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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 $n> 1$.

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    $\begingroup$ It is also equivalent to ask the analogue of the question in $\Bbb R^n$, if you’ve figured out what the antipodes look like after stereographic projection $\endgroup$
    – FShrike
    Aug 7 at 14:42
  • $\begingroup$ I don't have time for a full answer, but you could consider the stereographic projection $\mathbb{S}^n\setminus\{(0,\ldots,0,1)\}\rightarrow\mathbb{R}^n$, then look at where the intersection $H_\pm^{n+1}=\{x\in\mathbb{R}^{n+1}|\operatorname{sign}(x_0)=\pm 1\}$ with $\mathbb{S}^n$ lands, and then consider the primages of their intersection with $\mathbb{Q}^n$ and $\mathbb{R}^n\setminus\mathbb{Q}^n$ respectivly. $\endgroup$ Aug 7 at 14:58
  • $\begingroup$ @SamuelAdrianAntz I almost understand your comment, but it's not quite clear. I am a little suspicious of your last step because note that $D$ quite naturally bijects with its complement, so both are uncountable. I understand the stereographic projection to be a bijection, so the preimage of $\mathbb{Q}^n$ would be countable? Or have I misunderstood? $\endgroup$ Aug 7 at 15:19
  • $\begingroup$ @FShrike That's certainly a possible approach. But even if I could figure out where they antipodes go, I am not certain that this simplifies matters much (or enough for me). After all, $\mathbb{R}$ is still much simpler than $\mathbb{R}^n$, $n>1$. $\endgroup$ Aug 7 at 15:28

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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 $-y\notin A$.

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    $\begingroup$ So one can use effectively the same idea as for $n=1$. Fair enough. I convinced myself it would be much more difficult and didn't try it. Also, your profile description is fairly entertaining. $\endgroup$ Aug 7 at 21:26
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    $\begingroup$ As a minor correction, $A$ and $B$ are images, not preimages. $\endgroup$ Aug 7 at 21:33
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    $\begingroup$ Thanks a lot! :-) The mistake with $A$ and $B$ is also corrected now. $\endgroup$ Aug 7 at 21:40
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    $\begingroup$ No worries, and thanks for the answer. I'm sorry, but I think I have just observed another typo. I think the projection maps surject onto $D_1(0)$ in $\mathbb{R}^{n-1}$, the interior of the $(n-1)$-sphere. $\endgroup$ Aug 7 at 21:45
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    $\begingroup$ Oh, yes! Honestly, I'm really not that good in writing down my thoughts in the correct notation. $\endgroup$ Aug 7 at 21:47

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