Proving Sierpinski’s result without choice.

Solution 1:

The following theorem is due to Galvin: (see e.g., here )

Let $F:[\mathbb{R}]^2\to \{0,1\}$ be a coloring such that both $F^{-1}"\{0\}, F^{-1}"\{1\}$ have the Baire property. Then there is a perfect subset that is homogeneous with respect to $F$.

So if a model satisfies $\aleph_1\leq 2^{\aleph_0}$ + every set of reals has the Baire property + Galvin's theorem (which I believe uses ZF+DC), then that model will also have $2^{\aleph_0}\to(\aleph_1, \aleph_1)^2$.

In Shelah's model ($\mathcal{M}18$ in the Howard&Rubin book), every set has the BP and DC holds. Initially I wasn't sure whether $\aleph_1\leq 2^{\aleph_0}$ holds in that model. But as @ElliotGlazer kindly pointed out, since Shelah's model doesn't require an inaccessible to build, we can assume that there are no inaccessibles in $L$. This will imply $\aleph_1\leq 2^{\aleph_0}$ (because DC and $\aleph_1\not\leq 2^{\aleph_0}$ would imply that $\aleph_1$ is inaccessible in $L$).

So Sierpinski's coloring result does not hold in Shelah's model.