# Freiling axiom of symmetry

The Freiling theorem states:

Let $S = \{f: [0,1]\rightarrow A$ : A is a countable subset of $[0,1]\}$ . Consider the following statement:

For each $f$ in $S$, there exist $x$ and $y$ in $[0,1]$ such that $x$ is not in $f(y)$ and $y$ is not in $f(x)$.

I would like to know if is it possible to give a proof of the falsity of the CH (Continuum Hypotesis) using the Freiling theorem of Symmetry?

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It is possible, because it turns out that $AX(\aleph_1)\iff\lnot CH$. I might have time later today to write an answer. The proof itself appears Wikipedia. –  Asaf Karagila Feb 27 '12 at 8:55
The proof in Wikipedia doesn't take into account the concept of probability as formulated by Kolmogorov. Freiling gives an interpretation of his symmetry axiom based on probability of a dart to hit a number chosen among the class of the real numbers, but the probability (as mentioned on Wikipedia) is a concept referred to a measure, not to a class of numbers. So, ultimatley my question concerns the possibility to accept as true the Freiling axiom and consequently the validity or not of the CH. –  Riccardo.Alestra Feb 27 '12 at 11:36
I'm still not sure what the problem is. –  Asaf Karagila Feb 27 '12 at 13:03
The 'probability of Freiling' is not a probability defined in a Lebesgue measurable space. Even if we assume the Freiling probability to be correct from a formal point of view, is the Symmetry axiom alone enough to give a proof of the falsity of the CH? –  Riccardo.Alestra Feb 27 '12 at 13:47

Yes, it is possible to give a proof of $\neg\mathsf{CH}$ using Freiling's axiom (not theorem) of symmetry. My answer is CW because I'm essentially copying from Wikipedia the answer that Asaf already mentioned. I get the impression that this doesn't answer some question that the OP meant to ask, but it does answer the question that he did ask, so I thought it should be posted as such.
Suppose $\mathsf{CH}$ holds, so that $|[0,1]|=\omega_1$. Then letting $\sigma: [0,1] \to \omega_1$ be a bijection, the function $f:[0,1]\to\mathcal{P}([0,1])$ defined by $f(x)= \{y: \sigma(y) \preceq \sigma(x)\}$ demonstrates the failure of Freiling's axiom.