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In ZFC, the Freiling axiom of symmetry is equivalent to failure of the Continuum hypothesis. Under the hypothesis, countable initial segments of a well-order of the reals violate symmetry. When the continuum exceeds ${\aleph }_1$, the union of the countable values on an ${\aleph }_1$-sized set leaves a real outside; avoiding that real's countable value gives the two witnesses.

Let $f$ be given and choose $Y\subseteq \mathbb{R}$ of cardinality ${\aleph }_1$. Each $f(y)$ is countable, so the axiom of choice and infinite cardinal arithmetic giveSince the continuum is larger, choose $x$ outside this union. Then $x\notin f(y)$ for every $y\in Y$. The countable set $f(x)$ cannot contain $Y$, so choose $y\in Y\setminus f(x)$. ThusThe two points are distinct because $x\notin Y$. This proves the countable-valued free-pair criterion and the required direction of the Freiling axiom of symmetry without any measurability assumption on $f$.

The Freiling axiom of symmetry isThe values of $f$ are countable subsets of the real numbers, including finite ones. One may require $x\ne y$ without changing the axiom: apply the displayed version to $x\mapsto f(x)\cup \{x\}$.

The Generalized continuum hypothesis statesHere ${\kappa }^{+}$ is the successor cardinal; equivalently $2^{{\aleph }_{\alpha }}={\aleph }_{\alpha +1}$ for every ordinal $\alpha$.

The Freiling theorem givesFirst assume the Continuum hypothesis and fix a well-order $(r_{\alpha }{)}_{\alpha <{\omega }_1}$ of the real numbers. Define $f(r_{\alpha })=\{r_{\beta }:\beta \le \alpha \}$. Each value is countable. For any pair, one index is at most the other, so at least one of $x\in f(y)$ or $y\in f(x)$ holds. The same holds for $x=y$, since $x\in f(x)$. This violates the Freiling axiom of symmetry.

Conversely, assume $|\mathbb{R}|>{\aleph }_1$ and let $f$ assign a countable set to each real. Choose $X\subseteq \mathbb{R}$ of cardinality ${\aleph }_1$. By infinite cardinal arithmetic,Choose $y$ outside this union. Since $f(y)$ is countable and $X$ is uncountable, choose $x\in X\setminus f(y)$. Then $y\notin f(x)$ and $x\notin f(y)$, with $x\ne y$. This proves the Freiling axiom of symmetry and completes both implications.