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.
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.
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.
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 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.
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.
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.
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.
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.
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.
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 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.
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.
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.
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).
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.
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.
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.
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 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 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).
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.
metrizable topological space. Fact There is a Borel isomorphism between any two uncountable Polish spaces. Examples R, [0, 1], P(ω), ω2 = {countable infinite sequences of 0s and 1s}, ωω = {countable infinite sequences of natural numbers}
metrizable topological space. Fact There is a Borel isomorphism between any two uncountable Polish spaces. Examples R, [0, 1], P(ω), ω2 = {countable infinite sequences of 0s and 1s}, ωω = {countable infinite sequences of natural numbers} The cardinality of each of these examples is c = 2ℵ0 .
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.
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.
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 For every uncountable closed set C ⊆ R, |C| = 2ℵ0 .
introduced by Kurt G˝ odel in 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.”
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.
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 Iω denote the set of countable subsets of [0, 1]. AS: For every f : I → Iω, there are x, y ∈ I such that x ∈ f (y) and y ∈ f (x). Lemma If CH is true, then AS is false.
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 ∈ f (y) and y ∈ f (x). Lemma If CH is true, then AS is false. Proof. Suppose CH is true and fix an enumeration of I: xα | α < ω1 .
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 ∈ f (y) and y ∈ f (x). Lemma If CH is true, then AS is false. Proof. Suppose CH is true and fix an enumeration of I: xα | α < ω1 . Define f : I → Iω by f (xα) = {xβ | β < α}.
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 ∈ f (y) and y ∈ f (x). Lemma If CH is true, then AS is false. Proof. Suppose CH is true and fix an enumeration of I: xα | α < ω1 . 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.
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].
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.
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.
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.
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.
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.
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.
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.
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.
technique of forcing to construct a model S of ZF + DC in which: 1 All sets of real numbers are Lebesgue measurable. 2 All sets of real numbers have the perfect set property.
technique of forcing to construct a model S of ZF + DC in which: 1 All sets of real numbers are Lebesgue measurable. 2 All sets of real numbers have the perfect set property. The existence of an injective function f : ω1 → R is enough to construct a non-measurable set, so, in S, ℵ1 and 2ℵ0 are not even comparable.
technique of forcing to construct a model S of ZF + DC in which: 1 All sets of real numbers are Lebesgue measurable. 2 All sets of real numbers have the perfect set property. The existence of an injective function f : ω1 → R is enough to construct a non-measurable set, so, in S, ℵ1 and 2ℵ0 are not even comparable. However, 2 implies that, in S, every set of real numbers is either countable or of cardinality 2ℵ0 , so a form of CH is true in S.
technique of forcing to construct a model S of ZF + DC in which: 1 All sets of real numbers are Lebesgue measurable. 2 All sets of real numbers have the perfect set property. The existence of an injective function f : ω1 → R is enough to construct a non-measurable set, so, in S, ℵ1 and 2ℵ0 are not even comparable. However, 2 implies that, in S, every set of real numbers is either countable or of cardinality 2ℵ0 , so a form of CH is true in S. 1 implies that AS is true in S.
Chris Lambie-Hanson
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