Continuum Hypothesis III

Suppose you pick randomly a real number. What is the probability that it equals to 1? The probability is zero. Suppose [tex]X\subset [0,1][/tex] is a countable subset of the unit interval. What is the probability that a randomly picked real from the unit interval is in X? Again, it is zero, because the Lebesgue measure of X is zero, X being countable. Suppose now, that we are having a function

[tex]f\colon [0,1]\to \{\text{Countable subsets of }[0,1]\}.[/tex]

This means that f attaches a countable set of reals in the unit interval to each real in that interval. Let us consider the following statement

Feiling’s axiom. For every f (as above), there exist [tex]x\in [0,1][/tex] and [tex]y\in[0,1][/tex] such that [tex]x \notin f(y)[/tex] and [tex]y \notin f(x)[/tex].

Let us see what does it mean if Freiling’s axiom does not hold. If it does not hold, then there is a function f attaching countable sets of reals to reals with the property that given two random reals x and y, either [tex]y\in f(x)[/tex] or [tex]x\in f(y)[/tex]. This is supposed to be counterintuitive according to Freiling. Indeed, suppose we choose x first. Then because f(x) is countable, the probabilty for y to be in f(x) is zero. On the other hand, if y is chosen first, then the probability for x to be in f(y) is zero! But one of them happens certainly, provided Freiling’s axiom fails.

Freiling proposed this as an argument against Continuum Hypothesis, since

Theorem. The following are equivalent
1. Continuum Hypothesis
2. Freiling’s axiom fails.

In the abstract to his paper Axioms of Symmetry: Throwing Darts at the Real Number Line (see below for exact reference), Freiling writes boldly without hiding his believes:

We will give a simple philosophical “proof” of the negation of Cantor’s continuum hypothesis. […] We will assume the axioms of ZFC together with intuitively clear axioms […that] are justified by the symmetry in a thought experiment throwing darts at the real number line. We will in fact show why there must be an infinity of cardinalities between the integers and the reals. We will also show why Martin’s Axiom must be false, […] Following the philosophy — if you reject CH you are only two steps away from rejecting the axiom of choice.

The proof of Freiling’s Theorem is quite simple. At least in the direction CH => not-F.A.

Proof. Suppose CH holds. Let us construct a counter example function for Freiling’s axiom. Because by CH there is a bijection between [tex]\mathbb{R}[/tex] and [tex]\omega_1[/tex] (the first uncountable ordinal), it suffices to construct a function

[tex]f\colon \omega_1 \to \{\text{Countable subsets of }\omega_1\}.[/tex]

For every ordinal [tex]\alpha<\omega_1[/tex] let [tex]f(\alpha)=\{\beta<\omega_1\mid \beta\leqslant \alpha\}.[/tex] Now if we pick any two ordinals [tex]\alpha_1[/tex] and [tex]\alpha_2[/tex], then either [tex]\alpha_1\leqslant\alpha_2[/tex] or [tex]\alpha_2\leqslant \alpha_1[/tex] (or both). In each case either [tex]\alpha_1\in f(\alpha_2)[/tex] or [tex]\alpha_2\in f(\alpha_1)[/tex].

Suppose CH fails. Pick a subset [tex]A\subset [0,1][/tex] of size [tex]\aleph_1[/tex], which is smaller than the continuum because CH does not hold. Let [tex]C=\bigcup_{x\in A}f(x)[/tex], which is a [tex]\aleph_1[/tex]-union of countable sets, so is of size [tex]\aleph_1[/tex] as well. So there is [tex]y_0\in [0,1]\setminus C[/tex]. Now for all [tex]x\in A[/tex] we have [tex]y_0\notin f(x)[/tex]. The set [tex]f(y_0)[/tex] is countable, so there is also an element [tex]x_0\in A\setminus f(y_0)[/tex]. Clearly these satisfy Freiling’s axiom. [tex]\square[/tex]

Reference: Axioms of Symmetry: Throwing Darts at the Real Number Line, Chris Freiling, The Journal of Symbolic Logic, Vol. 51, No. 1 (Mar., 1986), pp. 190-200, Published by: Association for Symbolic Logic