PHIL 2000 Tools for Philosophers 1st Term 2016 Infinite Sets PHIL 2000 Tools for Philosophers 1st Term 2016

Topics for Discussion What are sets? Where are they? How do they relate to their members? Do they exist? How are they different from Venn diagrams? How many sets are there? Which axiom tells us this? How many empty sets are there?

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

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

A or B A implies C B implies C Therefore, C 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

The Axiom of Comprehension Basic idea of set theory: When you have some things, there is another thing, the collection of those things. For any condition C, there exists a set A such that: (For any x)(x is in A if and only if x satisfies C)

Bertrand Russell One of the founders of analytic philosophy (contemporary Anglophone philosophy). One of the greatest logicians of the 20th Century Showed that the basic idea of set theory can’t be right.

Russell’s Paradox Consider the condition: x is not a member of x

Russell’s Paradox According to the Axiom of Comprehension, there exists a set R such that: (For any x)(x is in R if and only if x satisfies “x is not a member of x”)

Russell’s Paradox According to the Axiom of Comprehension, there exists a set R such that: (For any x)(x is in R if and only if x is not a member of x)

Russell’s Paradox R = { x | x is not a member of x } Question: Is R a member of R?

Russell’s Paradox Let’s suppose: R is not a member of R. Then: R is a member of { x | x is not a member of x } Hence, R is a member of R

Russell’s Paradox Let’s suppose: R is a member of R. Then: R is a member of { x | x is not a member of x } Hence, R is not a member of R

Russell’s Paradox The Naïve Comprehension Schema leads to a contradiction. Therefore it is false. There are some properties with no corresponding set of things that have those properties.

Questions about Russell’s Paradox Is there something wrong with the condition? Does it make sense for a set to be a member of itself? Does it make sense for a set to not be a member of itself? Must set theory be false if its basic idea is?