Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 19 / 1 / ii

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 $f$.

Assume the Continuum hypothesis and enumerate $\mathbb{R}=\{r_{\alpha }:\alpha <{\omega }_1\}$. DefineEvery value is countable. For any two indices, say $\alpha \le \beta$, one has $r_{\alpha }\in f(r_{\beta })$, so the two required nonmembership conditions cannot both hold. Including the endpoint in each initial segment also rules out taking the two points equal. Therefore the free-pair assertion implies the negation of Continuum hypothesis. Cantor theorem and choice already give $2^{{\aleph }_0}\ge {\aleph }_1$, so

The Gödel constructible universe theorem gives $\mathrm{Con}(\mathrm{ZFC})\Rightarrow \mathrm{Con}(\mathrm{ZFC}+\mathrm{CH})$. The forcing independence theorem for the Continuum hypothesis gives $\mathrm{Con}(\mathrm{ZFC})\Rightarrow \mathrm{Con}(\mathrm{ZFC}+\neg \mathrm{CH})$. For example, the latter can be obtained by first passing to the constructible universe and then adding sufficiently many Cohen reals. By the equivalence in part (b), these are respectively models of the negation and affirmation of the free-pair assertion. Thus if ZFC is consistent, the assertion is independent of ZFC. The consistency qualification is essential: an inconsistent theory proves every sentence. These are syntactic relative-consistency implications, not a claim that bare consistency supplies a countable transitive model.