**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**