1
$\begingroup$

Let $n \geq 1$ be an integer. Let us call $S\subseteq \mathbb{R}^{n+1}$ an $n$-sphere if there is $x\in \mathbb{R}^{n+1}$ and $r\in \mathbb{R}$ with $r>0$ such that $$S = \{z\in \mathbb{R}^{n+1}: \|z-x\| = r\}, $$ where $\|\cdot\|$ denotes the Euclidean norm in $\mathbb{R}^{n+1}$.

Suppose that ${\cal S}\neq \emptyset$ is a set of $n$-spheres in $\mathbb{R}^{n+1}$.

Is there $M\subseteq \mathbb{R}^{n+1}$ such that $M$ intersects every member of ${\cal S}$, but for each $m\in M$ there is $S\in {\cal S}$ such that $S \cap (M\setminus \{m\}) = \emptyset$?

Note. The answer is positive if the set ${\cal S}$ is finite.

Improve this question
$\endgroup$
5
  • $\begingroup$ should it be $n$-spheres in $\mathbb{R}^{n+1}$? the use of the word radii seems to suggest this. $\endgroup$
    – Nick L
    Commented Sep 5, 2017 at 13:06
  • $\begingroup$ You are right - thanks! Will edit accordingly $\endgroup$
    – Dominic van der Zypen
    Commented Sep 5, 2017 at 13:09
  • $\begingroup$ Could you maybe give a complete definition (e.g. a sphere with radius $r$ and center $c$ is the set of all points ...). I think I have a counter-example, but it depends of course on what a "sphere" is for you. $\endgroup$
    – Dirk
    Commented Sep 5, 2017 at 13:10
  • $\begingroup$ @Dirk, thanks, I will do this in the next edit of the question, should be online in about 5 minutes $\endgroup$
    – Dominic van der Zypen
    Commented Sep 5, 2017 at 13:11
  • $\begingroup$ @DirkLiebhold I hope the formulation of the question is more solid now $\endgroup$
    – Dominic van der Zypen
    Commented Sep 5, 2017 at 13:18

1 Answer 1

Reset to default
6
$\begingroup$

Let $(S_i)_{i \in I}$ be a family of spheres of cardinality $|I| < \mathfrak{c}$ and let $S$ be a sphere distinct from each $S_i$. Each intersection $S \cap S_i$ is a (possibly degenerate) circle. Since there is a continuum of circles inside $S$, one can find a circle $C \subseteq S$ distinct from each $S \cap S_i$. Each intersection $C \cap S \cap S_i$ consists of at most two points. Since $|I| < \mathfrak{c} = |C|$, the family $(C \cap S \cap S_i)_{i \in I}$ does not cover $C$. In particular $S \setminus \bigcup_{i \in I} S_i$ is non-empty.

Using this and applying the method used by Martin Sleziak to answer your previous question Meeting a set of lines in $\mathbb{R}^n$ (or as in Will Brian's answer above), one gets a positive answer in ZFC.

Remark: it is annoying that you do not link to your previous very similar questions (here, here, here, and also here).

Improve this answer
$\endgroup$
1
  • $\begingroup$ I just realize this a moment ago! (Actually, I was going to do something a little more complicated, but yes, the point is that you don't need to rely on the smallness of a union of null or meager sets.) I will delete my answer now since it doesn't seem to contribute anything beyond what you say in yours -- the transfinite induction part seems to be covered in the links you give. $\endgroup$
    – Will Brian
    Commented Sep 5, 2017 at 14:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .