If $\cup (A_i \cup -A_i)=\mathbb S^d$, then is there an $x$ in $d$ sets?

Question

If for some collection of open sets $\cup_{i\in I} (A_i \cup^* -A_i)=\mathbb S^d$, then is there an $x\in \mathbb S^d$ and $i_1,\ldots i_d\in I$ for which $x\in A_{i_1}\cap \ldots \cap A_{i_d}$?

Important that the question implicitly requires $A_i\cap -A_i=\emptyset$, as $\cup^*$ denotes disjoint union. Also note that I want $x$ to be in $A_i$ and not in $-A_i$, otherwise a suitable version of this statement would follow from Ky Fan's theorem (which implies it even in this form for $d\le 2$).

Of course this problem is very similar to the Lyusternik-Shnirelman variant of the Borsuk-Ulam theorem, so I'm sure this has been studied, but I could not find it anywhere, what of course makes it less likely to hold, which I nonetheless believe.

Answer 1

Unfortunately, I think your conjecture is wrong: Embedd $S^3 \subset \mathbb C^2$ in the standard way. Consider the three sectors $$U_1 = \{z \in \mathbb C \ | \ z = re^{i\varphi}, r > 0, \varphi \in (0, 2\pi / 3)\}$$ $$U_2 = \{z \in \mathbb C \ | \ z = re^{i\varphi}, r > 0, \varphi \in (2\pi / 3, 4\pi/3)\}$$ $$U_3 = \{z \in \mathbb C \ | \ z = re^{i\varphi}, r > 0, \varphi \in (4\pi/3,2\pi )\}$$ and define $$A_i = \{(z_1,z_2) \in S^3 \ | \ z_1 \in U_i\}, i = 1,2,3$$ $$A_{3+i} = \{(z_1,z_2) \in S^3 \ | \ z_2 \in U_i\}, i = 1,2,3$$ Let us check that the six sets $A_1, A_2, A_3, A_4, A_5, A_6$ satisfy the conditions: They are obviously open. For each $(z_1, z_2) \in S^3$, one of $z_1,z_2$ is nonzero, say $z_1$. Then either $z_1$ or $-z_1$ lies in one of the sectors $U_1, U_2, U_3$. Hence the six sets cover the sphere. It is also clear that $A_i \cap (-A_i) = \emptyset$ for all $i$.

But now, finnally, the three sets $A_1, A_2, A_3$ and $A_4, A_5, A_6$ are pairwise disjoint, respectively. Hence there is no point which lies in $3$ of the sets $A_i$.