Freiling's Axiom of Symmetry Concretized 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.
 A: Theorem. The following are equivalent for a set $A$


*

*$A$ has the property of Freiling's axiom. That is, if $a\mapsto
X_a$ is any map from $A$ to the countable subsets of $A$, then
there are $a$ and $b$ with $a\notin X_b$ and $b\notin X_a$.

*$A$ has size at least $\aleph_2$.
Proof. ($1\to 2$) We prove the contrapositive. If $A$ had size
less than $\aleph_2$, then we can enumerate the elements of $A$ as
$A=\{\ a_\alpha\mid
\alpha<\omega_1\ \}$. For any $a\in A$, let $\alpha$ be least with $a=a_\alpha$, and map
$a\mapsto X_a=\{a_\beta\mid\beta<\alpha\}$. For any $a\neq b$, one
of them appears first before the other, and so either $a\in X_b$
or $b\in X_a$, contrary to statement $1$.
($2\to 1$) If $A$ has size at least $\aleph_2$, then suppose we
have any function $a\mapsto X_a$ where $X_a$ is a countable subset
of $A$. By applying the function $\omega_1$ many times, we may
find a subset $Y\subset A$ of size $\omega_1$, which is closed
under the map, in the sense that $a\in Y\to X_a\subset Y$. Now
pick any $b\notin Y$. Since $X_b$ is countable, there is some
$a\in Y$ with $a\notin X_b$, and since $b\notin Y$, it follows
that $b\notin X_a$, and so we have achieved Freiling's property.
QED
A: Take any set $A$ with $|A|>\aleph_1$ (e.g. $A=\omega_2$). Then it satisfies Freiling´s Axiom. For this let $f:A \to \mathcal{P}(A)$ with $|f(x)|=\aleph_0$ for $x\in A$. Fix a subset $B \subseteq A$ with $|B|=\aleph_1$. Note that $C:=\bigcup_{x \in B} f(x)$ has size $\aleph_1$ so there must be an $a \in A \setminus C$. Now since $f(a)$ is countable there must be a $b \in B \setminus f(a)$. But then $a \notin f(b)$ and $b \notin f(a)$.
A: 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 $\lambda$. 
