Dense subset of $S^{n-1}$ with antipodal complement Let $n \in \mathbb{N}$, and let $S^{n} \subseteq \mathbb{R}^{n+1}$ be the unit $n$-sphere.
Does there exist a dense subset $D$ of $S^n$ such that $\textbf{x} \in D \implies -\textbf{x} \notin D$?

This is true for $n=1$. Note that $S^1 \cong A := [0,1]/\mathbb{Z}$, where the equivalent question is: does there exists a dense subset $D'$ of $A$ such that $x \in D \implies x \pm \frac{1}{2} \notin D'$?
Consider the function $f : [0,1) \to \{0,1\}$,
$$f(x)= \begin{cases} 
      0 & x< \frac{1}{2},\text{ any decimal expansion of } x \text{ has finitely many 3s} \\
      1 & x < \frac{1}{2}, \;x \text{ has a decimal expansion with } \infty \text{ many 3s} \\
      1 & x = \frac{1}{2} \\
      1 & x > \frac{1}{2}, \text{ any decimal expansion of } x \text{ has finitely many 3s}\\
      0 & x > \frac{1}{2}, \;x \text{ has a decimal expansion with } \infty \text{ many 3s}
   \end{cases}
$$
Then $D' = f^{-1}(\{0\})$ is a suitable dense set.

However, I am not sure how to approach the cases $n >1$, since $S^n$ is no longer homeomorphic to a very simple object and my topological toolkit is quite unsophisticated. If the question can be answered with elementary methods, I welcome any hints. I am not even sure whether or not I expect such a set $D$ to exist for $n> 1$.
 A: This answer is an expansion of my comment, but I simplified it a bit without using the stereographic projection.
First of all, the obvious way of taking the intersection of $\mathbb{Q}^{n+1}$ with $\mathbb{S}^n\subset\mathbb{R}^{n+1}$ won't work, as it can even be empty as explained in this answer here. Consider the half-spaces $\mathbb{H}_\pm^{n+1}=\{x\in\mathbb{R}^{n+1}|\operatorname{sign}(x_0)=\pm 1\}$ and the (surjective) projections:
$$\operatorname{pr}_\pm\colon
\mathbb{H}_\pm^{n+1}\cap\mathbb{S}^n
\twoheadrightarrow\mathbb{D}^n,
(x_0,x_1,\ldots,x_n)\mapsto(x_1,\ldots,x_n)$$
flattening down both half-spheres.
The subsets $A=\operatorname{pr}_-^{-1}(\mathbb{D}^n\cap\mathbb{Q}^n)$ and $B=\operatorname{pr}_+^{-1}(\mathbb{D}^n\cap(\mathbb{R}^n\setminus\mathbb{Q}^n))$ are both dense (in the closure of their respective half-sphere) as preimages of dense sets under continuous maps. $A\cup B$ is therefore dense in $\mathbb{S}^n$ as the closure operator commutes with finite unions (See here).
For $x\in A$, we have $x_0<0$, therefore $-x\notin A$, as well as $x_1,\ldots,x_n\in\mathbb{Q}$, therefore $-x\notin B$.
For $y\in B$, we have $y_0>0$, therefore $-y\notin B$, as well as $y_1,\ldots,y_n\notin\mathbb{Q}$, therefore $-y\notin A$.
