9 February

In Class Assignment #1

Set: {2, 4, 6, 8, 10} MEMBERSUBSETNEITHER 2 {4, 6} {2, 10} {} 3 {{2, 4}, 6, 8, 10}

Set: {2, 4, 6, 8, 10} MEMBERSUBSETNEITHER 2 ✓ {4, 6} {2, 10} {} 3 {{2, 4}, 6, 8, 10}

Set: {2, 4, 6, 8, 10} MEMBERSUBSETNEITHER 2 ✓ {4, 6} ✓ {2, 10} {} 3 {{2, 4}, 6, 8, 10}

Set: {2, 4, 6, 8, 10} MEMBERSUBSETNEITHER 2 ✓ {4, 6} ✓ {2, 10} ✓ {} 3 {{2, 4}, 6, 8, 10}

Set: {2, 4, 6, 8, 10} MEMBERSUBSETNEITHER 2 ✓ {4, 6} ✓ {2, 10} ✓ {} ✓ 3 {{2, 4}, 6, 8, 10}

Set: {2, 4, 6, 8, 10} MEMBERSUBSETNEITHER 2 ✓ {4, 6} ✓ {2, 10} ✓ {} ✓ 3 ✓ {{2, 4}, 6, 8, 10}

Set: {2, 4, 6, 8, 10} MEMBERSUBSETNEITHER 2 ✓ {4, 6} ✓ {2, 10} ✓ {} ✓ 3 ✓ {{2, 4}, 6, 8, 10} ✓

In Class Assignment #2

{1, 1} = {1} Many students thought this was false. It’s not.

For any two sets A and B, A = B iff (for all x)(x ∈ A iff x ∈ B) Therefore, {1, 1} = {1} iff (for all x)(x ∈ {1, 1} iff x ∈ {1}) Therefore, if (for all x)(x ∈ {1, 1} iff x ∈ {1}), then {1, 1} = {1} To prove: (for all x)(x ∈ {1, 1} iff x ∈ {1})

Let x = 1. Then clearly: 1 ∈ {1, 1} And 1 ∈ {1} Therefore, 1 ∈ {1, 1} iff 1 ∈ {1} (A iff B is true when A and B are both true.)

To prove: (for all x)(x ∈ {1, 1} iff x ∈ {1}) Let x ≠ 1. Then clearly: x ∉ {1, 1} And x ∉ {1} Therefore, x ∈ {1, 1} iff x ∈ {1} (A iff B is true when A and B are both false.)

For every set S, S ∪ S = S ∩ S Proof: A ∪ B = {x: x ∈ A or x ∈ B} Therefore, S ∪ S = {x: x ∈ S or x ∈ S} (P or P) iff P Therefore, S ∪ S = {x: x ∈ S or x ∈ S} = {x: x ∈ S} = S

For every set S, S ∪ S = S ∩ S Proof, cont’d A ∩ B = {x: x ∈ A and x ∈ B} Therefore, S ∩ S = {x: x ∈ S and x ∈ S} (P and P) iff P Therefore, S ∩ S = {x: x ∈ S and x ∈ S} = {x: x ∈ S} = S And so: S ∪ S = S = S ∩ S

Write the Name in Extensive Notation {Hong Kong, London, New York} ∪ ({London, Sydney} ∩ {Sydney, Tokyo}) {Sydney} {Hong Kong, London, New York, Sydney}

Some Strange Answers {Hong Kong, London, New York, Sydney, Tokyo} {Sydney}

Problem 3b Write names for the following sets in intensive notation: {Michael} {x: Michael} {x: x is the first name of the person teaching this course} {x: x is a person named “Michael”}

Problem 4a Write the power set of the following sets: {{}}, i.e. {ø} Definition: power set of S = set of all subsets of S {} ⊆ {{}}{} ⊆ S, for every S {{}} ⊆ {{}}S ⊆ S, for every S So: P({{}}) = { {}, {{}} } = { ø, {ø} }

Remember If a set has N members, then its power set has 2 N members. So since {{}} has one member, its power set has 2 1 = 2 members. And indeed, { {}, {{}} } has 2 members.

Some Answers {}, {{}} {}, {}, {{}} { {}, ø } { {{}} } { {}, {{{}}} }

Write the power set of the following sets: {x: x is a dog} { ALL DOGS } { x: x is a dog } { {}, x: x is a dog } { {x: x is a dog} } { {}, {x: x is a dog} }

Write the power set of the following sets: {x: x is a dog} { dog } { {}, dog } { {}, {y: y is the set of dogs} } { {x: x has four legs}, {x: x growls}, {x: x runs}, {x: x is a dog}, … } { {small dogs}, {medium dogs}, {large dogs}, {hunting dogs}, ….}

Write a sentence that both uses and mentions the word ‘logic.’ A sentence obeys logic or does not obey logic. People study logic to train their logic. One must differentiate between the use of the word ‘logic’ and that of its token, “logic.”

Write a sentence that both uses and mentions the word ‘logic.’ Logic is one of the concepts covered in this course. Elementary logic course teachers use logic. ‘Logic’ has 5 letters and means rational. Logic is a subject and so is the name ‘logic.’

Some Clever Answers According to logic, ‘logic’ can’t both have 5 letters and not have 5 letters. I learned logic in logic class but I learned ‘logic’ in English class. Using ‘logic’ doesn’t require logic.

Homework #1

True or False TFIn binary notation, 1 = TFThere exists a set S that can be paired one-to-one with its power set. TFThe rational numbers are the same size as the power set of the natural numbers. TFAccording to standard set theory, the numerical size of the real numbers is the next highest number after the numerical size of the natural numbers.

True or False TFIn binary notation, 1 = TFThere exists a set S that can be paired one-to-one with its power set. TFThe rational numbers are the same size as the power set of the natural numbers. TFAccording to standard set theory, the numerical size of the real numbers is the next highest number after the numerical size of the natural numbers.

The Power Set Theorem Suppose that S is a set that CAN be paired 1-to-1 with P(S). Let’s call the members of S: a, b, c, … Let’s call the subsets of S: L, M, N, …

The Power Set Theorem a b c d e f … L M N O P Q … S P(S)

The Power Set Theorem a b c d e f … L M N O P Q … S P(S) x : x ϵ S & x ϵ x’s pair in P(S) K

The Power Set Theorem K ⊆ S So: K ϵ P(S) = {x: x ⊆ S} But K is not the pair of any member of S K ≠ L, K ≠ M, K ≠ N, K ≠ O… Therefore there is no 1-to-1 pairing between S and P(S)

True or False TFIn binary notation, 1 = TFThere exists a set S that can be paired one-to-one with its power set. TFThe rational numbers are the same size as the power set of the natural numbers. TFAccording to standard set theory, the numerical size of the real numbers is the next highest number after the numerical size of the natural numbers.

… 1 2 ½ ⅓ 3 4 … N R x : x ⊆ N P(N)

True or False TFIn binary notation, 1 = TFThere exists a set S that can be paired one-to-one with its power set. TFThe rational numbers are the same size as the power set of the natural numbers. TFAccording to standard set theory, the numerical size of the real numbers is the next highest number after the numerical size of the natural numbers.

Write names for the following sets in extensive notation {Hong Kong, London, New York} ∩ ({London, Sydney} ∪ {Sydney, Tokyo}) {London, Sydney, Tokyo} {London}

Write names for the following sets in extensive notation {x: x shaves everyone who doesn’t shave themselves} NAME: {}

Problem 3 Write two different names for the following set, both in intensive notation. (Hint: Michael Johnson is the instructor of this class.) {Michael Johnson}

Example Answers {x: x = Michael Johnson} {x: x is the instructor of this class} {x: x = Michael Johnson or x = Michael Johnson} {x: x = Michael Johnson and (P or not-P)} {y: y = Michael Johnson} {x: x assigned this homework} {x: x is named “Michael Johnson” & x is a philosopher at HKU} {x: x’s homepage is michaeljohnsonphilosophy.com}

Write the power set of {{{}}} Cheesy but acceptable answer: {x: x ⊆ {{{}}} }

Write the power set of {{{}}} How many members does {{{}}} have? How many subsets does it have? When a set S has only 1 member, its power set is a set containing the null set and S.

Write the power set of the following set. ({Hong Kong, London, New York} ∩ {London, Sydney}) ∪ {Sydney, Tokyo} {London} {London, Sydney, Tokyo}

LONDONSYDNEYTOKYO LONDONSYDNEY- LONDON-TOKYO LONDON-- -SYDNEYTOKYO -SYDNEY- --TOKYO ---

Barber Paradox

The Barber Paradox Once upon a time there was a village, and in this village lived a barber named B.

The Barber Paradox B shaved all the villagers who did not shave themselves, And B shaved none of the villagers who did shave themselves.

The Barber Paradox Question, did B shave B, or not?

Suppose B Shaved B 1. B shaved BAssumption 2. B did not shave any villager X where X shaved X Assumption 3. B did not shave B1,2 Logic

Suppose B Did Not Shave B 1. B did not shave BAssumption 2. B shaved every villager X where X did not shave X Assumption 3. B shaved B1,2 Logic

Contradictions with Assumptions We can derive a contradiction from the assumption that B shaved B. We can derive a contradiction from the assumption that B did not shave B.

The Law of Excluded Middle Everything is either true or not true. Either P or not-P, for any P. Either B shaved B or B did not shave B, there is no third option.

It’s the Law Either it’s Tuesday or it’s not Tuesday. Either it’s Wednesday or it’s not Wednesday. Either killing babies is good or killing babies is not good. Either this sandwich is good or it is not good.

Disjunction Elimination A or B A implies C B implies C Therefore, C

Example Either Michael is dead or he has no legs If Michael is dead, he can’t run the race. If Michael has no legs, he can’t run the race. Therefore, Michael can’t run the race.

Contradiction, No Assumptions B shaves B or B does not shave B [Law of Excluded Middle] If B shaves B, contradiction. If B does not shave B, contradiction. Therefore, contradiction

Contradictions Whenever we are confronted with a contradiction, we need to give up something that led us into the contradiction.

Give up Logic? For example, we used Logic in the proof that B shaved B if and only if B did not shave B. So we might consider giving up logic. A or B A implies C B implies C Therefore, C

No Barber In this instance, however, it makes more sense to give up our initial acquiescence to the story: We assumed that there was a village with a barber who shaved all and only the villagers who did not shave themselves.

The Barber Paradox The paradox shows us that there is no such barber, and that there cannot be.

Russell’s Paradox

Set Theoretic Rules Reduction: a ∈ {x: COND(x)} Therefore, COND(a) Abstraction: COND(a) Therefore, a ∈ {x: COND(x)}

Examples Reduction: Mt. Everest ∈ {x: x is a mountain} Therefore, Mt. Everest is a mountain. Abstraction: Mt. Everest is a mountain. Therefore, Mt. Everest ∈ {x: x is a mountain}

Self-Membered Sets It’s possible that some sets are members of themselves. Let S = {x: x is a set}. Since S is a set, S ∈ {x: x is a set} (by abstraction), and thus S ∈ S (by Def of S). Or consider H = {x: Michael hates x}. Maybe I even hate the set of things I hate. So H is in H.

Russell’s Paradox Set Most sets are non-self-membered. The set of mountains is not a mountain; the set of planets is not a planet; and so on. Define: R = {x: ¬x ∈ x}

Is R in R? 1. R ∈ RYes? 2. R ∈ {x: ¬x ∈ x}1, Def of R 3. ¬R ∈ R2, Reduction 4. ¬R ∈ RNo? 5. R ∈ {x: ¬x ∈ x}4, Abstraction 6. R ∈ R5, Def of R

Historical Importance Russell’s paradox was what caused Frege to stop doing mathematics and do philosophy of language instead.

Comparison with the Liar Russell thought that his paradox was of a kind with the liar, and that any solution to one should be a solution to the other. Basically, he saw both as arising from a sort of vicious circularity.

The von Neumann Heirarchy