TITLE: Freiling's Axiom of Symmetry Concretized
QUESTION [8 upvotes]: Freiling's Axiom of Symmetry says that for any function $f:[0,1]\to \mathcal{P}([0,1])$ such that for every $x\in [0,1]$ we have $|f(x)|=\aleph_0$, then there exist $y,z\in [0,1]$ such that $z\notin f(y)$ and $y\notin f(z)$.  This is equivalent to the negation of the continuum hypothesis (see this wiki page).
My question is whether one can produce an uncountable set $A$, using the axioms in ZFC, so that if we plug $A$ in place of $[0,1]$, it satisfies Freiling's Axiom.

REPLY [4 votes]: For any function $f:X\to P(X)$, a set $F\subseteq X$ is called free (or $f$-free) if $y\notin f(z)$ and $z\notin f(y)$ for all distinct $y,z$ in $F$. 
Hajnal's "free set theorem" says the following: If $f:\lambda \to [\lambda]^{<\mu}$ for some $\mu<\lambda$, then we cannot only find a 2-element free set but even a free set of size $\lambda$.  (Chapter 26 in the wonderful book by Komjáth and Totik is called "Set mappings". Hajnal's theorem is problem 8. For regular $\lambda$ this was proved earlier by Sierpiński.)
If you plug in $\lambda\ge\aleph_2$, $\mu=\aleph_1$, then you get: If all values of $f$ are at most countable, then there is free set of size $\lambda$.