and constitute a Boolean subalgebra of $2^\Omega$; furthermore, when restricted to ${\mathrm{BA}}(B)$, $\mu$ is a Boolean homomorphism. The atoms of ${\mathrm{BA}}(B)$ are the eigenspaces themselves. The $\mu$-image of ${\mathrm{BA}}(B)$ is a finite Boolean algebra of events in $(\Omega,{\ensuremath{\mathrm{CO}}})$. Its atoms are the $\mu$-images of the four eigenspaces. So each $\omega\in\Omega$ is in exactly one of those four $\mu$-images.

Fix some point $\omega\in\Omega$. It selects, for each of the 9 bases $B$, exactly one of the 4 vectors in $B$. Furthermore, when the same vector occurs in two bases, then it is selected from one if and only if it is selected from the other. This is because for $\vec v$ to be selected means that (writing $\langle \vec v\rangle$ for the subspace generated by $\vec v$) $\omega\in \mu\langle \vec v\rangle$, and this doesn’t depend on the rest of a basis.

But the point of the example of Cabello et al. is precisely that such a selection is impossible. Indeed, we would have 9 selections, one from each basis, but every selection would occur twice because, as indicated by the colors in [@W-KS], every vector occurs in exactly two bases. So 9 would have to be even.

The ${\mathbb C}^4$ in 