. . . ,−10,−5, 0, 5, 10, . . . }

[1]≡5=

{. . . ,−9,−4, 1, 6, 11, . . . }[2]≡5

= {. . . ,−8,−3, 2, 7, 12, . . . }[3]≡5

= {. . . ,−7,−2, 3, 8, 13, . . . }[4]≡5

= {. . . ,−6,−1, 4, 9, 14, . . . }[a]≡5

called residue classes modulo 5 (can be any n)

R≡5: Z ↔ Z a ≡5 b ⇐⇒ 5 | (a− b)

[0]≡5= {. . . ,−10,−5, 0, 5, 10, . . . }

[1]≡5= {. . . ,−9,−4, 1, 6, 11, . . . }

[2]≡5= {. . . ,−8,−3, 2, 7, 12, . . . }

[3]≡5= {. . . ,−7,−2, 3, 8, 13, . . . }

[4]≡5= {. . . ,−6,−1, 4, 9, 14, . . . }

R≡5: Z ↔ Z a ≡5 b ⇐⇒ 5 | (a− b)

[0]≡5= {. . . ,−10,−5, 0, 5, 10, . . . }

[1]≡5= {. . . ,−9,−4, 1, 6, 11, . . . }

[2]≡5= {. . . ,−8,−3, 2, 7, 12, . . . }

[3]≡5= {. . . ,−7,−2, 3, 8, 13, . . . }

[4]≡5= {. . . ,−6,−1, 4, 9, 14, . . . }

R≡5: Z ↔ Z a ≡5 b ⇐⇒ 5 | (a− b)

[0]≡5= {. . . ,−10,−5, 0, 5, 10, . . . }

[1]≡5= {. . . ,−9,−4, 1, 6, 11, . . . }

[2]≡5= {. . . ,−8,−3, 2, 7, 12, . . . }

[3]≡5= {. . . ,−7,−2, 3, 8, 13, . . . }

[4]≡5= {. . . ,−6,−1, 4, 9, 14, . . . }

R≡5: Z ↔ Z a ≡5 b ⇐⇒ 5 | (a− b)

[0]≡5= {. . . ,−10,−5, 0, 5, 10, . . . }

[1]≡5= {. . . ,−9,−4, 1, 6, 11, . . . }

[2]≡5= {. . . ,−8,−3, 2, 7, 12, . . . }

[3]≡5= {. . . ,−7,−2, 3, 8, 13, . . . }

[4]≡5= {. . . ,−6,−1, 4, 9, 14, . . . }

R≡5: Z ↔ Z a ≡5 b ⇐⇒ 5 | (a− b)

[0]≡5= {. . . ,−10,−5, 0, 5, 10, . . . }

[1]≡5= {. . . ,−9,−4, 1, 6, 11, . . . }

[2]≡5= {. . . ,−8,−3, 2, 7, 12, . . . }

[3]≡5= {. . . ,−7,−2, 3, 8, 13, . . . }

[4]≡5= {. . . ,−6,−1, 4, 9, 14, . . . }

Theorem.

The equivalence classes of R∼ are pairwise disjoint.

∀a, b ∈ A : ([a]∼ = [b]∼) ∨ ([a]∼ ∩ [b]∼ = ∅)The union of all equivalence classes is the whole A.⋃

a∈A[a]∼ = A

A is partitioned by R∼ into a disjoint union ofequivalence classes

Theorem.

∀a, b ∈ A : ([a]∼ = [b]∼) ∨ ([a]∼ ∩ [b]∼ = ∅)

The union of all equivalence classes is the whole A.⋃

a∈A[a]∼ = A

Theorem.

a∈A[a]∼ = A

Theorem.

a∈A[a]∼ = A

RelationsProof. For all a, b, we need:

([a]∼ = [b]∼) ∨ ([a]∼ ∩ [b]∼ = ∅)

Consider two cases: a ∼ b, a 6∼ b.

RelationsProof. For all a, b, we need:

([a]∼ = [b]∼) ∨ ([a]∼ ∩ [b]∼ = ∅)Consider two cases: a ∼ b, a 6∼ b.

RelationsCase a ∼ b. Take any x ∈ [a]∼.

(x ∼ a) ∧ (a ∼ b) =⇒ x ∼ b =⇒ x ∈ [b]∼

Hence [a]∼ ⊆ [b]∼. Similarly [b]∼ ⊆ [a]∼.

Therefore [a]∼ = [b]∼.

(x ∼ a) ∧ (a ∼ b) =⇒ x ∼ b =⇒ x ∈ [b]∼

RelationsCase a 6∼ b. Suppose ∃x : x ∈ [a]∼ ∩ [b]∼.

(x ∼ a) ∧ (x ∼ b) =⇒ a ∼ b — contradiction.

Therefore [a]∼ ∩ [b]∼ = ∅.

RelationsCase a ∼ b =⇒ [a]∼ = [b]∼

Case a 6∼ b =⇒ [a]∼ ∩ [b]∼ = ∅

Therefore ([a]∼ = [b]∼) ∨ ([a]∼ ∩ [b]∼ = ∅)Finally, for all a ∈ A: a ∼ a =⇒ a ∈ [a]∼ ⊆ A.

Hence A ⊆ ⋃

a∈A[a]∼ ⊆ A.

Therefore⋃

a∈A[a]∼ = A.

Case a 6∼ b =⇒ [a]∼ ∩ [b]∼ = ∅Therefore ([a]∼ = [b]∼) ∨ ([a]∼ ∩ [b]∼ = ∅)

Finally, for all a ∈ A: a ∼ a =⇒ a ∈ [a]∼ ⊆ A.

Hence A ⊆ ⋃

a∈A[a]∼ ⊆ A.

Therefore⋃

a∈A[a]∼ = A.

Hence A ⊆ ⋃

a∈A[a]∼ ⊆ A.

Therefore⋃

a∈A[a]∼ = A.

Hence A ⊆ ⋃

a∈A[a]∼ ⊆ A.

Therefore⋃

a∈A[a]∼ = A.

Hence A ⊆ ⋃

a∈A[a]∼ ⊆ A.

Therefore⋃

a∈A[a]∼ = A.

RelationsIf A finite, A/R∼ finite

If A has n elements, and if every [a]∼ has m elements,then m | n, and A/R∼ has n/m elements

If A infinite, A/R∼ can be finite or infinite

Set A is partially ordered

RelationsExamples:

R≤, R≥ : N ↔ N

Hasse diagram (illustration only):

RelationsExamples:

R≤, R≥ : N ↔ N

��

RelationsExamples:

R≤, R≥ : N ↔ N

��

RelationsRE : People ↔ People

xE y ⇐⇒ x is an descendant of y

(Everyone is his/her own descendant)

Lamech Bitenosh

Noah Naamah

Shem Ham Japheth

Lamech Bitenosh

Noah Naamah

Shem Ham Japheth

Lamech

Bitenosh

�Noah � Naamah

Japheth

RelationsR| : N ↔ N x | y ⇐⇒ ∃k : k · x = y

2 3 5 7

RelationsR| : N ↔ N x | y ⇐⇒ ∃k : k · x = y

1�2 �3 � 5 � 7

RelationsProve: R| is a partial order.

Proof. Let x ∈ N.

x | x ⇐⇒ ∃k : k · x = x ⇐⇒ T (since 1 · x = x)

Hence R| reflexive.

Let x, y ∈ N.

x | y =⇒ ∃k : k · x = y

y | x =⇒ ∃l : l · y = x

x = l · y = k · l · x =⇒ k · l = 1 =⇒k = l = 1 =⇒ x = y

Hence R| antisymmetric.

Proof. Let x ∈ N.

x | x ⇐⇒ ∃k : k · x = x ⇐⇒ T (since 1 · x = x)

Hence R| reflexive.

Let x, y ∈ N.

x | y =⇒ ∃k : k · x = y

y | x =⇒ ∃l : l · y = x

x = l · y = k · l · x =⇒ k · l = 1 =⇒k = l = 1 =⇒ x = y

Proof. Let x ∈ N.

x | x ⇐⇒ ∃k : k · x = x ⇐⇒ T (since 1 · x = x)

Hence R| reflexive.

Let x, y ∈ N.

x | y =⇒ ∃k : k · x = y

y | x =⇒ ∃l : l · y = x

x = l · y = k · l · x =⇒ k · l = 1 =⇒k = l = 1 =⇒ x = y

Proof. Let x ∈ N.

x | x ⇐⇒ ∃k : k · x = x ⇐⇒ T (since 1 · x = x)

Hence R| reflexive.

Let x, y ∈ N.

x | y =⇒ ∃k : k · x = y

y | x =⇒ ∃l : l · y = x

x = l · y = k · l · x =⇒ k · l = 1 =⇒k = l = 1 =⇒ x = y

Proof. Let x ∈ N.

x | x ⇐⇒ ∃k : k · x = x ⇐⇒ T (since 1 · x = x)

Hence R| reflexive.

Let x, y ∈ N.

x | y =⇒ ∃k : k · x = y

y | x =⇒ ∃l : l · y = x

x = l · y = k · l · x =⇒ k · l = 1 =⇒k = l = 1 =⇒ x = y

Proof. Let x ∈ N.

x | x ⇐⇒ ∃k : k · x = x ⇐⇒ T (since 1 · x = x)

Hence R| reflexive.

Let x, y ∈ N.

x | y =⇒ ∃k : k · x = y

y | x =⇒ ∃l : l · y = x

x = l · y = k · l · x =⇒ k · l = 1 =⇒k = l = 1 =⇒ x = y

Proof. Let x ∈ N.

x | x ⇐⇒ ∃k : k · x = x ⇐⇒ T (since 1 · x = x)

Hence R| reflexive.

Let x, y ∈ N.

x | y =⇒ ∃k : k · x = y

y | x =⇒ ∃l : l · y = x

x = l · y = k · l · x =⇒ k · l = 1 =⇒k = l = 1 =⇒ x = y

Proof. Let x ∈ N.

x | x ⇐⇒ ∃k : k · x = x ⇐⇒ T (since 1 · x = x)

Hence R| reflexive.

Let x, y ∈ N.

x | y =⇒ ∃k : k · x = y

y | x =⇒ ∃l : l · y = x

x = l · y = k · l · x =⇒ k · l = 1 =⇒k = l = 1 =⇒ x = y

Proof. Let x ∈ N.

x | x ⇐⇒ ∃k : k · x = x ⇐⇒ T (since 1 · x = x)

Hence R| reflexive.

Let x, y ∈ N.

x | y =⇒ ∃k : k · x = y

y | x =⇒ ∃l : l · y = x

x = l · y = k · l · x =⇒ k · l = 1 =⇒k = l = 1 =⇒ x = y

Hence R| antisymmetric.Discrete Mathematics I – p. 155/292

RelationsLet x, y, z ∈ N.

x | y =⇒ ∃k : k · x = y

y | z =⇒ ∃l : l · y = z

z = l · y = (k · l) · x =⇒ x | zHence R| transitive.

R| reflexive, antisymmetric and transitive, thereforeR| is a partial order.

x | y =⇒ ∃k : k · x = y

y | z =⇒ ∃l : l · y = z

x | y =⇒ ∃k : k · x = y

y | z =⇒ ∃l : l · y = z

x | y =⇒ ∃k : k · x = y

y | z =⇒ ∃l : l · y = z

z = l · y = (k · l) · x =⇒ x | z

Hence R| transitive.

x | y =⇒ ∃k : k · x = y

y | z =⇒ ∃l : l · y = z

x | y =⇒ ∃k : k · x = y

y | z =⇒ ∃l : l · y = z

RelationsR⊆ : P(S) ↔ P(S) A ⊆ B

S = {0, 1, 2}

�0 �

� 2�12 �

� 01�

RelationsR� : A ↔ A — a partial order

R� is called a total order, if for all a, b ∈ A,either a � b or b � a

Set A is called totally ordered

R� is called a total order, if for all a, b ∈ A,either a � b or b � a

Set A is called totally ordered

RelationsExamples:

R≤ R≥ — total (but still a partial order!)

RE R| R⊆ — not total

RelationsExamples:

R≤ R≥ — total (but still a partial order!)

RE R| R⊆ — not total

RelationsR� : A ↔ A — a partial order (need not be total)

a ∈ A

c ∈ A is an upper bound of a, if a � c

d ∈ A is a lower bound of a, if d � a

a ∈ A

Relationsa, b ∈ A

c ∈ A is an upper bound of a, b, if (a � c) ∧ (b � c)

d ∈ A is a lower bound of a, b, if (d � a) ∧ (d � b)

Relationsa, b ∈ A

c ∈ A is an upper bound of a, b, if (a � c) ∧ (b � c)

d ∈ A is a lower bound of a, b, if (d � a) ∧ (d � b)

Relationsa, b ∈ A

c ∈ A is the least upper bound of a, b, if

• c is an upper bound of a, b

• for all x ∈ A, (a � x) ∧ (b � x) ⇒ (c � x)

c = lub(a, b) (may not exist!)

Relationsa, b ∈ A

• for all x ∈ A, (a � x) ∧ (b � x) ⇒ (c � x)

Relationsa, b ∈ A

• for all x ∈ A, (a � x) ∧ (b � x) ⇒ (c � x)

Relationsa, b ∈ A

• for all x ∈ A, (a � x) ∧ (b � x) ⇒ (c � x)

Relationsa, b ∈ A

• for all x ∈ A, (a � x) ∧ (b � x) ⇒ (c � x)

Relationsa, b ∈ A

d ∈ A is the greatest lower bound of a, b, if

• d is a lower bound of a, b

• for all x ∈ A, (x � a) ∧ (x � b) ⇒ (x � d)

d = glb(a, b) (may not exist!)

Relationsa, b ∈ A

• for all x ∈ A, (x � a) ∧ (x � b) ⇒ (x � d)

Relationsa, b ∈ A

• for all x ∈ A, (x � a) ∧ (x � b) ⇒ (x � d)

Relationsa, b ∈ A

• for all x ∈ A, (x � a) ∧ (x � b) ⇒ (x � d)

Relationsa, b ∈ A

• for all x ∈ A, (x � a) ∧ (x � b) ⇒ (x � d)

RelationsExample:

RE : People ↔ PeoplexE y ⇐⇒ x is a descendant of y

lub(x, y) = youngest common ancestor(x, y)

glb(x, y) = oldest common descendant(x, y)

May not exist, e.g. if x, y are not relatives

RelationsExample:

RelationsR| : N ↔ N x | y ⇐⇒ x divides y

lub(a, b) = least common multiple(a, b)always exists

glb(a, b) = greatest common divisor(a, b)always exists

lub(a, b) =

least common multiple(a, b)always exists

lub(a, b) = least common multiple(a, b)

always exists

glb(a, b) =

greatest common divisor(a, b)always exists

glb(a, b) = greatest common divisor(a, b)

always exists

RelationsR≤ : N ↔ N

lub(a, b) = max(a, b) always exists

glb(a, b) = min(a, b) always exists

{lub(a, b), glb(a, b)} = {a, b}Same holds for any total order

lub(a, b) =

max(a, b) always exists

lub(a, b) = max(a, b)

always exists

glb(a, b) =

min(a, b) always exists

glb(a, b) = min(a, b)

always exists

{lub(a, b), glb(a, b)} = {a, b}

Same holds for any total order

RelationsR⊆ : P(S) ↔ P(S)

lub(A, B) = A ∪B always exists

glb(A, B) = A ∩B always exists

lub(A, B) =

A ∪B always exists

lub(A, B) = A ∪B

always exists

glb(A, B) =

A ∩B always exists

glb(A, B) = A ∩B

always exists

R� is called a lattice, if for all a, b ∈ A,lub(a, b) and glb(a, b) exist

(Sometimes A itself called a lattice)

RelationsExamples:

any total order (e.g. R≤, R≥)

R| R⊆ for any S

RelationsExamples:

R⊆ for any S

RelationsExamples:

R| R⊆ for any S

a ∈ A is maximal: ∀x ∈ A : (a � x) ⇒ (a = x)

b ∈ A is minimal: ∀x ∈ A : (x � b) ⇒ (b = x)

Can have many maximal/minimal elements

a ∈ A is the greatest: ∀x ∈ A : x � a

b ∈ A is the least: ∀x ∈ A : b � x

Can have at most one greatest/least element

RelationsExamples:

R⊆ : P(S) ↔ P(S)

∅ least S greatest

R≤ : N ↔ N 0 least no greatest

R≥ : N ↔ N no least 0 greatest

R| : N ↔ N 1 least 0 greatest

RelationsExamples:

R⊆ : P(S) ↔ P(S) ∅ least

S greatest

RelationsExamples:

R⊆ : P(S) ↔ P(S) ∅ least S greatest

RelationsExamples:

R≤ : N ↔ N

0 least no greatest

RelationsExamples:

R≤ : N ↔ N 0 least

no greatest

RelationsExamples:

R≥ : N ↔ N

no least 0 greatest

RelationsExamples:

R≥ : N ↔ N no least

0 greatest

RelationsExamples:

R| : N ↔ N

1 least 0 greatest

RelationsExamples:

R| : N ↔ N 1 least

0 greatest

RelationsExamples:

xE y ⇐⇒ x is a descendant of y

minimal elements: childless people

no least elements

no maximal elements (Adam and Eve? GE?)

no greatest elements

no least elements

RelationsR| : N \ {0, 1} ↔ N \ {0, 1}

minimal elements: prime numbers

no least elements

no maximal ⇒ no greatest

RelationsR| : N \ {0, 1} ↔ N \ {0, 1}minimal elements: prime numbers

no least elements

RelationsR⊆ : P(S) \ {∅, S} ↔ P(S) \ {∅, S}

minimal elements: singletons

no least elements

maximal elements: singleton complements

RelationsR⊆ : P(S) \ {∅, S} ↔ P(S) \ {∅, S}minimal elements: singletons

no least elements

RelationsA partially ordered set may have no greatest or leastelement (even if the set is finite)

A finite, totally ordered set must have the greatest andthe least elements

A finite, partially ordered set must have maximal andminimal elements (but may not have the greatest andthe least)

A maximal or minimal element may not be unique

If a greatest or least element exists, it is unique

Functions

FunctionsA function from set A to set B is a relationRf : A ↔ B, where for every a ∈ A, there is a uniqueb ∈ B, such that afb

��

Functionsf : A → B ⇐⇒

(Rf : A ↔ B) ∧ (∀a ∈ A : ∃!b ∈ B : afb)

f maps A into B

Instead of afb, write f(a) = b or f : a 7→ b

f maps a to b

(Rf : A ↔ B) ∧ (∀a ∈ A : ∃!b ∈ B : afb)

f maps A into B

f maps a to b

(Rf : A ↔ B) ∧ (∀a ∈ A : ∃!b ∈ B : afb)

f maps A into B

f maps a to b

(Rf : A ↔ B) ∧ (∀a ∈ A : ∃!b ∈ B : afb)

f maps A into B

f maps a to b

Functionsf : A → B

A is the domain of f Domf = A

B is the co-domain of f Codomf = B

f : a 7→ b a ∈ A b ∈ B

b is the image of a

a is the pre-image of b

f : a 7→ b a ∈ A b ∈ B

b is the image of a

f : a 7→ b a ∈ A b ∈ B

b is the image of a

f : a 7→ b a ∈ A b ∈ B

b is the image of a

FunctionsExamples:

Identity function idA : A → AidA = {(a, a) | a ∈ A} = R=A

FunctionsExamples:

Identity function idA : A → AidA = {(a, a) | a ∈ A} = R=A

��

Functionsf : N → N f = {(m, n) ∈ N2 | m2 = n}

= {(0, 0), (1, 1), (2, 4), (3, 9), (4, 16), . . .}

g : {0, 1, 2} → N g = {(0, 0), (1, 1), (2, 4)}g ⊆ f Domg ⊆ Domf Codomg ⊆ Codomf

Function g called a restriction of f

Function f called an extension of g

= {(0, 0), (1, 1), (2, 4), (3, 9), (4, 16), . . .}g : {0, 1, 2} → N g = {(0, 0), (1, 1), (2, 4)}

g ⊆ f Domg ⊆ Domf Codomg ⊆ Codomf

= {(0, 0), (1, 1), (2, 4), (3, 9), (4, 16), . . .}g : {0, 1, 2} → N g = {(0, 0), (1, 1), (2, 4)}g ⊆ f

Domg ⊆ Domf Codomg ⊆ Codomf

= {(0, 0), (1, 1), (2, 4), (3, 9), (4, 16), . . .}g : {0, 1, 2} → N g = {(0, 0), (1, 1), (2, 4)}g ⊆ f Domg ⊆ Domf Codomg ⊆ Codomf

FunctionsAn infinite sequence of elements of set A is anyfunction a : N → A

Integer sequence N → Z, Boolean sequence N → B,etc.

Instead of a(k) write ak: a = (a0, a1, a2, a3, . . . )

Examples:

a : N → N a = (0, 1, 4, 9, 16, . . . )

b : N → B b = (F, T, F, T, F, . . . )

Examples:

a : N → N a = (0, 1, 4, 9, 16, . . . )

b : N → B b = (F, T, F, T, F, . . . )

Examples:

a : N → N a = (0, 1, 4, 9, 16, . . . )

b : N → B b = (F, T, F, T, F, . . . )

Examples:

a : N → N a = (0, 1, 4, 9, 16, . . . )

b : N → B b = (F, T, F, T, F, . . . )

Examples:

a : N → N a = (0, 1, 4, 9, 16, . . . )

b : N → B b = (F, T, F, T, F, . . . )

FunctionsLet n ∈ N

Nn = {x ∈ N | x < n} = {0, 1, 2, . . . , n− 1}

A finite sequence of elements of set A is any functiona : Nn → A

Example: g = (0, 1, 4)

Number n is the sequence length

Nn = {x ∈ N | x < n} = {0, 1, 2, . . . , n− 1}A finite sequence of elements of set A is any functiona : Nn → A

Example: g = (0, 1, 4)

Functionsf : A → B g : B → C

The composition of f and g f ◦ g : A → C

Same as relation composition Rf◦g

Rf , Rg functions =⇒ Rf◦g a function

∀a ∈ A : (f ◦ g)(a) = g(f(a))

FunctionsExamples:

f : Z → Z n 7→ n + 1

g : Z → Z n 7→ n2

(f ◦ f)(n) = (n + 1) + 1 = n + 2

(f ◦ g)(n) = (n + 1)2 = n2 + 2n + 1

(g ◦ f)(n) = n2 + 1

(g ◦ g)(n) = (n2)2 = n4

Note (f ◦ g) 6= (g ◦ f)

FunctionsExamples:

f : Z → Z n 7→ n + 1

g : Z → Z n 7→ n2

(f ◦ f)(n) =

(n + 1) + 1 = n + 2

(f ◦ g)(n) = (n + 1)2 = n2 + 2n + 1

(g ◦ f)(n) = n2 + 1

(g ◦ g)(n) = (n2)2 = n4

Note (f ◦ g) 6= (g ◦ f)

FunctionsExamples:

f : Z → Z n 7→ n + 1

g : Z → Z n 7→ n2

(f ◦ f)(n) = (n + 1) + 1 = n + 2

(f ◦ g)(n) = (n + 1)2 = n2 + 2n + 1

(g ◦ f)(n) = n2 + 1

(g ◦ g)(n) = (n2)2 = n4

Note (f ◦ g) 6= (g ◦ f)

FunctionsExamples:

f : Z → Z n 7→ n + 1

g : Z → Z n 7→ n2

(f ◦ f)(n) = (n + 1) + 1 = n + 2

(f ◦ g)(n) =

(n + 1)2 = n2 + 2n + 1

(g ◦ f)(n) = n2 + 1

(g ◦ g)(n) = (n2)2 = n4

Note (f ◦ g) 6= (g ◦ f)

FunctionsExamples:

f : Z → Z n 7→ n + 1

g : Z → Z n 7→ n2

(f ◦ f)(n) = (n + 1) + 1 = n + 2

(f ◦ g)(n) = (n + 1)2 = n2 + 2n + 1

(g ◦ f)(n) = n2 + 1

(g ◦ g)(n) = (n2)2 = n4

Note (f ◦ g) 6= (g ◦ f)

FunctionsExamples:

f : Z → Z n 7→ n + 1

g : Z → Z n 7→ n2

(f ◦ f)(n) = (n + 1) + 1 = n + 2

(f ◦ g)(n) = (n + 1)2 = n2 + 2n + 1

(g ◦ f)(n) =

n2 + 1

(g ◦ g)(n) = (n2)2 = n4

Note (f ◦ g) 6= (g ◦ f)

FunctionsExamples:

f : Z → Z n 7→ n + 1

g : Z → Z n 7→ n2

(f ◦ f)(n) = (n + 1) + 1 = n + 2

(f ◦ g)(n) = (n + 1)2 = n2 + 2n + 1

(g ◦ f)(n) = n2 + 1

(g ◦ g)(n) = (n2)2 = n4

Note (f ◦ g) 6= (g ◦ f)

FunctionsExamples:

f : Z → Z n 7→ n + 1

g : Z → Z n 7→ n2

(f ◦ f)(n) = (n + 1) + 1 = n + 2

(f ◦ g)(n) = (n + 1)2 = n2 + 2n + 1

(g ◦ f)(n) = n2 + 1

(g ◦ g)(n) =

(n2)2 = n4

Note (f ◦ g) 6= (g ◦ f)

FunctionsExamples:

f : Z → Z n 7→ n + 1

g : Z → Z n 7→ n2

(f ◦ f)(n) = (n + 1) + 1 = n + 2

(f ◦ g)(n) = (n + 1)2 = n2 + 2n + 1

(g ◦ f)(n) = n2 + 1

(g ◦ g)(n) = (n2)2 = n4

Note (f ◦ g) 6= (g ◦ f)

FunctionsExamples:

f : Z → Z n 7→ n + 1

g : Z → Z n 7→ n2

(f ◦ f)(n) = (n + 1) + 1 = n + 2

(f ◦ g)(n) = (n + 1)2 = n2 + 2n + 1

(g ◦ f)(n) = n2 + 1

(g ◦ g)(n) = (n2)2 = n4

Note (f ◦ g) 6= (g ◦ f)

The (functional) inverse of f f−1 : B → A

Same as relation inverse Rf−1, but may not be afunction

(we say “functional inverse may not exist”)

∀a ∈ A, b ∈ B : (f(a) = b) ⇔ (f−1(b) = a)

FunctionsExamples:

f : Z → Z n 7→ n + 1

f−1(n) = n− 1

g : Z → Z n 7→ n2

g−1does not exist (i.e. Rg−1 is not a function)

FunctionsExamples:

f : Z → Z n 7→ n + 1

f−1(n) =

n− 1

g : Z → Z n 7→ n2

FunctionsExamples:

f : Z → Z n 7→ n + 1

f−1(n) = n− 1

g : Z → Z n 7→ n2

FunctionsExamples:

f : Z → Z n 7→ n + 1

f−1(n) = n− 1

g : Z → Z n 7→ n2

FunctionsExamples:

f : Z → Z n 7→ n + 1

f−1(n) = n− 1

g : Z → Z n 7→ n2

does not exist (i.e. Rg−1 is not a function)

FunctionsExamples:

f : Z → Z n 7→ n + 1

f−1(n) = n− 1

g : Z → Z n 7→ n2

g−1 does not exist (i.e. Rg−1 is not a function)

The range of f is the set of all elements in B that havea pre-image in A

f(A) = {b ∈ B | ∃a ∈ A : f(a) = b}

��

FunctionsExamples:

f : N → N n 7→ n + 1

f(N) = N \ {0} = {1, 2, 3, 4, . . . }g : {0, 1, 2} → N n 7→ n2

g({0, 1, 2}) = {0, 1, 4}

FunctionsExamples:

f : N → N n 7→ n + 1

f(N) =

N \ {0} = {1, 2, 3, 4, . . . }g : {0, 1, 2} → N n 7→ n2

g({0, 1, 2}) = {0, 1, 4}

FunctionsExamples:

f : N → N n 7→ n + 1

f(N) = N \ {0} = {1, 2, 3, 4, . . . }

g : {0, 1, 2} → N n 7→ n2

g({0, 1, 2}) = {0, 1, 4}

FunctionsExamples:

f : N → N n 7→ n + 1

f(N) = N \ {0} = {1, 2, 3, 4, . . . }g : {0, 1, 2} → N n 7→ n2

g({0, 1, 2}) = {0, 1, 4}

FunctionsExamples:

f : N → N n 7→ n + 1

f(N) = N \ {0} = {1, 2, 3, 4, . . . }g : {0, 1, 2} → N n 7→ n2

g({0, 1, 2}) =

{0, 1, 4}

FunctionsExamples:

f : N → N n 7→ n + 1

f(N) = N \ {0} = {1, 2, 3, 4, . . . }g : {0, 1, 2} → N n 7→ n2

g({0, 1, 2}) = {0, 1, 4}

FunctionsFunction called surjective, if its range is the wholeco-domain

f : A� B ⇐⇒ ∀b ∈ B : ∃a ∈ A : f(a) = b

Also say f maps A onto B

f : A� B ⇐⇒ ∀b ∈ B : ∃a ∈ A : f(a) = b

��

f : A� B ⇐⇒ ∀b ∈ B : ∃a ∈ A : f(a) = b

��

FunctionsFunction called injective, if it maps different elementsto different elements

f : A� B ⇐⇒ ∀x, y : (f(x) = f(y)) ⇒ (x = y)

Also say f maps A into B one-to-one

f : A� B ⇐⇒ ∀x, y : (f(x) = f(y)) ⇒ (x = y)

��

f : A� B ⇐⇒ ∀x, y : (f(x) = f(y)) ⇒ (x = y)

��

FunctionsExamples:

f : Cards � {♠,♥,♣,♦} f : x 7→ suit of x

Function f surjective, but not injective

g : N� N g : m 7→ m2

Function g injective, but not surjective

FunctionsExamples:

g : N� N g : m 7→ m2

FunctionsExamples:

g : N� N g : m 7→ m2

FunctionsExamples:

g : N� N g : m 7→ m2

FunctionsFunction called bijective, if it is both surjective andinjective

f : A��B ⇐⇒ ∀b ∈ B : ∃!a ∈ A : f(a) = b

Also say f is a one-to-one correspondence between Aand B

f : A��B ⇐⇒ ∀b ∈ B : ∃!a ∈ A : f(a) = b

��

f : A��B ⇐⇒ ∀b ∈ B : ∃!a ∈ A : f(a) = b

��

FunctionsExamples:

idA : A��A a 7→ a

Function idA bijective for any set A

id−1

A = idA

f : Z��Z n 7→ n + 5

g : Z��Z n 7→ −n

FunctionsExamples:

id−1

A = idA

f : Z��Z n 7→ n + 5

g : Z��Z n 7→ −n

FunctionsExamples:

id−1

A = idA

f : Z��Z n 7→ n + 5

g : Z��Z n 7→ −n

FunctionsExamples:

id−1

A = idA

f : Z��Z n 7→ n + 5

g : Z��Z n 7→ −n

FunctionsExamples:

id−1

A = idA

f : Z��Z n 7→ n + 5

g : Z��Z n 7→ −n

Functionsf : Z��Z n 7→ n + 5

Function f bijective

f−1 : Z��Z m 7→ m− 5

f ◦ f−1 = f−1 ◦ f = idZ

For any set A, bijective f : A��A is a permutation

f−1 : Z��Z m 7→ m− 5

f ◦ f−1 = f−1 ◦ f = idZ

f−1 : Z��Z m 7→ m− 5

f ◦ f−1 = f−1 ◦ f = idZ

f−1 : Z��Z m 7→ m− 5

f ◦ f−1 = f−1 ◦ f = idZ

f−1 : Z��Z m 7→ m− 5

f ◦ f−1 = f−1 ◦ f = idZ

Functionsg : Z��Z n 7→ −n

Function g bijective

g−1 : Z��Z g−1 = g

g ◦ g−1 = g−1 ◦ g = g ◦ g = idZ

For any set A, a permutation g : A��A with g−1 = gis an involution

g−1 : Z��Z g−1 = g

g ◦ g−1 = g−1 ◦ g = g ◦ g = idZ

g−1 : Z��Z g−1 = g

g ◦ g−1 = g−1 ◦ g = g ◦ g = idZ

g−1 : Z��Z g−1 = g

g ◦ g−1 = g−1 ◦ g = g ◦ g = idZ

g−1 : Z��Z g−1 = g

g ◦ g−1 = g−1 ◦ g = g ◦ g = idZ

If f , g surjective, then f ◦ g surjective

If f , g injective, then f ◦ g injective

If f , g bijective, then f ◦ g bijective

FunctionsProve: if f bijective, then f−1 bijective.

Proof. Consider Rf : A ↔ B, Rf−1 : B ↔ A

f bijective =⇒ ∀b ∈ B : ∃!a ∈ A : afb =⇒∀b ∈ B : ∃!a ∈ A : b(f−1)a =⇒ f−1 a function

f a function =⇒ ∀a ∈ A : ∃!b ∈ B : afb =⇒∀a ∈ A : ∃!b ∈ B : b(f−1)a =⇒ f−1 bijective

f bijective =⇒

∀b ∈ B : ∃!a ∈ A : afb =⇒∀b ∈ B : ∃!a ∈ A : b(f−1)a =⇒ f−1 a function

f bijective =⇒ ∀b ∈ B : ∃!a ∈ A : afb =⇒

∀b ∈ B : ∃!a ∈ A : b(f−1)a =⇒ f−1 a function

f bijective =⇒ ∀b ∈ B : ∃!a ∈ A : afb =⇒∀b ∈ B : ∃!a ∈ A : b(f−1)a =⇒

f−1 a function

f a function =⇒

∀a ∈ A : ∃!b ∈ B : afb =⇒∀a ∈ A : ∃!b ∈ B : b(f−1)a =⇒ f−1 bijective

f a function =⇒ ∀a ∈ A : ∃!b ∈ B : afb =⇒

∀a ∈ A : ∃!b ∈ B : b(f−1)a =⇒ f−1 bijective

f a function =⇒ ∀a ∈ A : ∃!b ∈ B : afb =⇒∀a ∈ A : ∃!b ∈ B : b(f−1)a =⇒

f−1 bijective

FunctionsProve: if both f and f−1 are functions, then both fand f−1 are bijective

Proof (sketch). Let f : A → B, f−1 : B → A.

f a function ⇒ f−1 surjective

f a function ⇒ f−1 injective

f = (f−1)−1 ⇒ f surjective and injective

To prove f : A��B, only need f−1 : B → A

f ◦ f−1 = idA f−1 ◦ f = idB

FunctionsS — any set A ⊆ S B = {F, T}

Indicator function of A χA : S → B

∀x ∈ S : χA(x) =

T if x ∈ A

F if x 6∈ A

Set of all subsets of S P(S) = {A | A ⊆ S}Set of all Boolean functions on S:

B(S) = {f | f : S → B}χ : P(S)��B(S) ∀A ⊆ S : A 7→ χA

Subsets of S�� Boolean functions on S

FunctionsS — any set A ⊆ S B = {F, T}Indicator function of A χA : S → B

∀x ∈ S : χA(x) =

T if x ∈ A

F if x 6∈ A

∀x ∈ S : χA(x) =

T if x ∈ A

F if x 6∈ A

∀x ∈ S : χA(x) =

T if x ∈ A

F if x 6∈ A

Set of all subsets of S P(S) = {A | A ⊆ S}

Set of all Boolean functions on S:B(S) = {f | f : S → B}

χ : P(S)��B(S) ∀A ⊆ S : A 7→ χA

∀x ∈ S : χA(x) =

T if x ∈ A

F if x 6∈ A

B(S) = {f | f : S → B}

χ : P(S)��B(S) ∀A ⊆ S : A 7→ χA

∀x ∈ S : χA(x) =

T if x ∈ A

F if x 6∈ A

∀x ∈ S : χA(x) =

T if x ∈ A

F if x 6∈ A

FunctionsSets A, B are called equinumerous, if there is abijection between A and B

A ∼= B ⇐⇒ ∃f : A��B

A, B ⊆ S =⇒ R∼= an equivalence on P(S)

A ∼= B ⇐⇒ ∃f : A��B

Functionsn ∈ N Nn = {x ∈ N | x < n}

N0 = ∅ N1 = {0} N2 = {0, 1} . . .

Nn = {0, 1, 2, . . . , n− 1}Set A called finite, if A ∼= Nn for some n ∈ N

Otherwise, the set is called infinite

Functionsn ∈ N Nn = {x ∈ N | x < n}N0 = ∅ N1 = {0} N2 = {0, 1} . . .

Nn = {0, 1, 2, . . . , n− 1}

Set A called finite, if A ∼= Nn for some n ∈ N

FunctionsFor every finite set A, there is a unique n ∈ N, suchthat A ∼= Nn

Proof.

Let f : A��Nk. Then f−1 : Nk��A.

Let g : A��Nl.

f−1 ◦ g : Nk��Nl. Therefore k = l.

Number n called the cardinality of A |A| = n

A, B finite, A ∼= B =⇒ |A| = |B|

Proof.

Let g : A��Nl.

A, B finite, A ∼= B =⇒ |A| = |B|

Proof.

Let g : A��Nl.

A, B finite, A ∼= B =⇒ |A| = |B|

Proof.

Let g : A��Nl.

A, B finite, A ∼= B =⇒ |A| = |B|

Proof.

Let g : A��Nl.

A, B finite, A ∼= B =⇒ |A| = |B|

Proof.

Let g : A��Nl.

Number n called the cardinality of A

|A| = n

A, B finite, A ∼= B =⇒ |A| = |B|

Proof.

Let g : A��Nl.

A, B finite, A ∼= B =⇒ |A| = |B|

Proof.

Let g : A��Nl.

A, B finite, A ∼= B =⇒ |A| = |B|

FunctionsN, Z, N2, N3, P(N), Neven — infinite

An infinite set is called countable, if it isequinumerous with N

In particular, N itself is countable

FunctionsProve: set N+ = N \ {0} is countable.

Proof. Let f : N → N+ ∀n : n 7→ n + 1

∀n ∈ N+ : n = (n− 1) + 1 = f(n− 1)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (m + 1 6= n + 1)

Hence f injective

f surjective and injective =⇒ f bijective

Proof. Let f : N → N+ ∀n : n 7→ n + 1

∀n ∈ N+ : n = (n− 1) + 1 = f(n− 1)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (m + 1 6= n + 1)

Hence f injective

Proof. Let f : N → N+ ∀n : n 7→ n + 1

∀n ∈ N+ : n = (n− 1) + 1 = f(n− 1)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (m + 1 6= n + 1)

Hence f injective

Proof. Let f : N → N+ ∀n : n 7→ n + 1

∀n ∈ N+ : n = (n− 1) + 1 = f(n− 1)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (m + 1 6= n + 1)

Hence f injective

Proof. Let f : N → N+ ∀n : n 7→ n + 1

∀n ∈ N+ : n = (n− 1) + 1 = f(n− 1)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (m + 1 6= n + 1)

Hence f injective

Proof. Let f : N → N+ ∀n : n 7→ n + 1

∀n ∈ N+ : n = (n− 1) + 1 = f(n− 1)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (m + 1 6= n + 1)

Hence f injective

Proof. Let f : N → N+ ∀n : n 7→ n + 1

∀n ∈ N+ : n = (n− 1) + 1 = f(n− 1)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (m + 1 6= n + 1)

Hence f injective

FunctionsN+ ⊆ N

0 1 2 3 4 5 6 7 · · ·l l l l l l l l1 2 3 4 5 6 7 8 · · ·

N+ ∼= N

A part is of the same size as the whole!

FunctionsN+ ⊆ N

0 1 2 3 4 5 6 7 · · ·l l l l l l l l1 2 3 4 5 6 7 8 · · ·

N+ ∼= N

FunctionsN+ ⊆ N

0 1 2 3 4 5 6 7 · · ·l l l l l l l l1 2 3 4 5 6 7 8 · · ·

N+ ∼= N

FunctionsN+ ⊆ N

0 1 2 3 4 5 6 7 · · ·l l l l l l l l1 2 3 4 5 6 7 8 · · ·

N+ ∼= N

FunctionsProve: set Neven = {0, 2, 4, 6, . . . } is countable.

Proof. Let f : N → Neven ∀n : n 7→ 2n

∀n ∈ Neven : n = 2 · n/2 = f(n/2)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (2m 6= 2n)

Hence f injective

∀n ∈ Neven : n = 2 · n/2 = f(n/2)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (2m 6= 2n)

Hence f injective

∀n ∈ Neven : n = 2 · n/2 = f(n/2)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (2m 6= 2n)

Hence f injective

∀n ∈ Neven : n = 2 · n/2 = f(n/2)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (2m 6= 2n)

Hence f injective

∀n ∈ Neven : n = 2 · n/2 = f(n/2)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (2m 6= 2n)

Hence f injective

∀n ∈ Neven : n = 2 · n/2 = f(n/2)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (2m 6= 2n)

Hence f injective

∀n ∈ Neven : n = 2 · n/2 = f(n/2)

Hence f surjective

∀m, n ∈ N : (m 6= n) ⇒ (2m 6= 2n)

Hence f injective

FunctionsNeven ⊆ N

0 1 2 3 4 5 6 7 · · ·l l l l l l l l0 2 4 6 8 10 12 14 · · ·

Neven∼= N

In general, any subset of a countable set is finite orcountable. Any quotient set of a countable set is finiteor countable.

0 1 2 3 4 5 6 7 · · ·l l l l l l l l0 2 4 6 8 10 12 14 · · ·

Neven∼= N

0 1 2 3 4 5 6 7 · · ·l l l l l l l l0 2 4 6 8 10 12 14 · · ·

Neven∼= N

0 1 2 3 4 5 6 7 · · ·l l l l l l l l0 2 4 6 8 10 12 14 · · ·

Neven∼= N

In general, any subset of a countable set is finite orcountable.

Any quotient set of a countable set is finiteor countable.

0 1 2 3 4 5 6 7 · · ·l l l l l l l l0 2 4 6 8 10 12 14 · · ·

Neven∼= N

FunctionsProve: set Z is countable.

Proof. Let f : N → Z

∀n : n 7→{

(n + 1)/2 if n odd−n/2 if n even

· · · −4 −3 −2 −1 0 1 2 3 4 · · ·l l l l l l l l l

· · · 8 6 4 2 0 1 3 5 7 · · ·f bijective

∀n : n 7→{

· · · −4 −3 −2 −1 0 1 2 3 4 · · ·l l l l l l l l l

· · · 8 6 4 2 0 1 3 5 7 · · ·f bijective

∀n : n 7→{

· · · −4 −3 −2 −1 0 1 2 3 4 · · ·l l l l l l l l l

· · · 8 6 4 2 0 1 3 5 7 · · ·f bijective

∀n : n 7→{

· · · −4 −3 −2 −1 0 1 2 3 4 · · ·l l l l l l l l l

· · · 8 6 4 2

1 3 5 7 · · ·f bijective

∀n : n 7→{

· · · −4 −3 −2 −1 0 1 2 3 4 · · ·l l l l l l l l l

· · · 8 6 4 2

3 5 7 · · ·f bijective

∀n : n 7→{

· · · −4 −3 −2 −1 0 1 2 3 4 · · ·l l l l l l l l l

· · · 8 6 4

3 5 7 · · ·f bijective

∀n : n 7→{

· · · −4 −3 −2 −1 0 1 2 3 4 · · ·l l l l l l l l l

· · · 8 6 4

2 0 1 3

5 7 · · ·f bijective

∀n : n 7→{

· · · −4 −3 −2 −1 0 1 2 3 4 · · ·l l l l l l l l l

· · · 8 6

4 2 0 1 3

5 7 · · ·f bijective

∀n : n 7→{

· · · −4 −3 −2 −1 0 1 2 3 4 · · ·l l l l l l l l l

· · · 8 6 4 2 0 1 3 5 7 · · ·

f bijective

∀n : n 7→{

· · · −4 −3 −2 −1 0 1 2 3 4 · · ·l l l l l l l l l

· · · 8 6 4 2 0 1 3 5 7 · · ·f bijective

FunctionsProve: Set N2 = N× N is countable.

Proof idea:

0 1 2 3 4

0 0 1 3 6 10

1 2 4 7 11 ·2 5 8 12 · ·3 9 13 · · ·4 14 · · · ·

Proof idea:

0 1 2 3 4

0 1 3 6 10

2 4 7 11 ·

5 8 12 · ·

9 13 · · ·

14 · · · ·

Proof idea:

0 1 2 3 4

1 3 6 10

2 4 7 11 ·

5 8 12 · ·

9 13 · · ·

14 · · · ·

Proof idea:

0 1 2 3 4

3 6 10

4 7 11 ·

5 8 12 · ·

9 13 · · ·

14 · · · ·

Proof idea:

0 1 2 3 4

0 0 1 3

7 11 ·

8 12 · ·

9 13 · · ·

14 · · · ·

Proof idea:

0 1 2 3 4

0 0 1 3 6

1 2 4 7

12 · ·

13 · · ·

14 · · · ·

Proof idea:

0 1 2 3 4

0 0 1 3 6 10

1 2 4 7 11

2 5 8 12

3 9 13

· · ·

· · · ·

Proof idea:

0 1 2 3 4

0 0 1 3 6 10

1 2 4 7 11 ·2 5 8 12 · ·3 9 13 · · ·4 14 · · · ·

FunctionsProve: Set N3 = N× N× N is countable.

Proof. N3 = (N× N)× N ∼= N× N ∼= N

In general, the Cartesian product of a finite number ofcountable sets is countable

Not true for an infinite Cartesian product!

Proof. N3 = (N× N)× N ∼= N× N ∼= N

FunctionsA digression on rational numbers

Fractions 1/2, −3/4, 355/113

Similar to Z2, but:

1/2 = 3/6 −5 = −10/2 = 30/(−6) . . .

Fractions 1/2, −3/4, 355/113

Similar to Z2, but:

1/2 = 3/6 −5 = −10/2 = 30/(−6) . . .

Fractions 1/2, −3/4, 355/113

Similar to Z2, but:

1/2 = 3/6 −5 = −10/2 = 30/(−6) . . .

Functionsa/b = c/d ⇐⇒ a · d = b · c b, d 6= 0

R∼ : Z2 ↔ Z2 (a, b) ∼ (c, d) ⇐⇒ a · d = b · cThe rational numbers: Q = Z2/R∼

R∼ : Z2 ↔ Z2 (a, b) ∼ (c, d) ⇐⇒ a · d = b · c

The rational numbers: Q = Z2/R∼

R∼ : Z2 ↔ Z2 (a, b) ∼ (c, d) ⇐⇒ a · d = b · cThe rational numbers: Q = Z2/R∼

FunctionsSet Q is infinite, and also dense: for any two distinctrationals, there is a rational in between

0 11/21/3

∀a, b ∈ Q :(

(a < b) ⇒ ∃x ∈ Q : a < x < b)

Still, set Q is countable

Z2 ∼= N =⇒ Q = Z2/R∼ ∼= N

1/21/3

∀a, b ∈ Q :(

(a < b) ⇒ ∃x ∈ Q : a < x < b)

Z2 ∼= N =⇒ Q = Z2/R∼ ∼= N

∀a, b ∈ Q :(

(a < b) ⇒ ∃x ∈ Q : a < x < b)

Z2 ∼= N =⇒ Q = Z2/R∼ ∼= N

∀a, b ∈ Q :(

(a < b) ⇒ ∃x ∈ Q : a < x < b)

Z2 ∼= N =⇒ Q = Z2/R∼ ∼= N

∀a, b ∈ Q :(

(a < b) ⇒ ∃x ∈ Q : a < x < b)

Z2 ∼= N =⇒ Q = Z2/R∼ ∼= N

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∀a, b ∈ Q :(

(a < b) ⇒ ∃x ∈ Q : a < x < b)

Z2 ∼= N =⇒ Q = Z2/R∼ ∼= N

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∀a, b ∈ Q :(

(a < b) ⇒ ∃x ∈ Q : a < x < b)

Z2 ∼= N =⇒ Q = Z2/R∼ ∼= N

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∀a, b ∈ Q :(

(a < b) ⇒ ∃x ∈ Q : a < x < b)

Z2 ∼= N =⇒ Q = Z2/R∼ ∼= N

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∀a, b ∈ Q :(

(a < b) ⇒ ∃x ∈ Q : a < x < b)

Z2 ∼= N =⇒ Q = Z2/R∼ ∼= N

FunctionsSets N, Z, Q countable: N ∼= Z ∼= Q

N ∼= N2 ∼= N3 ∼= N× Z×Q ∼= · · ·Are there any uncountable sets?

N ∼= N2 ∼= N3 ∼= N× Z×Q ∼= · · ·

Are there any uncountable sets?

N ∼= N2 ∼= N3 ∼= N× Z×Q ∼= · · ·Are there any uncountable sets?

FunctionsCantor’s Theorem. For all sets A, A 6∼= P(A).

Proof. Suppose ∃f : A��P(A).

Let D = {a ∈ A | a 6∈ f(a)}D ⊆ A =⇒ D ∈ P(A) =⇒ ∃d ∈ A : f(d) = D

d ∈ D — true or false?

Case d ∈ D =⇒ d 6∈ f(d) = DCase d 6∈ D = f(d) =⇒ d ∈ D

Contradiction! Bijection f cannot exist.

Let D = {a ∈ A | a 6∈ f(a)}

D ⊆ A =⇒ D ∈ P(A) =⇒ ∃d ∈ A : f(d) = D

Case d ∈ D =⇒ d 6∈ f(d) = D

Case d 6∈ D = f(d) =⇒ d ∈ D

FunctionsCorollary. Set P(N) is uncountable.

P(N) ∼= B(N) = {f : N → B} =⇒ B(N) 6∼= N

B = {F, T} ∼= {0, 1} = N2 ⊆ N

B(N) ∼= {f : N → N2} ⊆{f : N → N} = N× N× · · ·

B(N) uncountable =⇒ N× N× · · · uncountable

P(N) ∼= B(N) = {f : N → B} =⇒ B(N) 6∼= N

B = {F, T} ∼= {0, 1} = N2 ⊆ N

B(N) ∼= {f : N → N2} ⊆{f : N → N} = N× N× · · ·

P(N) ∼= B(N) = {f : N → B} =⇒ B(N) 6∼= N

B = {F, T} ∼= {0, 1} = N2 ⊆ N

B(N) ∼= {f : N → N2} ⊆{f : N → N} = N× N× · · ·

P(N) ∼= B(N) = {f : N → B} =⇒ B(N) 6∼= N

B = {F, T} ∼= {0, 1} = N2 ⊆ N

B(N) ∼= {f : N → N2} ⊆{f : N → N} = N× N× · · ·

P(N) ∼= B(N) = {f : N → B} =⇒ B(N) 6∼= N

B = {F, T} ∼= {0, 1} = N2 ⊆ N

B(N) ∼= {f : N → N2} ⊆{f : N → N} = N× N× · · ·

FunctionsA digression on real numbers

Rationals (20, −3/5, . . . ), irrationals (√

2, π, . . . )

Definition: Dedekind cuts of Q

Q = Q0 ∪Q1 ∀x ∈ Q0, y ∈ Q1 : x < y

Every Dedekind cut defines a real number

2, π, . . . )

Q = Q0 ∪Q1 ∀x ∈ Q0, y ∈ Q1 : x < y

2, π, . . . )

Q = Q0 ∪Q1 ∀x ∈ Q0, y ∈ Q1 : x < y

2, π, . . . )

Q = Q0 ∪Q1 ∀x ∈ Q0, y ∈ Q1 : x < y

2, π, . . . )

Q = Q0 ∪Q1 ∀x ∈ Q0, y ∈ Q1 : x < y

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2, π, . . . )

Q = Q0 ∪Q1 ∀x ∈ Q0, y ∈ Q1 : x < y

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FunctionsThe real numbers R = {all Dedekind cuts of Q}

Irrationals are “gaps between rationals”

Example: π = 3.1415926536 . . .

Number π defined by

Q0 = {3, 3.1, 3.14, 3.141, 3.1415, 3.141592, . . .}Q1 = {4, 3.2, 3.15, 3.142, 3.1416, 3.141593, . . .}We approximate real by rationals using a positionalnumber system (decimal, binary, etc.)

FunctionsThe real numbers R = {all Dedekind cuts of Q}Irrationals are “gaps between rationals”

Example: π = 3.1415926536 . . .

Q0 = {3, 3.1, 3.14, 3.141, 3.1415, 3.141592, . . .}Q1 = {4, 3.2, 3.15, 3.142, 3.1416, 3.141593, . . .}

We approximate real by rationals using a positionalnumber system (decimal, binary, etc.)

Example: π = 3.1415926536 . . .

FunctionsIs set R countable?

Consider [0, 1] = {a ∈ R : 0 ≤ a < 1}Decimal representation of a: sequence N → N10

For example, π − 3 = .141592

f(0) = 1 f(1) = 4 f(2) = 1 f(3) = 5 . . .

(For simplicity, ignore .1415000 . . . = .1414999 . . .)

Consider [0, 1] = {a ∈ R : 0 ≤ a < 1}

Decimal representation of a: sequence N → N10

For example, π − 3 = .141592

f(0) = 1 f(1) = 4 f(2) = 1 f(3) = 5 . . .

For example, π − 3 = .141592

f(0) = 1 f(1) = 4 f(2) = 1 f(3) = 5 . . .

For example, π − 3 = .141592

f(0) = 1 f(1) = 4 f(2) = 1 f(3) = 5 . . .

For example, π − 3 = .141592

f(0) = 1 f(1) = 4 f(2) = 1 f(3) = 5 . . .

For example, π − 3 = .141592

f(0) = 1 f(1) = 4 f(2) = 1 f(3) = 5 . . .

FunctionsB = {F, T} ∼= {0, 1} = N2 ⊆ N10 =⇒

B(N) = {f : N → B} ∼={f : N → N2} ⊆ {f : N → N10} ∼= [0, 1]

B(N) uncountable, hence [0, 1] uncountable

Therefore, R uncountable

FunctionsB = {F, T} ∼= {0, 1} = N2 ⊆ N10 =⇒B(N) = {f : N → B} ∼=

{f : N → N2} ⊆ {f : N → N10} ∼= [0, 1]

FunctionsFinite sets:

n elements, n− 1 gaps

Infinite sets:

R are gaps in Q, but R is much bigger than Q

Weird!

Infinite sets:

Weird!

Infinite sets:

Weird!

Infinite sets:

Weird!

Infinite sets:

Weird!

Infinite sets:

Weird!

FunctionsN, P(N), P(P(N)), . . . — uncountable

All of different cardinalities — and there any manymore. . .

So much more they don’t even form a set!

Induction

InductionNatural numbers: N = {0, 1, 2, 3, 4, 5, 6, 7, . . . }

God created the natural numbers, all the restis the work of man.

L. Kronecker (1823–1891)

InductionNatural numbers: N = {0, 1, 2, 3, 4, 5, 6, 7, . . . }

God created the natural numbers, all the restis the work of man.

L. Kronecker (1823–1891)

InductionThe only possible definition of N is self-referential:

• 0 ∈ N

• for all x ∈ N next(x) ∈ N

• everything else 6∈ N

This is an inductive definition

• 0 ∈ N

InductionStructure of inductive definition:

• induction base

• inductive step(s)

• completeness (sometimes implicit)

• induction base

InductionFor example, a queue:

• the empty set ∅ is a queue;

• a queue with a new person behind is still a queue

• every queue is formed in this way

If we know what “behind” means, all works!

InductionAnother example: Boolean statements

• F , T are statements

• if A, B are statements, then ¬A, A ∧B, A ∨B,A ⇒ B, A ⇔ B are statements

• there are no other statements

InductionIn general:

• induction base: initial objects

• inductive step(s): ways to make new objects

• completeness: no other objects allowed!

InductionInductive proofs: “the domino principle”

Need to prove ∀x ∈ N : P (x) for some P

• induction base: P (0)

• inductive step: ∀x ∈ N : P (x) ⇒ P (next(x))(here P (x) is the induction hypothesis)

• by completeness, P (x) true for all x ∈ N

InductionExample: plane colouring

Consider n lines in the plane.

Can always colour regions like a chessboard.

InductionProof.

Induction base: n = 0.

Paint the plane white.

InductionProof.

Induction base: n = 0. Paint the plane white.

InductionProof.

Induction base: n = 0. Paint the plane white.

InductionInductive step. Suppose can colour for n lines.

Need to color for n + 1 lines.

Add another line, invert all colours on one side.

By induction, can colour for all n.

InductionProve: Any postage ≥ 8p can be paid by 3p and 5pstamps.

Proof. Induction base: 8 = 3 + 5.

Inductive step. Suppose can pay n (n ≥ 8).

We need: can pay n + 1

Case 1: have used a 5. Replace 5 → 3 + 3.

Case 2: have only used 3s. Since n ≥ 8, there are atleast three 3s. Replace 3 + 3 + 3 → 5 + 5.

In both cases have paid n + 1.

By induction, can pay any n ≥ 8.

Proof.

Induction base: 8 = 3 + 5.

Proof. Induction base:

8 = 3 + 5.

Case 2: have only used 3s.

Since n ≥ 8, there are atleast three 3s. Replace 3 + 3 + 3 → 5 + 5.

Case 2: have only used 3s. Since n ≥ 8, there are atleast three 3s.

Replace 3 + 3 + 3 → 5 + 5.

InductionProve: For any finite set A,

|A| = n =⇒ |P(A)| = 2n

|A| = 0 =⇒ A = ∅ =⇒P(A) = {∅} =⇒ |P(A)| = 1 = 20

Inductive step. Suppose it holds for a given n:for all B |B| = n =⇒ |P(B)| = 2n

We need:for all A |A| = n + 1 =⇒ |P(A)| = 2n+1

|A| = n =⇒ |P(A)| = 2n

Proof.

Induction base:

|A| = 0 =⇒ A = ∅ =⇒P(A) = {∅} =⇒ |P(A)| = 1 = 20

|A| = n =⇒ |P(A)| = 2n

|A| = 0 =⇒ A = ∅ =⇒P(A) = {∅} =⇒ |P(A)| = 1 = 20

|A| = n =⇒ |P(A)| = 2n

|A| = 0 =⇒

A = ∅ =⇒P(A) = {∅} =⇒ |P(A)| = 1 = 20

|A| = n =⇒ |P(A)| = 2n

|A| = 0 =⇒ A = ∅ =⇒

P(A) = {∅} =⇒ |P(A)| = 1 = 20

|A| = n =⇒ |P(A)| = 2n

|A| = 0 =⇒ A = ∅ =⇒P(A) = {∅} =⇒

|P(A)| = 1 = 20

|A| = n =⇒ |P(A)| = 2n

|A| = 0 =⇒ A = ∅ =⇒P(A) = {∅} =⇒ |P(A)| = 1 = 20

|A| = n =⇒ |P(A)| = 2n

|A| = 0 =⇒ A = ∅ =⇒P(A) = {∅} =⇒ |P(A)| = 1 = 20

InductionLet a ∈ A B = A \ {a}.

We have |B| = n, therefore |P(B)| = 2n.

Let P = P(B) Q = {X ∪ {a} | X ∈ P}P(A) = P ∪Q P ∩Q = ∅ |P | = |Q| = 2n

Hence |P(A)| = |P |+ |Q| = 2n + 2n = 2n · 2 = 2n+1

By induction, statement true for all A

Hence |P(A)| = |P |+ |Q| = 2n + 2n = 2n · 2 = 2n+1

Let P = P(B)

Q = {X ∪ {a} | X ∈ P}P(A) = P ∪Q P ∩Q = ∅ |P | = |Q| = 2n

Hence |P(A)| = |P |+ |Q| = 2n + 2n = 2n · 2 = 2n+1

Let P = P(B) Q = {X ∪ {a} | X ∈ P}

P(A) = P ∪Q P ∩Q = ∅ |P | = |Q| = 2n

Hence |P(A)| = |P |+ |Q| = 2n + 2n = 2n · 2 = 2n+1

Let P = P(B) Q = {X ∪ {a} | X ∈ P}P(A) = P ∪Q

P ∩Q = ∅ |P | = |Q| = 2n

Hence |P(A)| = |P |+ |Q| = 2n + 2n = 2n · 2 = 2n+1

Let P = P(B) Q = {X ∪ {a} | X ∈ P}P(A) = P ∪Q P ∩Q = ∅

|P | = |Q| = 2n

Hence |P(A)| = |P |+ |Q| = 2n + 2n = 2n · 2 = 2n+1

InductionConsider induction on n ∈ N

Inductive step: P (n) ⇒ P (n + 1)

Suppose can only prove:(

P (0) ∧ P (1) ∧ · · · ∧ P (n− 1))

⇒ P (n)

(Also covers T ⇒ P (0))

So-called “strong” induction(in fact, the implication has become weaker!)

Does P (n) still hold for all n?

P (0) ∧ P (1) ∧ · · · ∧ P (n− 1))

⇒ P (n)

P (0) ∧ P (1) ∧ · · · ∧ P (n− 1))

⇒ P (n)

P (0) ∧ P (1) ∧ · · · ∧ P (n− 1))

⇒ P (n)

P (0) ∧ P (1) ∧ · · · ∧ P (n− 1))

⇒ P (n)

P (0) ∧ P (1) ∧ · · · ∧ P (n− 1))

⇒ P (n)

InductionLet Q(n) ⇐⇒ P (0) ∧ P (1) ∧ · · · ∧ P (n)

We need: ∀n ∈ N : Q(n) ( =⇒ ∀n ∈ N : P (n))

T ⇒ P (0) ⇐⇒ P (0) ⇐⇒ Q(0) induction base

Q(n) ⇐⇒ P (0)∧P (1)∧ · · · ∧P (n) =⇒ P (n+1)(

P (0) ∧ · · · ∧ P (n))

∧ P (n + 1) ⇐⇒ Q(n + 1)

Hence Q(n) ⇒ Q(n + 1) inductive step

By induction, ∀n : Q(n), therefore ∀n : P (n)

We need: ∀n ∈ N : Q(n) ( =⇒ ∀n ∈ N : P (n))

Q(n) ⇐⇒ P (0)∧P (1)∧ · · · ∧P (n) =⇒ P (n+1)(

P (0) ∧ · · · ∧ P (n))

∧ P (n + 1) ⇐⇒ Q(n + 1)

We need: ∀n ∈ N : Q(n) ( =⇒ ∀n ∈ N : P (n))

T ⇒ P (0) ⇐⇒ P (0) ⇐⇒ Q(0)

induction base

Q(n) ⇐⇒ P (0)∧P (1)∧ · · · ∧P (n) =⇒ P (n+1)(

P (0) ∧ · · · ∧ P (n))

∧ P (n + 1) ⇐⇒ Q(n + 1)

We need: ∀n ∈ N : Q(n) ( =⇒ ∀n ∈ N : P (n))

Q(n) ⇐⇒ P (0)∧P (1)∧ · · · ∧P (n) =⇒ P (n+1)(

P (0) ∧ · · · ∧ P (n))

∧ P (n + 1) ⇐⇒ Q(n + 1)

We need: ∀n ∈ N : Q(n) ( =⇒ ∀n ∈ N : P (n))

Q(n) ⇐⇒ P (0)∧P (1)∧ · · · ∧P (n) =⇒ P (n+1)

P (0) ∧ · · · ∧ P (n))

∧ P (n + 1) ⇐⇒ Q(n + 1)

We need: ∀n ∈ N : Q(n) ( =⇒ ∀n ∈ N : P (n))

Q(n) ⇐⇒ P (0)∧P (1)∧ · · · ∧P (n) =⇒ P (n+1)(

P (0) ∧ · · · ∧ P (n))

∧ P (n + 1) ⇐⇒ Q(n + 1)

We need: ∀n ∈ N : Q(n) ( =⇒ ∀n ∈ N : P (n))

Q(n) ⇐⇒ P (0)∧P (1)∧ · · · ∧P (n) =⇒ P (n+1)(

P (0) ∧ · · · ∧ P (n))

∧ P (n + 1) ⇐⇒ Q(n + 1)

Hence Q(n) ⇒ Q(n + 1)

inductive step

We need: ∀n ∈ N : Q(n) ( =⇒ ∀n ∈ N : P (n))

Q(n) ⇐⇒ P (0)∧P (1)∧ · · · ∧P (n) =⇒ P (n+1)(

P (0) ∧ · · · ∧ P (n))

∧ P (n + 1) ⇐⇒ Q(n + 1)

We need: ∀n ∈ N : Q(n) ( =⇒ ∀n ∈ N : P (n))

Q(n) ⇐⇒ P (0)∧P (1)∧ · · · ∧P (n) =⇒ P (n+1)(

P (0) ∧ · · · ∧ P (n))

∧ P (n + 1) ⇐⇒ Q(n + 1)

InductionProve: ∀n ∈ N : n ≥ 2 ⇒ n is divisible by a prime

Proof. Induction base: 2 | 2 prime

Inductive step. Suppose true ∀m < n. True for n?

Case 1: n prime n | nCase 2: ∃m < n : m | nThen ∃p : (p prime) ∧ (p | m)

(p | m) ∧ (m | n) =⇒ p | nIn both cases n divisible by a prime

By induction, all n ≥ 2 divisible by a prime

Proof.

Induction base: 2 | 2 prime

2 | 2 prime

Case 1: n prime n | n

Case 2: ∃m < n : m | nThen ∃p : (p prime) ∧ (p | m)

Case 1: n prime n | nCase 2: ∃m < n : m | n

Then ∃p : (p prime) ∧ (p | m)

(p | m) ∧ (m | n) =⇒ p | n

In both cases n divisible by a prime

Graphs

The Königsberg Bridges (L. Euler, 1707–1783)

A tour crossing every bridge exactly once?

Graphs

The Königsberg Bridges graph

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4 • nodes (islands) 7 ◦ nodes (bridges)

Graphs

Wolf, goat and cabbage

Farmer wants to take W , G, C across the riverCan only take one item at a time

W eats G, G eats C — farmer must keep an eye

Graphs

� FWC

Graphs

� FWC

Graphs

� FWC

FWG �

Graphs

� FWC

FWG �

Graphs

� FWC

FWG �

Graphs

� FWC

FWG �

Graphs

� FWC

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Graphs

� FWC

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Graphs

� FWC

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Graphs

Houses and wells

Each of 3 houses needs a path to each of 3 wells

Paths must not cross

Graphs

Houses and wells

Graphs

Houses and wells

Graphs

Houses and wells

Graphs

Houses and wells

Graphs

V — finite set, elements called nodes R⇀ : V ↔ V

R⇀ called irreflexive, if ∀a ∈ A : ¬(a ⇀ a)

R⇀ called symmetric, if ∀a, b ∈ A : a ⇀ b ⇒ b ⇀ a

A graph is an irreflexive, symmetric relationE = R⇀ : V ↔ V

Nodes u, v called adjacent, if u ⇀ v

Pairs in E called edges

Common notation: graph G = (V, E)

Graphs

The complete graph on V : K(V ) = (V, E)where E = {(u, v) ∈ V 2 | u 6= v}

The complete graph on n nodes: K(n)

Graphs

Different graphs may be “similar”

Two graphs are called isomorphic, if there is abijection between their node sets, which preserves theedges

G1 = (V1, E1) G2 = (V2, E2)

G1∼= G2 ⇐⇒ ∃f : V1��V2 :

∀u, v ∈ V1 : (u, v) ∈ E1 ⇔ (f(u), f(v)) ∈ E2

Bijection f is the isomorphism between G1, G2

Graphs

G1 = (V1, E1) G2 = (V2, E2)

G1∼= G2 ⇐⇒ ∃f : V1��V2 :

∀u, v ∈ V1 : (u, v) ∈ E1 ⇔ (f(u), f(v)) ∈ E2

Graphs

G1 = (V1, E1) G2 = (V2, E2)

G1∼= G2 ⇐⇒ ∃f : V1��V2 :

∀u, v ∈ V1 : (u, v) ∈ E1 ⇔ (f(u), f(v)) ∈ E2

Graphs

G1 = (V1, E1) G2 = (V2, E2)

G1∼= G2 ⇐⇒ ∃f : V1��V2 :

∀u, v ∈ V1 : (u, v) ∈ E1 ⇔ (f(u), f(v)) ∈ E2

Graphs

G1 = (V1, E1) G2 = (V2, E2)

G1∼= G2 ⇐⇒ ∃f : V1��V2 :

∀u, v ∈ V1 : (u, v) ∈ E1 ⇔ (f(u), f(v)) ∈ E2

Graphs

Example:

Graphs

Example:

�2 �1

�2 �1�

Graphs

G = (V, E) V = V1 ∪ V2 V1 ∩ V2 = ∅

G called bipartite (or two-coloured), if

• for all u, v ∈ V1, (u, v) 6∈ E

• for all u, v ∈ V2, (u, v) 6∈ E

Sets V1, V2 called colour classes of G

Graphs

G = (V, E) V = V1 ∪ V2 V1 ∩ V2 = ∅G called bipartite (or two-coloured), if

• for all u, v ∈ V1, (u, v) 6∈ E

• for all u, v ∈ V2, (u, v) 6∈ E

Graphs

G = (V, E) V = V1 ∪ V2 V1 ∩ V2 = ∅G called bipartite (or two-coloured), if

• for all u, v ∈ V1, (u, v) 6∈ E

• for all u, v ∈ V2, (u, v) 6∈ E

Graphs

Example:

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The Königsberg Bridges graph is bipartite

Graphs

Example:

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The Königsberg Bridges graph is bipartite

Graphs

Example:

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The wolf/goat/cabbage graph is bipartite

Graphs

Example:

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The wolf/goat/cabbage graph is bipartite

Graphs

V1 ∩ V2 = ∅

Bipartite graph with all possible edges between V1, V2

is called complete bipartite

K(V1, V2) = (V1 ∪ V2, (V1 × V2) ∪ (V2 × V1))

The complete bipartite graph on m + n nodes:K(m, n)

Can also define n-partite (n-coloured), and completen-partite graph

Graphs

V1 ∩ V2 = ∅Bipartite graph with all possible edges between V1, V2

K(V1, V2) = (V1 ∪ V2, (V1 × V2) ∪ (V2 × V1))

Graphs

K(V1, V2) = (V1 ∪ V2, (V1 × V2) ∪ (V2 × V1))

Graphs

K(V1, V2) = (V1 ∪ V2, (V1 × V2) ∪ (V2 × V1))

Graphs

K(V1, V2) = (V1 ∪ V2, (V1 × V2) ∪ (V2 × V1))

Graphs

Example:

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The houses/wells graph is complete bipartite

K({H1, H2, H3}, {W1, W2, W3})

Graphs

Example:

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K({H1, H2, H3}, {W1, W2, W3})

Graphs

Example:

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K({H1, H2, H3}, {W1, W2, W3})

Graphs

G = (V, E)

A walk: sequence (u, u1, . . . , uk−1, v), such that

(u ⇀ u1) ∧ (u1 ⇀ u2) ∧ · · · ∧ (uk−1 ⇀ v)

Nodes u, v connected by a walk: u# v

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

A tour is a walk from a node to itself: u# u

Graphs

G = (V, E)

(u ⇀ u1) ∧ (u1 ⇀ u2) ∧ · · · ∧ (uk−1 ⇀ v)

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

Graphs

G = (V, E)

(u ⇀ u1) ∧ (u1 ⇀ u2) ∧ · · · ∧ (uk−1 ⇀ v)

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

Graphs

G = (V, E)

(u ⇀ u1) ∧ (u1 ⇀ u2) ∧ · · · ∧ (uk−1 ⇀ v)

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

Graphs

G = (V, E)

(u ⇀ u1) ∧ (u1 ⇀ u2) ∧ · · · ∧ (uk−1 ⇀ v)

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

Graphs

G = (V, E)

(u ⇀ u1) ∧ (u1 ⇀ u2) ∧ · · · ∧ (uk−1 ⇀ v)

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

Graphs

0# 5 : 0 ⇀ 3 ⇀ 1 ⇀ 4 ⇀ 6 ⇀ 3 ⇀ 0 ⇀ 2 ⇀ 5

A tour: 0 ⇀ 3 ⇀ 1 ⇀ 4 ⇀ 6 ⇀ 3 ⇀ 5 ⇀ 2 ⇀ 0

Graphs

0# 5 : 0 ⇀ 3 ⇀ 1 ⇀ 4 ⇀ 6 ⇀ 3 ⇀ 0 ⇀ 2 ⇀ 5

A tour: 0 ⇀ 3 ⇀ 1 ⇀ 4 ⇀ 6 ⇀ 3 ⇀ 5 ⇀ 2 ⇀ 0

Graphs

0# 5 : 0 ⇀ 3 ⇀ 1 ⇀ 4 ⇀ 6 ⇀ 3 ⇀ 0 ⇀ 2 ⇀ 5

A tour: 0 ⇀ 3 ⇀ 1 ⇀ 4 ⇀ 6 ⇀ 3 ⇀ 5 ⇀ 2 ⇀ 0

Graphs

For all v ∈ V , (v) is a walk v # v of length 0

For all u, v ∈ V , (u# v) ⇒ (v # u)

For all u, v, w ∈ V , (u# v) ∧ (v # w) ⇒ (u# w)

Therefore R# : V ↔ V is an equivalence relation

Classes of R# called connected components

A graph is connected, if every two nodes areconnected

Graphs

For all u, v ∈ V , (u# v) ⇒ (v # u)

Graphs

For all u, v ∈ V , (u# v) ⇒ (v # u)

Graphs

For all u, v ∈ V , (u# v) ⇒ (v # u)

Graphs

For all u, v ∈ V , (u# v) ⇒ (v # u)

Graphs

For all u, v ∈ V , (u# v) ⇒ (v # u)

Graphs

G = (V, E)

A path is a walk where all nodes are distinct

Nodes u, v connected by a path: u v

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

∀i, j ∈ Nk+1 : i 6= j ⇒ ui 6= uj

A cycle is a tour of length ≥ 3 where all nodes exceptthe final are distinct: u v ⇀ u

A graph without cycles called acyclic

Graphs

G = (V, E)

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

∀i, j ∈ Nk+1 : i 6= j ⇒ ui 6= uj

Graphs

G = (V, E)

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

∀i, j ∈ Nk+1 : i 6= j ⇒ ui 6= uj

Graphs

G = (V, E)

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

∀i, j ∈ Nk+1 : i 6= j ⇒ ui 6= uj

Graphs

G = (V, E)

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

∀i, j ∈ Nk+1 : i 6= j ⇒ ui 6= uj

Graphs

G = (V, E)

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

∀i, j ∈ Nk+1 : i 6= j ⇒ ui 6= uj

Graphs

G = (V, E)

u = u0 ⇀ u1 ⇀ u2 ⇀ . . . ⇀ uk−1 ⇀ uk = v

∀i, j ∈ Nk+1 : i 6= j ⇒ ui 6= uj

Graphs

0 5 : 0 ⇀ 2 ⇀ 7 ⇀ 10 ⇀ 8 ⇀ 3 ⇀ 5

A cycle: 3 ⇀ 8 ⇀ 10 ⇀ 9 ⇀ 4 ⇀ 6 ⇀ 3

Graphs

0 5 : 0 ⇀ 2 ⇀ 7 ⇀ 10 ⇀ 8 ⇀ 3 ⇀ 5

A cycle: 3 ⇀ 8 ⇀ 10 ⇀ 9 ⇀ 4 ⇀ 6 ⇀ 3

Graphs

0 5 : 0 ⇀ 2 ⇀ 7 ⇀ 10 ⇀ 8 ⇀ 3 ⇀ 5

A cycle: 3 ⇀ 8 ⇀ 10 ⇀ 9 ⇀ 4 ⇀ 6 ⇀ 3

Graphs

G = (V, E)

R : V ↔ V — equivalence relation?

Prove: For all u, v ∈ V , there is a path u v, iffthere is a walk u# v.

R , R# : V ↔ V R = R#

Proof. (u v) ⇒ (u# v): trivial

(u# v) ⇒ (u v): induction on walk length

Graphs

G = (V, E)

R , R# : V ↔ V R = R#

Graphs

G = (V, E)

R , R# : V ↔ V R = R#

Graphs

G = (V, E)

R , R# : V ↔ V R = R#

Graphs

G = (V, E)

R , R# : V ↔ V R = R#

Proof.

(u v) ⇒ (u# v): trivial

Graphs

G = (V, E)

R , R# : V ↔ V R = R#

Graphs

G = (V, E)

R , R# : V ↔ V R = R#

Graphs

Induction base: (u) is both u# u and u u

Inductive step: consider walk u# w ⇀ v

Assume statement holds for u# w: path u w

Case 1: path u w does not visit v

Then u w ⇀ v is a path

Case 2: path u w visits v: u v w ⇀ v

Take initial u v

Graphs

Take initial u v

Graphs

Take initial u v

Graphs

Take initial u v

Graphs

Take initial u v

Graphs

Take initial u v

Graphs

Take initial u v

Graphs

G = (V, E) v ∈ V

The degree of node v is the number of nodes adjacentto v

deg(v) = |{u ∈ V | v ⇀ u}|

deg(a) = deg(d) = 2

deg(b) = deg(c) = 4

deg(e) = deg(f) = 4

Graphs

G = (V, E) v ∈ V

deg(v) = |{u ∈ V | v ⇀ u}|

deg(a) = deg(d) = 2

deg(b) = deg(c) = 4

deg(e) = deg(f) = 4

Graphs

G = (V, E) v ∈ V

deg(v) = |{u ∈ V | v ⇀ u}|

deg(a) = deg(d) = 2

deg(b) = deg(c) = 4

deg(e) = deg(f) = 4

Graphs

G = (V, E) v ∈ V

deg(v) = |{u ∈ V | v ⇀ u}|

deg(a) = deg(d) = 2

deg(b) = deg(c) = 4

deg(e) = deg(f) = 4

Graphs

G = (V, E) v ∈ V

deg(v) = |{u ∈ V | v ⇀ u}|

deg(a) = deg(d) = 2

deg(b) = deg(c) = 4

deg(e) = deg(f) = 4

Graphs

An Euler tour of graph G is a tour which visits everyedge in E exactly once

a ⇀ b ⇀ c ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ c ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b ⇀ f ⇀ a

Graphs

⇀ b ⇀ c ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b ⇀ f ⇀ a

Graphs

a ⇀ b

⇀ c ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ c

⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ c ⇀ f

⇀ e ⇀ d ⇀ c ⇀ e ⇀ b ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ c ⇀ f ⇀ e

⇀ d ⇀ c ⇀ e ⇀ b ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ c ⇀ f ⇀ e ⇀ d

⇀ c ⇀ e ⇀ b ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ c ⇀ f ⇀ e ⇀ d ⇀ c

⇀ e ⇀ b ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ c ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e

⇀ b ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ c ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b

⇀ f ⇀ a

Graphs

a ⇀ b ⇀ c ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b ⇀ f

Graphs

a ⇀ b ⇀ c ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b ⇀ f ⇀ a

Graphs

G = (V, E)

Theorem: Graph G has an Euler tour iff

• G is connected

• every node in V has even degree

Gives an efficient test for Euler tour existence

Graphs

G = (V, E)

• G is connected

Graphs

G = (V, E)

• G is connected

Graphs

G = (V, E)

• G is connected

Graphs

G = (V, E)

• G is connected

Graphs

Proof.

G has Euler tour ⇒ G connected: trivial

G has Euler tour ⇒ every node has even degree

Consider v ∈ V

Suppose v visited k times by Euler tour

Every visit uses 2 new edges (in and out)

Hence deg(v) = 2k

Graphs

Proof.

Consider v ∈ V

Hence deg(v) = 2k

Graphs

Proof.

Consider v ∈ V

Hence deg(v) = 2k

Graphs

Proof.

Consider v ∈ V

Hence deg(v) = 2k

Graphs

Proof.

Consider v ∈ V

Hence deg(v) = 2k

Graphs

Proof.

Consider v ∈ V

Hence deg(v) = 2k

Graphs

G connected ∧ every node has even degree⇒ G has an Euler tour

Take any v0 ∈ V

Consider any walk v0 ⇀ v1 ⇀ . . . ⇀ vk 6= v0

Node vk has an odd number of visited edges

deg(vk) is even ⇒ vk has an unvisited edge

Extend walk: v0 ⇀ v1 ⇀ . . . ⇀ vk ⇀ vk+1

Repeat until v0 ⇀ v1 ⇀ . . . ⇀ vm = v0

Graphs

Take any v0 ∈ V

Graphs

Take any v0 ∈ V

Graphs

Take any v0 ∈ V

Graphs

Take any v0 ∈ V

Graphs

Take any v0 ∈ V

Graphs

Take any v0 ∈ V

Graphs

Consider tour v0 ⇀ v1 ⇀ . . . ⇀ vm = v0

Suppose some vi has unvisited edge to vm+1

By symmetry, let vi = v0

Extend walk: v0 ⇀ . . . ⇀ vk ⇀ vm = v0 ⇀ vm+1

Repeat until every vi has no unvisited edges

G connected =⇒ all edges in E visited

Therefore, G has an Euler tour

Graphs

Example:

a ⇀ b ⇀ c ⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b

Graphs

Example:

⇀ b ⇀ c ⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b

Graphs

Example:

a ⇀ b

⇀ c ⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b

Graphs

Example:

a ⇀ b ⇀ c

⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b

Graphs

Example:

a ⇀ b ⇀ c ⇀ f

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b

Graphs

Example:

a ⇀ b ⇀ c ⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b

Graphs

Example:

a ⇀ b ⇀ c ⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b

⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b

Graphs

Example:

a ⇀ b ⇀ c ⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f

⇀ e ⇀ d ⇀ c ⇀ e ⇀ b

Graphs

Example:

a ⇀ b ⇀ c ⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f ⇀ e

⇀ d ⇀ c ⇀ e ⇀ b

Graphs

Example:

a ⇀ b ⇀ c ⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f ⇀ e ⇀ d

⇀ c ⇀ e ⇀ b

Graphs

Example:

a ⇀ b ⇀ c ⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f ⇀ e ⇀ d ⇀ c

⇀ e ⇀ b

Graphs

Example:

a ⇀ b ⇀ c ⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e

Graphs

Example:

a ⇀ b ⇀ c ⇀ f ⇀ a

b ⇀ c ⇀ f ⇀ a ⇀ b ⇀ f ⇀ e ⇀ d ⇀ c ⇀ e ⇀ b

Graphs

A Hamiltonian cycle of graph G is a cycle whichvisits every node in V exactly once

a ⇀ b ⇀ e ⇀ d ⇀ c ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ e ⇀ d ⇀ c ⇀ f ⇀ a

Graphs

⇀ b ⇀ e ⇀ d ⇀ c ⇀ f ⇀ a

Graphs

a ⇀ b

⇀ e ⇀ d ⇀ c ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ e

⇀ d ⇀ c ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ e ⇀ d

⇀ c ⇀ f ⇀ a

Graphs

a ⇀ b ⇀ e ⇀ d ⇀ c

⇀ f ⇀ a

Graphs

a ⇀ b ⇀ e ⇀ d ⇀ c ⇀ f

Graphs

a ⇀ b ⇀ e ⇀ d ⇀ c ⇀ f ⇀ a

Graphs

Exercise: find an efficient test for existence ofHamiltonian cycle. . .

. . . and claim your $1 000 000!

See www.claymath.org for details

Graphs

. . . and claim your $1 000 000!

Graphs

. . . and claim your $1 000 000!

Graphs

G = (V, E) G′ = (V ′, E ′)

G′ is a subgraph of G, if V ′ ⊆ V , E ′ ⊆ E

G′ ⊆ G

Graphs

G = (V, E) G′ = (V ′, E ′)

G′ ⊆ G

Graphs

G = (V, E) G′ = (V ′, E ′)

G′ ⊆ G

Graphs

G = (V, E) G′ = (V ′, E ′)

G′ ⊆ G

Graphs

G = (V, E) G′ = (V ′, E ′)

G′ is a spanning subgraph of G, if V ′ = V , E ′ ⊆ E

G′ v G

Graphs

G = (V, E) G′ = (V ′, E ′)

G′ v G

Graphs

G = (V, E) G′ = (V ′, E ′)

G′ v G

Graphs

G = (V, E) G′ = (V ′, E ′)

G′ v G

Graphs

V — a finite set G(V ) — set of all graphs on V

R⊆ : G(V ) ↔ G(V )

∀G : G ⊆ G G v G

∀G, G′ : (G′ ⊆ G) ∧ (G ⊆ G′) ⇒ G = G′

∀G, G′, G′′ : (G′′ ⊆ G′) ∧ (G′ ⊆ G) ⇒ G′′ ⊆ G

Therefore, R⊆ is a partial order

Similarly, Rv is a partial order

Graphs

R⊆ : G(V ) ↔ G(V )

∀G : G ⊆ G G v G

∀G, G′ : (G′ ⊆ G) ∧ (G ⊆ G′) ⇒ G = G′

∀G, G′, G′′ : (G′′ ⊆ G′) ∧ (G′ ⊆ G) ⇒ G′′ ⊆ G

Graphs

R⊆ : G(V ) ↔ G(V )

∀G : G ⊆ G G v G

∀G, G′ : (G′ ⊆ G) ∧ (G ⊆ G′) ⇒ G = G′

∀G, G′, G′′ : (G′′ ⊆ G′) ∧ (G′ ⊆ G) ⇒ G′′ ⊆ G

Graphs

R⊆ : G(V ) ↔ G(V )

∀G : G ⊆ G G v G

∀G, G′ : (G′ ⊆ G) ∧ (G ⊆ G′) ⇒ G = G′

∀G, G′, G′′ : (G′′ ⊆ G′) ∧ (G′ ⊆ G) ⇒ G′′ ⊆ G

Graphs

R⊆ : G(V ) ↔ G(V )

∀G : G ⊆ G G v G

∀G, G′ : (G′ ⊆ G) ∧ (G ⊆ G′) ⇒ G = G′

∀G, G′, G′′ : (G′′ ⊆ G′) ∧ (G′ ⊆ G) ⇒ G′′ ⊆ G

Graphs

R⊆ : G(V ) ↔ G(V )

∀G : G ⊆ G G v G

∀G, G′ : (G′ ⊆ G) ∧ (G ⊆ G′) ⇒ G = G′

∀G, G′, G′′ : (G′′ ⊆ G′) ∧ (G′ ⊆ G) ⇒ G′′ ⊆ G

Graphs

R⊆ : G(V ) ↔ G(V )

∀G : G ⊆ G G v G

∀G, G′ : (G′ ⊆ G) ∧ (G ⊆ G′) ⇒ G = G′

∀G, G′, G′′ : (G′′ ⊆ G′) ∧ (G′ ⊆ G) ⇒ G′′ ⊆ G

Graphs

Recall: a graph is

• connected, if every two nodes connected

• acyclic, if there is no cycle

A connected acyclic graph is called a tree

Graphs

Recall: a graph is

Graphs

Recall: a graph is

Graphs

Recall: a graph is

Graphs

Recall: a graph is

� ��

��

Graphs

G = (V, E) — a tree

Prove: |V | = |E|+ 1.

Proof. Induction base: V = {v}, E = ∅|E| = 0 |V | = 1 = |E|+ 1

Graphs

Prove: |V | = |E|+ 1.

Graphs

Prove: |V | = |E|+ 1.

Proof. Induction base: V = {v}, E = ∅

|E| = 0 |V | = 1 = |E|+ 1

Graphs

Prove: |V | = |E|+ 1.

Graphs

Inductive step: assume statement holds for all propersubgraphs of G

Take any edge (u, v) ∈ E.

Let G′ = (V, E \ {(u, v), (v, u)}).

Consider R : V ↔ V in G′

Graphs

Let G′ = (V, E \ {(u, v), (v, u)}).

Graphs

Let G′ = (V, E \ {(u, v), (v, u)}).

Graphs

Let G′ = (V, E \ {(u, v), (v, u)}).

�u � v

Graphs

Let G′ = (V, E \ {(u, v), (v, u)}).

�u � v

Graphs

Suppose u v in G′

G′u v

Then u v ⇀ u a cycle in G — contradiction

Therefore u 6 v in G′

Graphs

Suppose u v in G′

�u � v

Graphs

Suppose u v in G′

�u � v

Graphs

Suppose u v in G′

�u � v

Graphs

Vu = [u] Vv = [v] (in G′)

Eu = {(x, y) | x, y ∈ Vu} Ev = {(x, y) | x, y ∈ Vv}Gu = (Vu, Eu) Gv = (Vv, Ev)

Gu Gvu v

G connected =⇒ Gu, Gv connected

G acyclic =⇒ Gu, Gv acyclic

Graphs

Vu = [u] Vv = [v] (in G′)

Eu = {(x, y) | x, y ∈ Vu} Ev = {(x, y) | x, y ∈ Vv}

Gu = (Vu, Eu) Gv = (Vv, Ev)

Gu Gvu v

Graphs

Vu = [u] Vv = [v] (in G′)

Gu Gvu v

Graphs

Vu = [u] Vv = [v] (in G′)

�u � v

Graphs

Vu = [u] Vv = [v] (in G′)

�u � v

Graphs

Vu = [u] Vv = [v] (in G′)

�u � v

Graphs

By induction hypothesis:|Vu| = |Eu|+ 1 |Vv| = |Ev|+ 1

|V | = |Vu|+ |Vv| = (|Eu|+ 1) + (|Ev|+ 1) =

(|Eu|+ |Ev|+ 1) + 1 = |E|+ 1

Graphs

|V | = |Vu|+ |Vv| = (|Eu|+ 1) + (|Ev|+ 1) =

(|Eu|+ |Ev|+ 1) + 1 = |E|+ 1

Graphs

|V | = |Vu|+ |Vv| = (|Eu|+ 1) + (|Ev|+ 1) =

(|Eu|+ |Ev|+ 1) + 1 = |E|+ 1

Graphs

Corollary: G has a node of degree 1.

Proof. G connected ⇒ no nodes of degree 0.

Suppose all degrees ≥ 2. |E| ≥ 2 · |V |/2 = |V |But |E| = |V | − 1. Hence assumption false.

Therefore G has a node of degree 1.

A node of degree 1 in a tree is called a leaf.

Graphs

Suppose all degrees ≥ 2.

|E| ≥ 2 · |V |/2 = |V |But |E| = |V | − 1. Hence assumption false.

Graphs

Suppose all degrees ≥ 2. |E| ≥ 2 · |V |/2 = |V |

But |E| = |V | − 1. Hence assumption false.

Graphs

G = (V, E) G′ = (V ′, E ′)

Recall: G′ is a spanning subgraph of G, if V ′ = V ,E ′ ⊆ E

Rv : G(V ) ↔ G(V )

Consider restricting Rv to the set of all

• connected graphs in G(V )

• acyclic graphs in G(V )

Graphs

G = (V, E) G′ = (V ′, E ′)

Rv : G(V ) ↔ G(V )

Graphs

G = (V, E) G′ = (V ′, E ′)

Rv : G(V ) ↔ G(V )

Graphs

G = (V, E) G′ = (V ′, E ′)

Rv : G(V ) ↔ G(V )

Graphs

G = (V, E) G′ = (V ′, E ′)

Rv : G(V ) ↔ G(V )

Graphs

G = (V, E) G′ = (V ′, E ′)

Rv : G(V ) ↔ G(V )

Graphs

G = (V, E)

Prove: G is a tree iff G is v-minimal in the set of allconnected graphs on V

Proof. G connected

Need to prove: G acyclic iff G v-minimal

Equivalent to: G has a cycle iff G not v-minimal

Graphs

G = (V, E)

Proof. G connected

Graphs

G = (V, E)

Proof. G connected

Graphs

G = (V, E)

Proof. G connected

Graphs

G = (V, E)

Proof. G connected

Graphs

G has a cycle ⇒ G not v-minimal

Suppose G has a cycle

Remove any edge from cycle

Remaining graph connected

Hence G not v-minimal

Graphs

G not v-minimal ⇒ G has a cycle

Suppose G not v-minimal

For some u, v ∈ V , removing edge u ⇀ v does notdisconnect the graph

Therefore there is another path u v

Hence G has a cycle

Graphs

Hence G has a cycle

Graphs

Hence G has a cycle

Graphs

Hence G has a cycle

Graphs

Hence G has a cycle

Graphs

G = (V, E)

Prove: G is a tree iff G is v-maximal in the set of allacyclic graphs on V

Proof. G acyclic

Need to prove: G connected iff G v-maximal

Equivalent to:G disconnected iff G not v-maximal

Graphs

G = (V, E)

Proof. G acyclic

Graphs

G = (V, E)

Proof. G acyclic

Graphs

G = (V, E)

Proof. G acyclic

Graphs

G = (V, E)

Proof. G acyclic

Graphs

G disconnected ⇒ G not v-maximal

Suppose G disconnected

Add any edge between two connected components

Resulting graph acyclic

Hence G not v-maximal

Graphs

G not v-maximal ⇒ G disconnected

Suppose G not v-maximal

For some u, v ∈ V , adding edge u ⇀ v does notcreate cycle

Therefore u, v are in different connected components

Hence G disconnected

Graphs

A graph is called planar, if it can be drawn on theplane without edge crossings.

Examples: any tree, any cycle

Complete graphs:K(1), K(2), K(3), K(4). Not K(5).

Complete bipartite graphs: K(2, 3). Not K(3, 3).

Graphs

Examples: any tree

, any cycle

Graphs

Complete graphs:K(1), K(2), K(3), K(4).

Not K(5).

Graphs

Complete bipartite graphs: K(2, 3).

Not K(3, 3).

Graphs

G = (V, E) How to test if G is planar?

Subdivision: Let u ⇀ v. Add new node x

Replace u ⇀ v by u ⇀ x ⇀ v

G non-planar ⇒ new graph non-planar

Graphs

Subdivision: Let u ⇀ v.

Add new node x

Graphs

Only K(5) and K(3, 3) are “really” non-planar.

Theorem (Kuratowski). A graph is planar iff it has nosubgraph obtained from K(5) or K(3, 3) bysubdivisions.

Proof: difficult.

Graphs

Proof: difficult.

Graphs

Proof: difficult.

Graphs

Recall: if G = (V, E) a tree, then |V | = |E|+ 1.

Generalisation: G = (V, E) — planar

Drawing of G partitions the plane into faces

Let F be the set of all faces

Graphs

Examples:

G is a tree: |V | = |E|+ 1 |F | = 1

G has one cycle: |V | = |E| |F | = 2

Theorem (Euler). For any planar graph G,|V | − |E|+ |F | = 2.

Proof: induction.

Graphs

Examples:

G is a tree: |V | = |E|+ 1 |F | = 1

Proof: induction.

Graphs

Examples:

G is a tree: |V | = |E|+ 1 |F | = 1

Proof: induction.

Graphs

Examples:

G is a tree: |V | = |E|+ 1 |F | = 1

Proof: induction.

Discrete Mathematics I - Computer Science Departmenttiskin/teach/dm1/o.pdf · Discrete maths in...

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