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2 February

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Discussion Questions What’s the difference between Michael and {Michael}? What would happen if we said Michael = {Michael}?

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Barber Paradox

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

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

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

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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

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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

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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.

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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.

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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.

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

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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.

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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

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

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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

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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.

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

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Russell’s Paradox

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

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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}

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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.

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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}

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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

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

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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.

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The von Neumann Heirarchy

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Cantor’s Diagonal Proof

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Numbers vs. Numerals

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Decimal Representations A decimal representation of a real number consists of two parts: A finite string S 1 of Arabic numerals. An infinite string S 2 of Arabic numerals. It looks like this: S 1. S 2

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We can’t actually write out any decimal representations, since we can’t write infinite strings of numerals. But we can write out abbreviations of some decimal representations. 1/4 = 0.25 1/7 = 0.142857 π = ? _______

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We will prove that there cannot be a list of all the decimal representations between ‘0.0’ and ‘1.0’. A list is something with a first member, then a second member, then a third member and so on, perhaps continuing forever.

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Choose an Arbitrary List 1.‘8’‘4’‘3’‘0’ … 2.‘2’‘5’‘6’‘2’‘5’‘6’‘2’‘5’… 3.‘7’‘9’‘2’‘5’‘1’‘0’‘7’‘2’… 4.‘9’‘8’‘0’‘6’‘4’‘2’‘8’‘1’… 5.‘3’ … 6.‘4’‘3’‘7’ ‘1’‘0’‘2’‘0’… 7.‘8’ ‘1’‘3’‘2’‘9’ ‘6’… 8.‘1’‘6’‘1’‘6’‘1’‘6’‘1’‘6’… …

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Find the Diagonal 1.‘8’‘4’‘3’‘0’ … 2.‘2’‘5’‘6’‘2’‘5’‘6’‘2’‘5’… 3.‘7’‘9’‘2’‘5’‘1’‘0’‘7’‘2’… 4.‘9’‘8’‘0’‘6’‘4’‘2’‘8’‘1’… 5.‘3’ … 6.‘4’‘3’‘7’ ‘1’‘0’‘2’‘0’… 7.‘8’ ‘1’‘3’‘2’‘9’ ‘6’… 8.‘1’‘6’‘1’‘6’‘1’‘6’‘1’‘6’… …

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Diagonal = 0.85263096… Add move each numeral ‘1 up’– so ‘8’ becomes ‘9’, ‘5’ becomes ‘6’, etc. New Representation = 0.96374107…

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New Number Not on the List ‘9’‘6’‘3’‘7’‘4’‘1’‘0’‘7’… 1.‘8’‘4’‘3’‘0’ … 2.‘2’‘5’‘6’‘2’‘5’‘6’‘2’‘5’… 3.‘7’‘9’‘2’‘5’‘1’‘0’‘7’‘2’… 4.‘9’‘8’‘0’‘6’‘4’‘2’‘8’‘1’… 5.‘3’ … 6.‘4’‘3’‘7’ ‘1’‘0’‘2’‘0’… 7.‘8’ ‘1’‘3’‘2’‘9’ ‘6’… 8.‘1’‘6’‘1’‘6’‘1’‘6’‘1’‘6’… …

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Doesn’t Help to Add It In! ‘9’‘6’‘3’‘7’‘4’‘1’‘0’‘7’… 1.‘8’‘4’‘3’‘0’ … 2.‘2’‘5’‘6’‘2’‘5’‘6’‘2’‘5’… 3.‘7’‘9’‘2’‘5’‘1’‘0’‘7’‘2’… 4.‘9’‘8’‘0’‘6’‘4’‘2’‘8’‘1’… 5.‘3’ … 6.‘4’‘3’‘7’ ‘1’‘0’‘2’‘0’… 7.‘8’ ‘1’‘3’‘2’‘9’ ‘6’… 8.‘1’‘6’‘1’‘6’‘1’‘6’‘1’‘6’… …

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Discussion Questions Does this prove you can’t list all the real numbers? How do we fix the proof? Can you use a similar proof to show that the rational numbers aren’t countable? Can you list the powerset of the natural numbers?

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