The Continuum Hypothesis, the Axiom of Choice, and Lebesgue Measurability

Chris Lambie-Hanson

CMU Graduate Student Seminar

11 October 2011 1 For all x ∈ S, x ≤ x. 2 For all x, y ∈ S, if x ≤ y and y ≤ x, then x = y. 3 For all x, y, z ∈ S, if x ≤ y and y ≤ z, then x ≤ z. 4 For all x, y ∈ S, x ≤ y or y ≤ x. A linear order ≤ on a set S is a well-order if, for every nonempty X ⊆ S, there is a ≤-least element in X , i.e. there is x ∈ X such that, for all y ∈ X , x ≤ y.

Deﬁnitions

If S is a set, then a linear order on S is a binary relation ≤ such that 2 For all x, y ∈ S, if x ≤ y and y ≤ x, then x = y. 3 For all x, y, z ∈ S, if x ≤ y and y ≤ z, then x ≤ z. 4 For all x, y ∈ S, x ≤ y or y ≤ x. A linear order ≤ on a set S is a well-order if, for every nonempty X ⊆ S, there is a ≤-least element in X , i.e. there is x ∈ X such that, for all y ∈ X , x ≤ y.

Deﬁnitions

If S is a set, then a linear order on S is a binary relation ≤ such that 1 For all x ∈ S, x ≤ x. 3 For all x, y, z ∈ S, if x ≤ y and y ≤ z, then x ≤ z. 4 For all x, y ∈ S, x ≤ y or y ≤ x. A linear order ≤ on a set S is a well-order if, for every nonempty X ⊆ S, there is a ≤-least element in X , i.e. there is x ∈ X such that, for all y ∈ X , x ≤ y.

Deﬁnitions

If S is a set, then a linear order on S is a binary relation ≤ such that 1 For all x ∈ S, x ≤ x. 2 For all x, y ∈ S, if x ≤ y and y ≤ x, then x = y. 4 For all x, y ∈ S, x ≤ y or y ≤ x. A linear order ≤ on a set S is a well-order if, for every nonempty X ⊆ S, there is a ≤-least element in X , i.e. there is x ∈ X such that, for all y ∈ X , x ≤ y.

Deﬁnitions

If S is a set, then a linear order on S is a binary relation ≤ such that 1 For all x ∈ S, x ≤ x. 2 For all x, y ∈ S, if x ≤ y and y ≤ x, then x = y. 3 For all x, y, z ∈ S, if x ≤ y and y ≤ z, then x ≤ z. A linear order ≤ on a set S is a well-order if, for every nonempty X ⊆ S, there is a ≤-least element in X , i.e. there is x ∈ X such that, for all y ∈ X , x ≤ y.

Deﬁnitions

If S is a set, then a linear order on S is a binary relation ≤ such that 1 For all x ∈ S, x ≤ x. 2 For all x, y ∈ S, if x ≤ y and y ≤ x, then x = y. 3 For all x, y, z ∈ S, if x ≤ y and y ≤ z, then x ≤ z. 4 For all x, y ∈ S, x ≤ y or y ≤ x. A linear order ≤ on a set S is a well-order if, for every nonempty X ⊆ S, there is a ≤-least element in X , i.e. there is x ∈ X such that, for all y ∈ X , x ≤ y. The Axiom of Choice is equivalent to the Well-ordering Theorem, which asserts that every set can be well-ordered.

The Axiom of Choice

The Axiom of Choice is the assertion: For every family of nonempty sets F, there is a function g such that dom(g) = F and, for every X ∈ F, g(X ) ∈ X . The Axiom of Choice

The Axiom of Choice is the assertion: For every family of nonempty sets F, there is a function g such that dom(g) = F and, for every X ∈ F, g(X ) ∈ X .

The Axiom of Choice is equivalent to the Well-ordering Theorem, which asserts that every set can be well-ordered. • A natural number n is an ordinal describing a well-ordering with n elements. • The ordinal describing the order type of the natural numbers is ω. • The ordinal describing the order type of the natural numbers plus one element larger than all of the natural numbers is ω + 1. An ordinal α is actually a set of ordinals, well-ordered by ∈, of order-type α. If α = β + 1 = β ∪ {β}, then α is called a successor ordinal. Otherwise, it is called a limit ordinal.

Ordinals, Informally

An ordinal number describes the order type of a well-ordering. • The ordinal describing the order type of the natural numbers is ω. • The ordinal describing the order type of the natural numbers plus one element larger than all of the natural numbers is ω + 1. An ordinal α is actually a set of ordinals, well-ordered by ∈, of order-type α. If α = β + 1 = β ∪ {β}, then α is called a successor ordinal. Otherwise, it is called a limit ordinal.

Ordinals, Informally

An ordinal number describes the order type of a well-ordering. • A natural number n is an ordinal describing a well-ordering with n elements. • The ordinal describing the order type of the natural numbers plus one element larger than all of the natural numbers is ω + 1. An ordinal α is actually a set of ordinals, well-ordered by ∈, of order-type α. If α = β + 1 = β ∪ {β}, then α is called a successor ordinal. Otherwise, it is called a limit ordinal.

Ordinals, Informally

An ordinal number describes the order type of a well-ordering. • A natural number n is an ordinal describing a well-ordering with n elements. • The ordinal describing the order type of the natural numbers is ω. An ordinal α is actually a set of ordinals, well-ordered by ∈, of order-type α. If α = β + 1 = β ∪ {β}, then α is called a successor ordinal. Otherwise, it is called a limit ordinal.

Ordinals, Informally

An ordinal number describes the order type of a well-ordering. • A natural number n is an ordinal describing a well-ordering with n elements. • The ordinal describing the order type of the natural numbers is ω. • The ordinal describing the order type of the natural numbers plus one element larger than all of the natural numbers is ω + 1. An ordinal α is actually a set of ordinals, well-ordered by ∈, of order-type α. If α = β + 1 = β ∪ {β}, then α is called a successor ordinal. Otherwise, it is called a limit ordinal. The cardinality of A, written |A|, can be thought of as the equivalence class of all sets for which bijections with A exist. For two sets A and B, we say |A| |B| if there is an injective function f : A → B. Under the Axiom of Choice, a cardinality is often identiﬁed with the smallest ordinal of that cardinality. Thus, under AC, cardinalities are well-ordered by . If α is an ordinal, the α-th inﬁnite cardinal is written ℵα. The smallest ordinal of cardinality ℵα is written ωα.

Cardinals

Two sets A and B have the same cardinality if there is a bijection f : A → B. For two sets A and B, we say |A| |B| if there is an injective function f : A → B. Under the Axiom of Choice, a cardinality is often identiﬁed with the smallest ordinal of that cardinality. Thus, under AC, cardinalities are well-ordered by . If α is an ordinal, the α-th inﬁnite cardinal is written ℵα. The smallest ordinal of cardinality ℵα is written ωα.

Cardinals

Two sets A and B have the same cardinality if there is a bijection f : A → B. The cardinality of A, written |A|, can be thought of as the equivalence class of all sets for which bijections with A exist. Under the Axiom of Choice, a cardinality is often identiﬁed with the smallest ordinal of that cardinality. Thus, under AC, cardinalities are well-ordered by . If α is an ordinal, the α-th inﬁnite cardinal is written ℵα. The smallest ordinal of cardinality ℵα is written ωα.

Cardinals

Two sets A and B have the same cardinality if there is a bijection f : A → B. The cardinality of A, written |A|, can be thought of as the equivalence class of all sets for which bijections with A exist. For two sets A and B, we say |A| |B| if there is an injective function f : A → B. Under the Axiom of Choice, a cardinality is often identiﬁed with the smallest ordinal of that cardinality. Thus, under AC, cardinalities are well-ordered by . Cardinals

Two sets A and B have the same cardinality if there is a bijection f : A → B. The cardinality of A, written |A|, can be thought of as the equivalence class of all sets for which bijections with A exist. For two sets A and B, we say |A| |B| if there is an injective function f : A → B. Under the Axiom of Choice, a cardinality is often identiﬁed with the smallest ordinal of that cardinality. Thus, under AC, cardinalities are well-ordered by . If α is an ordinal, the α-th inﬁnite cardinal is written ℵα. The smallest ordinal of cardinality ℵα is written ωα. Every natural number n is a cardinal number.

ω0 = ω, and |ω| = ℵ0.

ω1 is the set of all countable ordinals, and |ω1| = ℵ1.

ω2 is the set of all ordinals of cardinality ≤ ℵ1, and |ω2| = ℵ2.

Cardinals

A set is called countable if it is either ﬁnite or has the same cardinality as the set of natural numbers. ω0 = ω, and |ω| = ℵ0.

ω1 is the set of all countable ordinals, and |ω1| = ℵ1.

ω2 is the set of all ordinals of cardinality ≤ ℵ1, and |ω2| = ℵ2.

Cardinals

A set is called countable if it is either ﬁnite or has the same cardinality as the set of natural numbers.

Every natural number n is a cardinal number. ω1 is the set of all countable ordinals, and |ω1| = ℵ1.

ω2 is the set of all ordinals of cardinality ≤ ℵ1, and |ω2| = ℵ2.

Cardinals

A set is called countable if it is either ﬁnite or has the same cardinality as the set of natural numbers.

Every natural number n is a cardinal number.

ω0 = ω, and |ω| = ℵ0. ω2 is the set of all ordinals of cardinality ≤ ℵ1, and |ω2| = ℵ2.

Cardinals

A set is called countable if it is either ﬁnite or has the same cardinality as the set of natural numbers.

Every natural number n is a cardinal number.

ω0 = ω, and |ω| = ℵ0.

ω1 is the set of all countable ordinals, and |ω1| = ℵ1. Cardinals

A set is called countable if it is either ﬁnite or has the same cardinality as the set of natural numbers.

Every natural number n is a cardinal number.

ω0 = ω, and |ω| = ℵ0.

ω1 is the set of all countable ordinals, and |ω1| = ℵ1.

ω2 is the set of all ordinals of cardinality ≤ ℵ1, and |ω2| = ℵ2. Theorem For every set A, |A| ≺ |P(A)|.

It is clear that |A| |P(A)|, so it suﬃces to prove that there is no bijection f : A → P(A).

Cantor’s Theorem

If A is a set, then the power set of A, written P(A), is the set of all subsets of A. It is clear that |A| |P(A)|, so it suﬃces to prove that there is no bijection f : A → P(A).

Cantor’s Theorem

If A is a set, then the power set of A, written P(A), is the set of all subsets of A. Theorem For every set A, |A| ≺ |P(A)|. Cantor’s Theorem

If A is a set, then the power set of A, written P(A), is the set of all subsets of A. Theorem For every set A, |A| ≺ |P(A)|.

It is clear that |A| |P(A)|, so it suffices to prove that there is no bijection f : A → P(A). Hypergame is a two-player game with alternating moves played as follows: 1 Player 1 names a finite game. 2 Player 2 makes the first move in that finite game. 3 The players continue playing that finite game until it ends.

Hypergame

We say a finite game is a two-player game with alternating moves which is guaranteed to end in a finite number of moves. 1 Player 1 names a finite game. 2 Player 2 makes the first move in that finite game. 3 The players continue playing that finite game until it ends.

Hypergame

We say a ﬁnite game is a two-player game with alternating moves which is guaranteed to end in a ﬁnite number of moves.

Hypergame is a two-player game with alternating moves played as follows: 2 Player 2 makes the first move in that finite game. 3 The players continue playing that finite game until it ends.

Hypergame

We say a ﬁnite game is a two-player game with alternating moves which is guaranteed to end in a ﬁnite number of moves.

Hypergame is a two-player game with alternating moves played as follows: 1 Player 1 names a ﬁnite game. 3 The players continue playing that ﬁnite game until it ends.

Hypergame

We say a ﬁnite game is a two-player game with alternating moves which is guaranteed to end in a ﬁnite number of moves.

Hypergame is a two-player game with alternating moves played as follows: 1 Player 1 names a finite game. 2 Player 2 makes the first move in that finite game. Hypergame

We say a ﬁnite game is a two-player game with alternating moves which is guaranteed to end in a ﬁnite number of moves.

Hypergame is a two-player game with alternating moves played as follows: 1 Player 1 names a finite game. 2 Player 2 makes the first move in that finite game. 3 The players continue playing that finite game until it ends. I: “Let’s play chess!” II:f3 I: e6 II:g4 I: Qh4#

Hypergame

Example: II:f3 I: e6 II:g4 I: Qh4#

Hypergame

Example: I: “Let’s play chess!” I: e6 II:g4 I: Qh4#

Hypergame

Example: I: “Let’s play chess!” II:f3 II:g4 I: Qh4#

Hypergame

Example: I: “Let’s play chess!” II:f3 I: e6 I: Qh4#

Hypergame

Example: I: “Let’s play chess!” II:f3 I: e6 II:g4 Hypergame

Example: I: “Let’s play chess!” II:f3 I: e6 II:g4 I: Qh4# Clearly, yes, but... I: “Let’s play hypergame!” II:“Let’s play hypergame!” I: “Let’s play hypergame!” II:“Let’s play hypergame!” ...

Hypergame

Is Hypergame a ﬁnite game? I: “Let’s play hypergame!” II:“Let’s play hypergame!” I: “Let’s play hypergame!” II:“Let’s play hypergame!” ...

Hypergame

Is Hypergame a ﬁnite game? Clearly, yes, but... II:“Let’s play hypergame!” I: “Let’s play hypergame!” II:“Let’s play hypergame!” ...

Hypergame

Is Hypergame a ﬁnite game? Clearly, yes, but... I: “Let’s play hypergame!” I: “Let’s play hypergame!” II:“Let’s play hypergame!” ...

Hypergame

Is Hypergame a ﬁnite game? Clearly, yes, but... I: “Let’s play hypergame!” II:“Let’s play hypergame!” II:“Let’s play hypergame!” ...

Hypergame

Is Hypergame a ﬁnite game? Clearly, yes, but... I: “Let’s play hypergame!” II:“Let’s play hypergame!” I: “Let’s play hypergame!” Hypergame

Is Hypergame a ﬁnite game? Clearly, yes, but... I: “Let’s play hypergame!” II:“Let’s play hypergame!” I: “Let’s play hypergame!” II:“Let’s play hypergame!” ... Suppose for sake of contradiction that there is an inﬁnite set A and a bijection f : A → P(A). We say that (a0, a1, a2,...) is a path in A if:

1 a0 ∈ A

2 For all n, an+1 ∈ f (an) Let X = {a ∈ A | there is no infinite path in A starting with a}. Since f is a bijection, there is x ∈ A such that f (x) = X . There can be no infinite path in A starting with x, as the second element of such a path would have to be in X . Thus, x ∈ f (x). But then (x, x, x,...) is an infinite path in A starting with x. Contradiction.

Proof of Cantor’s Theorem We say that (a0, a1, a2,...) is a path in A if:

1 a0 ∈ A

Proof of Cantor’s Theorem

Suppose for sake of contradiction that there is an infinite set A and a bijection f : A → P(A). Let X = {a ∈ A | there is no infinite path in A starting with a}. Since f is a bijection, there is x ∈ A such that f (x) = X . There can be no infinite path in A starting with x, as the second element of such a path would have to be in X . Thus, x ∈ f (x). But then (x, x, x,...) is an infinite path in A starting with x. Contradiction.

Proof of Cantor’s Theorem

Suppose for sake of contradiction that there is an inﬁnite set A and a bijection f : A → P(A). We say that (a0, a1, a2,...) is a path in A if:

1 a0 ∈ A

2 For all n, an+1 ∈ f (an) Since f is a bijection, there is x ∈ A such that f (x) = X . There can be no inﬁnite path in A starting with x, as the second element of such a path would have to be in X . Thus, x ∈ f (x). But then (x, x, x,...) is an inﬁnite path in A starting with x. Contradiction.

Proof of Cantor’s Theorem

Suppose for sake of contradiction that there is an inﬁnite set A and a bijection f : A → P(A). We say that (a0, a1, a2,...) is a path in A if:

1 a0 ∈ A

2 For all n, an+1 ∈ f (an) Let X = {a ∈ A | there is no infinite path in A starting with a}. There can be no infinite path in A starting with x, as the second element of such a path would have to be in X . Thus, x ∈ f (x). But then (x, x, x,...) is an infinite path in A starting with x. Contradiction.

Proof of Cantor’s Theorem

Suppose for sake of contradiction that there is an inﬁnite set A and a bijection f : A → P(A). We say that (a0, a1, a2,...) is a path in A if:

1 a0 ∈ A

2 For all n, an+1 ∈ f (an) Let X = {a ∈ A | there is no inﬁnite path in A starting with a}. Since f is a bijection, there is x ∈ A such that f (x) = X . But then (x, x, x,...) is an inﬁnite path in A starting with x. Contradiction.

Proof of Cantor’s Theorem

Suppose for sake of contradiction that there is an inﬁnite set A and a bijection f : A → P(A). We say that (a0, a1, a2,...) is a path in A if:

1 a0 ∈ A

Suppose for sake of contradiction that there is an inﬁnite set A and a bijection f : A → P(A). We say that (a0, a1, a2,...) is a path in A if:

1 a0 ∈ A

Fact There is a Borel isomorphism between any two uncountable Polish spaces. Examples ω R, [0, 1], P(ω), 2 = {countable inﬁnite sequences of 0s and 1s}, ωω = {countable inﬁnite sequences of natural numbers} The cardinality of each of these examples is c = 2ℵ0 .

Real numbers Fact There is a Borel isomorphism between any two uncountable Polish spaces. Examples ω R, [0, 1], P(ω), 2 = {countable inﬁnite sequences of 0s and 1s}, ωω = {countable inﬁnite sequences of natural numbers} The cardinality of each of these examples is c = 2ℵ0 .

Real numbers

Definition A Polish space is a separable, completely metrizable topological space. Examples ω R, [0, 1], P(ω), 2 = {countable infinite sequences of 0s and 1s}, ωω = {countable infinite sequences of natural numbers} The cardinality of each of these examples is c = 2ℵ0 .

Real numbers

Deﬁnition A Polish space is a separable, completely metrizable topological space.

Fact There is a Borel isomorphism between any two uncountable Polish spaces. The cardinality of each of these examples is c = 2ℵ0 .

Real numbers

Deﬁnition A Polish space is a separable, completely metrizable topological space.

• There is a surjection f : ω1 → R. • There is an injection g : R → ω1. ℵ • Every uncountable subset of R has cardinality 2 0 . Hilbert’s 1st problem.

The Continuum Hypothesis

ℵ CH: 2 0 = ℵ1. • There is a surjection f : ω1 → R. • There is an injection g : R → ω1. ℵ • Every uncountable subset of R has cardinality 2 0 . Hilbert’s 1st problem.

The Continuum Hypothesis

ℵ CH: 2 0 = ℵ1.

Alternative formulations: • There is an injection g : R → ω1. ℵ • Every uncountable subset of R has cardinality 2 0 . Hilbert’s 1st problem.

The Continuum Hypothesis

ℵ CH: 2 0 = ℵ1.

Alternative formulations:

• There is a surjection f : ω1 → R. ℵ • Every uncountable subset of R has cardinality 2 0 . Hilbert’s 1st problem.

The Continuum Hypothesis

ℵ CH: 2 0 = ℵ1.

Alternative formulations:

• There is a surjection f : ω1 → R. • There is an injection g : R → ω1. Hilbert’s 1st problem.

The Continuum Hypothesis

ℵ CH: 2 0 = ℵ1.

Alternative formulations:

• There is a surjection f : ω1 → R. • There is an injection g : R → ω1. ℵ • Every uncountable subset of R has cardinality 2 0 . The Continuum Hypothesis

ℵ CH: 2 0 = ℵ1.

Alternative formulations:

• There is a surjection f : ω1 → R. • There is an injection g : R → ω1. ℵ • Every uncountable subset of R has cardinality 2 0 . Hilbert’s 1st problem. Figure: Georg Cantor (1845-1918) Deﬁnition If X is a topological space, then a set S ⊆ X is a perfect set if it is closed and has no isolated points.

Fact If S is a nonempty perfect subset of a Polish space, then |S| = 2ℵ0 . Deﬁnition A subset A of a Polish space has the perfect set property if either A is countable or A contains a nonempty perfect subset.

Theorem If X is a Polish space, then every closed set C ⊆ X can be written uniquely as the disjoint union of a perfect set and a countable set. Corollary ℵ For every uncountable closed set C ⊆ R, |C| = 2 0 .

Cantor-Bendixson Theorem Fact If S is a nonempty perfect subset of a Polish space, then |S| = 2ℵ0 . Deﬁnition A subset A of a Polish space has the perfect set property if either A is countable or A contains a nonempty perfect subset.

Cantor-Bendixson Theorem Deﬁnition If X is a topological space, then a set S ⊆ X is a perfect set if it is closed and has no isolated points. Deﬁnition A subset A of a Polish space has the perfect set property if either A is countable or A contains a nonempty perfect subset.

Cantor-Bendixson Theorem Deﬁnition If X is a topological space, then a set S ⊆ X is a perfect set if it is closed and has no isolated points.

Fact If S is a nonempty perfect subset of a Polish space, then |S| = 2ℵ0 . Theorem If X is a Polish space, then every closed set C ⊆ X can be written uniquely as the disjoint union of a perfect set and a countable set. Corollary ℵ For every uncountable closed set C ⊆ R, |C| = 2 0 .

Cantor-Bendixson Theorem Deﬁnition If X is a topological space, then a set S ⊆ X is a perfect set if it is closed and has no isolated points.

Theorem If X is a Polish space, then every closed set C ⊆ X can be written uniquely as the disjoint union of a perfect set and a countable set. Cantor-Bendixson Theorem Deﬁnition If X is a topological space, then a set S ⊆ X is a perfect set if it is closed and has no isolated points.

CH is true in L.

Shelah: “L looks like the head of a gay chapter of the Ku Klux Klan - a case worthy of study, but probably not representative.”

G˝odel’sConstructible Universe

The constructible universe, denoted L, was introduced by Kurt G˝odelin 1938. It is a model of ZFC. CH is true in L.

Shelah: “L looks like the head of a gay chapter of the Ku Klux Klan - a case worthy of study, but probably not representative.”

G˝odel’sConstructible Universe

The constructible universe, denoted L, was introduced by Kurt G˝odelin 1938. It is a model of ZFC.

L is, in a sense, the smallest model of ZFC containing all of the ordinals. Shelah: “L looks like the head of a gay chapter of the Ku Klux Klan - a case worthy of study, but probably not representative.”

G˝odel’sConstructible Universe

The constructible universe, denoted L, was introduced by Kurt G˝odelin 1938. It is a model of ZFC.

L is, in a sense, the smallest model of ZFC containing all of the ordinals.

CH is true in L. G˝odel’sConstructible Universe

The constructible universe, denoted L, was introduced by Kurt G˝odelin 1938. It is a model of ZFC.

L is, in a sense, the smallest model of ZFC containing all of the ordinals.

CH is true in L.

Shelah: “L looks like the head of a gay chapter of the Ku Klux Klan - a case worthy of study, but probably not representative.” In 1963, Paul Cohen introduced the technique of forcing, which allows for the construction of new models of ZFC.

”Adding sets, but very gently.”

Cohen used forcing to construct a model of ZFC in which CH is false by adding ℵ2-many new subsets of ω.

This “settled” Hilbert’s ﬁrst problem.

Forcing ”Adding sets, but very gently.”

Cohen used forcing to construct a model of ZFC in which CH is false by adding ℵ2-many new subsets of ω.

This “settled” Hilbert’s ﬁrst problem.

Forcing

In 1963, Paul Cohen introduced the technique of forcing, which allows for the construction of new models of ZFC. Cohen used forcing to construct a model of ZFC in which CH is false by adding ℵ2-many new subsets of ω.

This “settled” Hilbert’s ﬁrst problem.

Forcing

In 1963, Paul Cohen introduced the technique of forcing, which allows for the construction of new models of ZFC.

”Adding sets, but very gently.” This “settled” Hilbert’s ﬁrst problem.

Forcing

In 1963, Paul Cohen introduced the technique of forcing, which allows for the construction of new models of ZFC.

”Adding sets, but very gently.”

Cohen used forcing to construct a model of ZFC in which CH is false by adding ℵ2-many new subsets of ω. Forcing

In 1963, Paul Cohen introduced the technique of forcing, which allows for the construction of new models of ZFC.

”Adding sets, but very gently.”

Cohen used forcing to construct a model of ZFC in which CH is false by adding ℵ2-many new subsets of ω.

This “settled” Hilbert’s first problem. Let I = [0, 1], and let Iω denote the set of countable subsets of [0, 1]. AS: For every f : I → Iω, there are x, y ∈ I such that x 6∈ f (y) and y 6∈ f (x). Lemma If CH is true, then AS is false. Proof. Suppose CH is true and fix an enumeration of I: hxα | α < ω1i. Define f : I → Iω by f (xα) = {xβ | β < α}. Then, for any α < β < ω1, xα ∈ f (xβ), so f is a counterexample to AS.

The Axiom of Symmetry AS: For every f : I → Iω, there are x, y ∈ I such that x 6∈ f (y) and y 6∈ f (x). Lemma If CH is true, then AS is false. Proof. Suppose CH is true and ﬁx an enumeration of I: hxα | α < ω1i. Deﬁne f : I → Iω by f (xα) = {xβ | β < α}. Then, for any α < β < ω1, xα ∈ f (xβ), so f is a counterexample to AS.

The Axiom of Symmetry

Let I = [0, 1], and let Iω denote the set of countable subsets of [0, 1]. Lemma If CH is true, then AS is false. Proof. Suppose CH is true and ﬁx an enumeration of I: hxα | α < ω1i. Deﬁne f : I → Iω by f (xα) = {xβ | β < α}. Then, for any α < β < ω1, xα ∈ f (xβ), so f is a counterexample to AS.

The Axiom of Symmetry

Let I = [0, 1], and let Iω denote the set of countable subsets of [0, 1]. AS: For every f : I → Iω, there are x, y ∈ I such that x 6∈ f (y) and y 6∈ f (x). Proof. Suppose CH is true and ﬁx an enumeration of I: hxα | α < ω1i. Deﬁne f : I → Iω by f (xα) = {xβ | β < α}. Then, for any α < β < ω1, xα ∈ f (xβ), so f is a counterexample to AS.

The Axiom of Symmetry

Let I = [0, 1], and let Iω denote the set of countable subsets of [0, 1]. AS: For every f : I → Iω, there are x, y ∈ I such that x 6∈ f (y) and y 6∈ f (x). Lemma If CH is true, then AS is false. Deﬁne f : I → Iω by f (xα) = {xβ | β < α}. Then, for any α < β < ω1, xα ∈ f (xβ), so f is a counterexample to AS.

The Axiom of Symmetry

Let I = [0, 1], and let Iω denote the set of countable subsets of [0, 1]. AS: For every f : I → Iω, there are x, y ∈ I such that x 6∈ f (y) and y 6∈ f (x). Lemma If CH is true, then AS is false. Proof. Suppose CH is true and fix an enumeration of I: hxα | α < ω1i. Define f : I → Iω by f (xα) = {xβ | β < α}. Then, for any α < β < ω1, xα ∈ f (xβ), so f is a counterexample to AS. Freiling’s Dartboard Argument: Fix an f : I → Iω. Throw two darts at the interval [0, 1]. If the first dart is thrown and lands at point x, then, since f (x) is countable and thus has measure 0, the probability that the second dart lands in f (x) is 0. Since this is true no matter where the first dart lands, it should be true even before the first dart is thrown. By symmetry of the order in which the darts are thrown, if y denotes the point where the second dart lands, the probability that the first dart lands in f (y) is 0. So, almost surely, the two darts will land at two points witnessing that AS holds for f .

The Axiom of Symmetry

AS is actually equivalent to the negation of CH and was introduced by Chris Freiling as an argument against CH. Fix an f : I → Iω. Throw two darts at the interval [0, 1]. If the first dart is thrown and lands at point x, then, since f (x) is countable and thus has measure 0, the probability that the second dart lands in f (x) is 0. Since this is true no matter where the first dart lands, it should be true even before the first dart is thrown. By symmetry of the order in which the darts are thrown, if y denotes the point where the second dart lands, the probability that the first dart lands in f (y) is 0. So, almost surely, the two darts will land at two points witnessing that AS holds for f .

The Axiom of Symmetry

AS is actually equivalent to the negation of CH and was introduced by Chris Freiling as an argument against CH.

Freiling’s Dartboard Argument: Throw two darts at the interval [0, 1]. If the first dart is thrown and lands at point x, then, since f (x) is countable and thus has measure 0, the probability that the second dart lands in f (x) is 0. Since this is true no matter where the first dart lands, it should be true even before the first dart is thrown. By symmetry of the order in which the darts are thrown, if y denotes the point where the second dart lands, the probability that the first dart lands in f (y) is 0. So, almost surely, the two darts will land at two points witnessing that AS holds for f .

The Axiom of Symmetry

AS is actually equivalent to the negation of CH and was introduced by Chris Freiling as an argument against CH.

Freiling’s Dartboard Argument: Fix an f : I → Iω. If the first dart is thrown and lands at point x, then, since f (x) is countable and thus has measure 0, the probability that the second dart lands in f (x) is 0. Since this is true no matter where the first dart lands, it should be true even before the first dart is thrown. By symmetry of the order in which the darts are thrown, if y denotes the point where the second dart lands, the probability that the first dart lands in f (y) is 0. So, almost surely, the two darts will land at two points witnessing that AS holds for f .

The Axiom of Symmetry

AS is actually equivalent to the negation of CH and was introduced by Chris Freiling as an argument against CH.

Freiling’s Dartboard Argument: Fix an f : I → Iω. Throw two darts at the interval [0, 1]. Since this is true no matter where the first dart lands, it should be true even before the first dart is thrown. By symmetry of the order in which the darts are thrown, if y denotes the point where the second dart lands, the probability that the first dart lands in f (y) is 0. So, almost surely, the two darts will land at two points witnessing that AS holds for f .

The Axiom of Symmetry

AS is actually equivalent to the negation of CH and was introduced by Chris Freiling as an argument against CH.

Freiling’s Dartboard Argument: Fix an f : I → Iω. Throw two darts at the interval [0, 1]. If the ﬁrst dart is thrown and lands at point x, then, since f (x) is countable and thus has measure 0, the probability that the second dart lands in f (x) is 0. By symmetry of the order in which the darts are thrown, if y denotes the point where the second dart lands, the probability that the ﬁrst dart lands in f (y) is 0. So, almost surely, the two darts will land at two points witnessing that AS holds for f .

The Axiom of Symmetry

AS is actually equivalent to the negation of CH and was introduced by Chris Freiling as an argument against CH.

The Axiom of Symmetry

AS is actually equivalent to the negation of CH and was introduced by Chris Freiling as an argument against CH.

Freiling’s Dartboard Argument: Fix an f : I → Iω. Throw two darts at the interval [0, 1]. If the first dart is thrown and lands at point x, then, since f (x) is countable and thus has measure 0, the probability that the second dart lands in f (x) is 0. Since this is true no matter where the first dart lands, it should be true even before the first dart is thrown. By symmetry of the order in which the darts are thrown, if y denotes the point where the second dart lands, the probability that the first dart lands in f (y) is 0. So, almost surely, the two darts will land at two points witnessing that AS holds for f . One objection involves Lebesgue measurability: Freiling’s argument seems to appeal to an intuition that {(x, y) | x ∈ f (y)} and {(x, y) | y ∈ f (x)} both have measure 0. This intuition seems to require that there is a very well-behaved measure on the reals, which is precluded by the Axiom of Choice. In fact, if f is a counterexample to AS, then both {(x, y) | x ∈ f (y)} and {(x, y) | y ∈ f (x)} are non-measurable.

The other objection to Freiling’s argument also features the Axiom of Choice in a prominent role. These two objections suggest that AS might really be more about Choice than about CH.

Problems with Freiling’s Arguments

Two main objections can be raised. Freiling’s argument seems to appeal to an intuition that {(x, y) | x ∈ f (y)} and {(x, y) | y ∈ f (x)} both have measure 0. This intuition seems to require that there is a very well-behaved measure on the reals, which is precluded by the Axiom of Choice. In fact, if f is a counterexample to AS, then both {(x, y) | x ∈ f (y)} and {(x, y) | y ∈ f (x)} are non-measurable.

The other objection to Freiling’s argument also features the Axiom of Choice in a prominent role. These two objections suggest that AS might really be more about Choice than about CH.

Problems with Freiling’s Arguments