Freiling's axiom of symmetry and CH - need some help

Question

Reading this there is a proof with

(1) Suppose $2^\kappa = \kappa^+$. Then there exists a bijection $\sigma : \kappa^+ \to \mathcal{P}(\kappa)$. Setting $f : \mathcal{P}(\kappa) \to \mathcal{P}(\mathcal{P}(\kappa))$ defined via $\sigma(\alpha) \mapsto \{\sigma(\beta) : \beta \preceq \alpha\}$, it is easy to see that this demonstrates the failure of Freiling's axiom.

I don't see it easily. I don't see how such a function demonstrates the failure of the axiom - I guess I don't understand the relation in $\sigma$ and the domain and codomain of f are different than [0,1] and countable subsets of the reals.

I also found this and I think I lack the knowledge about countable ordinals and would like some suggestions on resources for that.

(2) $\text{Proof:} (\Rightarrow) $ Suppose the Continuum Hypothesis is true. Then $\text{card}((0,1)) = \aleph_1$. Well-order the elements of $(0,1)$ as $r_1, r_2, r_3, \ldots$ and let $f(r_i) = \{r_j : j \leq i\}$. Then $f$ is in $\mathcal{A}$, and for $j \leq i$, $r_j$ is in $f(r_i)$, while for $j > i$, $r_i$ is in $f(r_j)$. Furthermore, $f(r_i)$ is countable for all $i$ (having cardinality less than $\aleph_1$) $\Rightarrow$ Freiling’s Axiom of Symmetry is false.

I guess it's almost the same thing as (1) although I'd like to know what's the difference and how they translate the domain and codomain of f in the first proof into the functions in Freiling's axiom.

Any insights, explanations, or references would be greatly appreciated.

Answer 1

The continuum hypothesis implies the failure of Freiling's axiom of symmetry and indeed it is equivalent to the failure of this axiom.

To see this, assume first that CH holds. What this means is that there is a well ordering $\preceq$ of the real numbers in order type $\omega_1$, the first uncountable ordinal. One important thing to say about such an order type is that every element has only countably many things below it. That is the essential nature of $\omega_1$, namely, it is an uncountable well order, but the predecessors of any point in it forms a countable set. Thus, the function $f(x)=\{ y\mid y\preceq x\}$ maps every real $x$ to a countable set of reals, as desired for Freiling's axiom. But for any two reals $x,y$, either $x\preceq y$ or $y\preceq x$, and so either $x\in f(y)$ or $y\in f(x)$, contrary to Freiling's axiom.

You didn't ask about it, but one can also prove the converse. Namely, if CH fails, then Freiling's axiom holds. To see this, suppose that CH fails. Fix any function $f$ from $\mathbb{R}$ to countable sets of reals. Fix any set $S$ of $\omega_1$ many distinct reals, and apply $f$ in all possible ways so as to close under the function. The set $S^+$ that you get still has size $\omega_1$, and since CH fails, there must be a real $x\notin S^+$. But $f(x)$ is a countable set, whereas $S^+$ is uncountable, so there is some $y\in S^+$ with $y\notin f(x)$. But $f(y)\subseteq S^+$ by design, and so $x\notin f(y)$. So we have found the independent pair, as desired to witness this instance of Freiling's axiom.

It is exactly the analogous argument at higher cardinals $\kappa$. Namely, $\text{AX}_\kappa$ is equivalent to the failure of GCH at $\kappa$, with $\kappa^+<2^\kappa$. For the one direction, if the GCH holds at $\kappa$, there is an enumeration of $P(\kappa)$ in order type $\kappa^+$, and all proper initial segments have size $\kappa$. So the function mapping any point to its predecessors is a counterexample to $\text{AX}_\kappa$. Conversely, if GCH fails at $\kappa$, fix $\kappa^+$ many subsets of $\kappa$, and then close under a given function. If $x$ is not amongst them, then $f(x)$ rules out only $\kappa$ many points, so there is some $y$ in $S^+$ not in $f(x)$, and since $f(y)\subseteq S^+$ it follows that also $x\notin f(y)$.

I've realized that these arguments are different than the arguments that appear on Wikipedia, and I think I like my arguments better.