# Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis

## Abstract

**:**

## 1. Introduction

## 2. Brown on the Continuum Hypothesis

#### 2.1. Brown’s Version

- (CA)
- m($\bigcup _{n=1}^{\infty}$ A
_{n}) = $\sum _{n=1}^{\infty}$m(A_{n}) , provided the sets A_{n}are disjoint and measurable.

- (i)
- Randomness. Each throw selects a point randomly: Any two sub-intervals of equal length are equally likely to receive the dart. More generally: If we specify any measurable subset E of [0, 1], the probability Pr(x ∈ E) that the dart will hit a point x in E is given by the Lebesgue measure m(E).
- (ii)
- Independence. The two throws are independent: neither has any influence on the other. Formally, if E and F are any two (measurable) subsets of [0, 1], then Pr(x ∈ E/y ∈ F) = Pr(x ∈ E) and Pr(y ∈ F / x ∈ E) = Pr(y ∈ F).
- (iii)
- Symmetry. Brown writes [2]: “The independence and randomness of the darts guarantees the symmetry of the throws. Consequently, either dart may be considered the first throw.” Brown means: In determining the probability of any outcome for the darts taken singly or as a pair, we may freely suppose that either dart is the first throw.

_{q}= {x ∈ [0, 1]/x < q}, the initial segment of [0, 1] that precedes q in the well-ordering. An important fact about well-ordered sets is that no well-ordered set is order-isomorphic to any of its initial segments.3

- (CH)
- The cardinality of the real numbers, and hence of [0, 1], is ℵ
_{1}: the least uncountable cardinal number.

_{q}is a countable set.4

_{q}. By Symmetry, we may regard the throw that lands on q as having occurred first, giving us the value of q and fixing the countable set S

_{q}. By Independence and Randomness, the probability that the other dart lands in S

_{q}is Pr(p < q) = m(S

_{q}) = 0, since S

_{q}is a countable set. In effect, the first dart fixes a set of measure zero (namely, S

_{q}), and the second dart throw is thus a near-miraculous event: the selection of a point in that measure-zero set. There is a parallel result if q < p, since m(S

_{p}) = 0 as well. Thus, every time a pair of darts is thrown, an event of zero probability will occur. While logically possible, this is absurd. The upshot is that we should abandon CH.

_{q}) = 0 for each q. If we employ the usual ordering relation ≤ instead, then the initial segment {x/0 ≤ x ≤ q} would have measure q, for each q, and there is no paradoxical conclusion.

_{q}, as fixed prior to the second throw, so that we may identify Pr(p < q) with m(S

_{q}). He clarifies [2]: “A prediction based on either throw cannot be dismissed in the way we might dismiss someone who said of a license number on a passing car: ‘Wow, there was only a one in a million chance of that happening.’ We are rightly impressed only if the number is fixed independently of the outcome (i.e., predicted before the result is known).”

#### 2.2. Freiling’s Version

- (FSA)
- (∀f: ℝ → ℝ
_{ℵ0})∃x ∃y(y ∉ f(x) & x ∉ f(y))

_{ℵ0}is the set of all countable subsets of ℝ. FSA says: If f assigns a countable set of real numbers to each real number, then we can find two real numbers x and y such that x is not in f(y) and y is not in f(x). Freiling’s refutation of CH now breaks into two steps: a short preliminary proof that FSA is equivalent to ~CH, and a probabilistic argument for FSA. Since there is no question about the validity of the first step, let us look at his probabilistic argument.

_{ℵ0}. For any real number x, f(x) is a countable set and thus m(f(x)) = 0. Select two points p and q in [0, 1] randomly and independently–as in Brown’s version, we may assume this is done by tossing darts. If FSA is false, then either p ∈ f(q) or q ∈ f(p). Without loss of generality, suppose p ∈ f(q). The assumption of independence implies that we may consider the choice of q as prior and that we can identify Pr(p ∈ f(q)) with m(f(q)).5 In defense of this crucial presumption that one may regard q as selected first, Freiling takes the same stance as Brown: “the real number line does not really know which dart was thrown first or second.” Since Pr(p ∈ f(q)) = m(f(q)) = 0, every time we select points p and q at random, an event of zero probability occurs. That is absurd, so we should accept FSA. The parallel with Brown’s argument is obvious: Brown works with the instance of FSA where f(x) = {y/y < x}. Notice, however, that Freiling’s argument does not depend upon the choice of f. Thus, if we can refute Brown’s argument, we also refute Freiling’s version of the argument.

## 3. Critique of the Argument

_{q})? As we saw, Brown justifies this step by appealing to his principle of Symmetry for the dart tosses: “either dart could be considered the first throw.” Neither dart toss has any causal influence on the other, so why should the order of the tosses matter? We may proceed as if the throw that hits the <-larger number, q, comes first, so that S

_{q}may be regarded as fixed.

_{p}, and this probability is appropriately given by the Lebesgue measure of $\overline{{S}_{p}}$ (see Figure 1). But m($\overline{{S}_{p}}$) = 1, so we now have Pr(p < q) = 1, rather than 0! Clearly, there is something wrong with both Brown’s argument, and this one.

_{q}) is irrelevant: we must have Pr(p < q) = 1. Similarly, we could specify that the lesser value is always designated the ‘first toss’, p, in which case we once again have Pr(p < q) = 1. And so on. Let us rule out these possibilities as incompatible with Brown’s description of the problem.

_{q}) (=0), and m($\overline{{S}_{p}}$) (=1). The correct Symmetry principle, however, is not that we may consider either dart to have been tossed first, but rather that the probability Pr(p < q) (and of course all probabilities of outcomes for the pair of darts) should be invariant under permutation of the order of tosses.

_{q}has one-dimensional measure 0. Indeed, that suggests that the two-dimensional measure of S must be 0, by appeal to Fubini’s Theorem, which tells us how to compute iterated integrals ‘by slices’. However, it is not obvious that the requirements for the application of Fubini’s Theorem are met in our example.

- (i)
- Each of the cross-sections S
_{q}and S_{p}has a well-defined one-dimensional measure of 0. - (ii)
- The probability distribution for the pair of throws is given by the product measure.
- (iii)
- S has a well-defined two-dimensional measure (e.g., m(S) = Pr(p < q) = ½).

...not meant to be a mathematical statement of the Lebesgue measurability of a certain type of set. Rather, it is an expression of an obvious, almost physical intuition concerning the inherently nonmathematical notions of prediction, accuracy, and time independence.

_{q}) = 0. The conditional probability of seeing a lower value is indeed zero. If you update your probability for second < first by Bayesian conditionalization, then you must expect (with probability 1) to see a higher value for the second dart. This is puzzling because it is reasoning to a foregone conclusion. Conditional upon any observed value, you believe with probability 1 that the value selected by the next dart will be larger. This is very strange, given the symmetry of the situation. Indeed, you half expect that you will be wrong (well, not precisely half, since Pr(p < q) is undefined). To my mind, this anomaly is the core of Brown’s reductio argument.

## 4. The Double Lottery

The refutation of CH made use of a principle to the effect that when picking out an initial segment, we end up with a set of lower cardinality. We can use this fact to get apparently paradoxical results from smaller well-ordered sets. For instance, pick a pair of natural numbers at random. Let them be m and n. Suppose m is chosen at random. What is the probability that m is less than n? It is zero. Similarly, the probability that n is less than m is also zero. Does this refute the view that the cardinality of the natural numbers is ℵ_{0}? The answer is No, but we should reject the claim that this argument is parallel to Freiling’s. The conclusion this argument actually justifies is that we cannot talk about the probability of picking natural numbers at random… We cannot throw darts at the natural numbers in the same way we can throw them at the reals between [0, 1].

_{q}= {p/p < q} for the dart throw [S

_{q}= {p/p < q} for the lottery] has lower cardinality than that of the full outcome space, and each such initial segment is a set of measure 0.11 The only difference is that CH is needed for the latter result in the Brown-Freiling case, while CA must be dropped if we are to accept Randomness in the de Finetti lottery.

_{q}= {p/p < q}, is 0. So Pr(p < q) = 0. So, each time we draw a pair of tickets, a near-miraculous (probability 0) event occurs—one that is foreseeable. This, it is alleged, is absurd.13

- (i)
- Each of the cross-sections S
_{q}= {p ∈ ℕ/p < q} has one-dimensional measure 0. (Any two tickets are equally good; each ticket must have zero probability of winning.) - (ii)
- The probability distribution for the pair of tickets is given by the product measure.
- (iii)
- S has a well-defined two-dimensional measure: μ(S) = Pr(p < q) = ½. (There is no reason why either ticket number should have a higher chance of being larger.)

## 5. Symmetry and Finitely Additive Measures

#### 5.1. Paradoxical Sets and Finitely Additive Measures

_{k}such that σ

_{k}(i) = j, so once again we have the equiprobability of any two singleton sets. For this choice for G, we get additional interesting relationships: Even = {2, 4, 6, ...} and Odd = {1, 3, 5, ...} are equiprobable.15

`τ`, x) ∈ X

_{1}(A) = E and θ

_{2}(B) = E for θ

_{1}, θ

_{2}in G. Then E is “equiprobable” with each of A, B and thus ‘twice as probable’ as itself! This problem arises most notoriously in the Banach-Tarski paradox. Happily, it does not arise in the de Finetti lottery, an easy consequence of the simple test for paradoxicality that I state next. This test, developed by Tarski and others, is sufficient for all of the examples considered in this paper. My (abbreviated) presentation is based on Wagon [9].

_{1}, ..., A

_{n}, B

_{1}, ..., B

_{m}of E and θ

_{1}, ..., θ

_{n}, σ

_{1}, ..., σ

_{m}∈ G such that E = ∪θ

_{i}(A

_{i}) and E = ∪σ

_{j}(B

_{j}). That is: E has two disjoint subsets ∪A

_{i}and ∪B

_{j}, each of which may be taken apart and rearranged via symmetry mappings in G to cover all of E (Wagon [9]). This definition is a generalization of the worrisome case just raised in connection with the de Finetti example. The existence of a G-paradoxical subset implies that G cannot consistently represent a set of equiprobability relationships among subsets of X. An important theorem of Tarski asserts that the existence or nonexistence of such subsets is equivalent to the nonexistence or existence of a finitely additive measure.

**Theorem**

**5.1**

^{n}and G is any sub-group of isometries, i.e., distance-preserving bijections. We have already seen an example of such a group in the de Finetti single-ticket lottery: the set of translations by a fixed integer. This is a group of isometries on the real line ℝ

^{1}: if θ

_{d}(x) = x + d, then the distance between θ

_{d}(x) and θ

_{d}(y) is the same as the distance between x and y. Two elegant results tell us exactly when a group G of isometries on ℝ

^{n}generates a G-paradoxical subset.16

**Theorem**

**5.2**

- (1)
- For any E ⊆ ℝ
^{n}, E ≠ >, there is a finitely additive, G-invariant measure μ on $\mathcal{P}$(ℝ^{n}) such that μ(E) = 1. - (2)
- No nonempty subset E of ℝ
^{n}is G-paradoxical. - (3)
- No nonempty subset E of ℝ
^{n}contains two disjoint subsets A, B such that A = σ(E) and B = τ(E) for σ, τ ∈ G.

**Theorem**

**5.3**

- (1)
- Some nonempty subset E of X is such that A = σ(E) and B = τ(E) are disjoint subsets of E.
- (2)
- There is some x ∈ X such that whenever w
_{1}and w_{2}are ‘words’ in σ, τ beginning with σ and τ, respectively, then w_{1}(x) ≠ w_{2}(x).

^{−1}, τ, τ

^{−1}} with no trivial pairings (occurrences of σσ

^{−1}, σ

^{−1}σ, τ τ

^{−1}or ττ

^{−1}). Two examples: τστσ

^{−1}, στ

^{−1}σ. A word thus corresponds to a finite sequence of successive applications of the mappings σ, σ

^{−1}, τ, τ

^{−1}(with no trivial pairings along the way).

^{n}, Theorems 5.2 and 5.3 combine for a simple test for the existence of G-paradoxical subsets of ℝ

^{n}(equivalently: a test of whether G defines a coherent set of equiprobability relationships).

**Test for**

**G-paradoxical subsets**.

^{n}such that w

_{1}(x) ≠ w

_{2}(x) where w

_{1}and w

_{2}are finitely many successive (non-trivial) applications of σ, σ

^{-1}, τ, τ

^{-1}beginning with σ for w

_{1}, and τ for w

_{2}.

_{d}: θ

_{d}(x) = x + d, d ∈ Z}. If we let X = ℝ

^{1}, we can apply the test. Since G is commutative, it is immediately clear from the test that there is no G-paradoxical subset. Tarski’s Theorem then tells us that there is a finitely additive measure on X with μ(ℕ) = 1.17

#### 5.2. Application to Examples

^{2}.

- Translations: θ(x, y) = (x + a, y + b), where a, b ∈ ℝ.
- Reflection over the line x=y: ρ(x, y) = (y, x).

^{2}, and we may apply the test for G-paradoxical subsets of ℝ

^{2}. The application is straightforward: the test shows that there are no G-paradoxical subsets.19

^{2}with μ(E) = 1. Clearly, μ(S) = μ(T) = ½, and the restriction of μ to E is the probability measure that we seek. So we have a coherent representation of all symmetries in the example–though with a finitely additive measure, instead of the two-dimensional (countably additive) Lebesgue measure.

^{2}and the same group G of isometries, but this time we restrict our attention to subsets of F = ℕ × ℕ. Our result tells us that, without paradox, we can maintain that any two subsets of F related by translation are equiprobable, and also that S = {(p, q)/p, q ∈ ℕ and p < q} and T = {(p, q)/p, q ∈ ℕ and p > q} are equiprobable. Tarski’s Theorem now ensures that there is a finitely additive G-invariant measure η on ℝ

^{2}with η(F) = 1, η(S) = η(T) = ½. The restriction of η to F is the probability measure that we seek. Again, we have a coherent representation of all of the symmetries in the problem – though once again, not with the usual product measure.

Double dart throw | Double lottery | |

1) Pr*(x ∈ E) = 0 for countable E | 1) Pr*(x ∈ E) = 0 for finite E | |

2) Pr*(p < q) = ½ | 2) Pr*(p < q) = ½ | |

3) Pr*(second < first / first = q) = 0.20 | 3) Pr*(second < first/ first = q) = 0.21 |

## 6. Sequential Throws and Non-Conglomerability

- (CNG)
- Let < = {h
_{n}: n = 1, ...} be an exhaustive partition of a countable space X. Then a probability distribution P is conglomerable for events in an algebra 𝒜 if the following holds for all A ∈ 𝒜: - If c
_{1}≤ P(A/h_{n}) ≤ c_{2}(n = 1, 2, ...), then c_{1}≤ P(A) ≤ c_{2}.

- (1)
- The ordinary dynamic Dutch Book argument for conditionalization fails in this setting; and
- (2)
- There is a dynamic Dutch Book argument against an update policy that yields a foregone conclusion.

_{1}, Pr

_{t1}(second < first) = Pr*(second < first) = ½. But you (and the bookie) know that at t

_{2}, you will find out the value of first and update via conditionalization to Pr

_{t2}(second < first) = Pr*(second < first/first = q) = 0, regardless of the value q. So, at t

_{1}, the following bets are acceptable (the second is better than fair):

- Pay $1 for a bet that pays [$2 if second < first, $0 otherwise];
- After inspecting the first ticket at t
_{2}: pay $1.99 for new bet [$2 if first < second, $0 otherwise].

_{1}.

_{1}, second < first and first < second are equiprobable. But you (and the bookie) know that at t

_{2}, you will learn the value of first and update to Pr

_{t2}(second < first) = Pr(second < first/first = q) = 0, regardless of the value q. So, at t

_{1}, the following bets are acceptable (the second is better than fair):

- Win $1 if second < first; lose $1 if first < second.
- After inspecting the first ticket at t
_{2}: pay $1.99 for new bet [$2 if first < second, $0 otherwise].

## 7. Conclusion

- Outright success: the symmetries are consistent (non-paradoxical) and can be incorporated into an adequate probability model (representing all features of the problem).
- Intermediate case: the symmetries are consistent, but cannot be represented by an adequate probability model.
- Outright paradox: the symmetries themselves lead to inconsistency.

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1 | For the remainder of this paper, all citations of Brown refer to [2]. |

2 | Strictly speaking, it is the one-dimensional Lebesgue measure on ℝ (rather than on [0, 1]) that is characterized by translation invariance; see Rudin [8]. We are considering the interval [0, 1] with a restricted Lebesgue measure, and we understand Brown’s assumption as arising from a more general intuition of translation invariance on ℝ. Note: the uniqueness of the Lebesgue measure depends upon countable additivity. Banach showed that there are finitely additive translation invariant measures on ℝ other than the Lebesgue measure (Wagon [9]). |

3 | That is: if W is a set well-ordered by <, and S _{q} = {x ∈ W / x < q}, then there is no order-preserving bijection between W and S_{q}. See Halmos [10]. |

4 | This follows from CH together with the well-ordering (and hence the Axiom of Choice). Since the cardinality of the full well-ordered set [0, 1] is ℵ _{1} (the least uncountable cardinal number), each of its initial segments is countable. |

5 | Actually, with m(f(q) ∩ [0, 1]), and this should be understood in what follows. |

6 | In fact, both sets are non-measurable, as I shall explain shortly. |

7 | Essentially this argument, first made by Sierpinski, appears in Hauser [7]. Freiling is clearly aware of (and even cites) Sierpinski’s result. I shall return to this point below. |

8 | |

9 | |

10 | For the most part, I set this approach aside due to complexities. |

11 | If S = {(p, q) / p, q ∈ ℕ and p < q}, then S _{q} is finite for any choice of q. |

12 | In fact, we’d have something stronger: an argument against adopting a finitely additive probability measure over any infinite set. This point emerges from the discussion in section 6. |

13 | A parallel argument can be run if we use a non-standard measure: we can foresee that each time we draw a pair of tickets, a miraculous (infinitesimal probability) event occurs. |

14 | This follows from the definition of the product measure. Any finite union of rectangles inside S will have measure 0, while any finite union of rectangles covering S will have measure 1. |

15 | |

16 | A more complete statement is that G has no free subsemigroup of rank 2, but this version is not needed for present purposes. |

17 | The other way of representing the symmetries, via finite permutations, leads to the same result by an application of Theorems 5.1 and 5.3. |

18 | Note: the identity mapping counts as a translation. |

19 | To illustrate the proof: suppose σ(x, y) = θ _{a,b}(x, y) = (x+a, y+b) and τ(x, y) = ρθ_{c,d}(x, y) = (y + d, x + c). Then ττσ(x, y) = σττ(x, y) = (x+a+c+d, y+b+c+d). So it is not the case that w_{1}(x, y) ≠ w_{2}(x, y) for any two words starting in σ and τ. Other cases can be handled similarly (or even more easily). |

20 | This comes from translation invariance of μ. For any countable F ⊆ ℝ and uncountable H ⊆ ℝ, Pr*(F/H) = 0. |

21 | This comes from translation invariance of η. For any finite F ⊆ ℕ and infinite H ⊆ ℕ, Pr*(F/H) = 0. |

22 | For more discussion of this type of bet and its relationship to equiprobability, see Bartha [17]. |

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Bartha, P.
Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis. *Symmetry* **2011**, *3*, 636-652.
https://doi.org/10.3390/sym3030636

**AMA Style**

Bartha P.
Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis. *Symmetry*. 2011; 3(3):636-652.
https://doi.org/10.3390/sym3030636

**Chicago/Turabian Style**

Bartha, Paul.
2011. "Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis" *Symmetry* 3, no. 3: 636-652.
https://doi.org/10.3390/sym3030636