A Question Regarding the Origin of the Axiom of Symmetry

Question

It is my understanding that Chris Freiling's "Axiom of Symmetry" is based on a counterexample to CH given by Sierpinski in his book "Hypothese de continu". Since I neither read nor speak French, I was wondering whether anyone who was familiar with his counterexample would be kind enough to provide me with the English translation. Thanks in advance for your help. It is greatly appreciated. Also, is there a translation of Sierpinski's book available?

Answer 1

In view of the comments, this appears to be his

Proposition $P_1$: $\Bbb R^2$ is the union of two sets, of which one has countable intersection with each line parallel to the $y$-axis, and the other countable intersection with each line parallel to the $x$-axis.

(I’ve modernized the language slightly as well as translating it.) His first result in Chapter $1$ is that this is equivalent to $\mathsf{CH}$.

This is clearly equivalent to the result stated in the Simms paper that you mention in the comments. Its equivalence to the negation of the axiom of symmetry isn’t in Sierpiński, so far as I can tell at a quick look, but the equivalence is very easily established.

Let $S$ be as in that result. For $x\in\Bbb R$ let

$$f(x)=\{y\in\Bbb R:\langle y,x\rangle\in S\text{ or }\langle x,y\rangle\notin S\}\;,$$

a countable subset of $\Bbb R$. For distinct $x,y\in\Bbb R$ we have

$$y\notin f(x)\quad\text{ iff }\quad\langle y,x\rangle\notin S\text{ and }\langle x,y\rangle\in S\tag{1}$$

and

$$x\notin f(y)\quad\text{ iff }\quad\langle x,y\rangle\notin S\text{ and }\langle y,x\rangle\in S\;;\tag{2}$$

clearly $(1)$ and $(2)$ are incompatible, so Freiling’s axiom of symmetry fails.

Now suppose that a function $f$ that assigns to each real number a countable set of reals is a counterexample to Freiling’s axiom of symmetry: there do not exist distinct $x,y\in\Bbb R$ such that $x\notin f(y)$ and $y\notin f(x)$. Without loss of generality $x\in f(x)$ for each $x\in\Bbb R$. Let

$$S=\bigcup_{x\in\Bbb R}\big(\{x\}\times f(x)\big)$$

and

$$T=\bigcup_{x\in\Bbb R}\big(f(x)\times\{x\}\big)\;.$$

If $\langle x,y\rangle\notin S$, then $y\notin f(x)$, so $x\in f(y)$, and $\langle x,y\rangle\in T$. Similarly, if $\langle x,y\rangle\notin T$, then $x\notin f(y)$, so $y\in f(x)$, and $\langle x,y\rangle\in S$. $S$ and $T$ are now two sets as in Proposition $P_1$, so $\mathsf{CH}$ holds by Sierpiński’s result.