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](http://en.wikipedia.org/wiki/Freiling%27s_axiom_of_symmetry)).


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.



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


1. $A$ 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 $b\notin X\_a$.
2. $A$ 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**


