3
$\begingroup$

Let $X$ be a normed space and $S(x, \varepsilon)=\{y\in X: \|x-y\|=\varepsilon\}$​ denote the sphere centered at $x\in X$​ with radius $\varepsilon>0​.$ Consider two different spheres $S_1=S(x_1, \varepsilon_1)$​ and $S_2=S(x_2, \varepsilon_2)$ with non-empty intersection​. What can we say about the set $S_1\cap S_2$​? Is it true that this set is always homeomorphic to a sphere or a ball of "one dimension less" (or even more dimensions less)?

What we know for sure: A paper by Jussi Vaisala contains a short proof of the following:

Lemma 2.2: The intersection of two spheres in a finite dimensional normed space $X$​ of dimension $n\geq 3$​ is a connected set. Furthermore, if $X$​ is strictly convex, then this intersection is either a singleton or homeomorphic to an $(n-2)​$-sphere.

My guess is that the second claim of the Lemma still holds for $n=2​$, but was omitted from it because of the intersection not being connected. So, this should pretty much settle my question for finite dimensional strictly convex spaces.

Without the assumption of $X$​ being strictly convex we may have some more possibilities. Consider $X=\mathbb{R}^3​$ equipped with the $\|⋅\|_\infty$​ norm. The spheres are in fact cubes in $\mathbb{R}^3​$​ and the intersection of two cubes may additionally be (homeomorphic to) a square, which is the closed unit ball of $(\mathbb{R}^2, \|⋅\|_\infty)​$, or (homeomorphic to) a line segment, which is the closed unit ball of $(\mathbb{R}, \|⋅\|_\infty)$​.

So, what can we say if we remove the strict convexity assumption? Is there a counterexample where the intersection is neither a sphere or a ball? I couldn't find any. Additionally what can be said for infinite dimensional normed spaces?

I couldn't find any literature regarding this topic and probably for a good reason since Vaisala mentions that while $S_1\cap S_2$ being connected was known by Novikoff in 1955, $S_1\cap S_2$​ being homeomorphic to the $(n-2)​$-sphere seemed to be a new result [sic] at the time he published it (2010).

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Is there a counterexample in dimension 2 of the first claim of the lemma ? $\endgroup$ – Jean Marie Nov 19 '16 at 19:49
  • $\begingroup$ Yes, just take the usual norm and two circles that are not tangent. Then their intersection is a two-point set (which has two connected components). I think it should be easy to show that no matter what norm you take in $\mathbb{R}^2$, you will always find spheres which intersect in exactly two points, thus every norm in $\mathbb{R}^2$ should give a counterexample. $\endgroup$ – tree detective Nov 19 '16 at 20:02
  • 1
    $\begingroup$ Maybe you should also consider annuli, i.e. sets of the form $A(x,\varepsilon_1,\varepsilon_2)=\{y\in X\mid\varepsilon_1\leq\|x-y\|\leq\varepsilon_2\}$? Note that in $(\mathbb R^3,\|\cdot\|_\infty)$, the intersection of $S((0,0,2),3)$ and $S((0,0,-2),3)$ is homeomorphic to such an annulus in $(\mathbb R^2,\|\cdot\|_\infty)$, namely to $A((0,0),1,2)$. $\endgroup$ – Dejan Govc Nov 19 '16 at 20:04
  • 1
    $\begingroup$ Yes, you are right! I completely overlooked it, it could be neither a ball nor a sphere after all. So we should rephrase the question: are sphere, ball or annulus the only possibilities? $\endgroup$ – tree detective Nov 19 '16 at 20:27
  • 1
    $\begingroup$ @treedetective: Here's an example that's neither a ball nor a sphere nor an annulus. If we intersect $S((1,0,0),1)$ and $S((0,1,0),1)$ in $(\mathbb R^3,\|\cdot\|_\infty)$, we obtain two squares connected by a pair of line segments, namely $([0,1]\times [0,1]\times\{-1,1\})\cup(\{(0,0),(1,1)\}\times[-1,1])$. $\endgroup$ – Dejan Govc Nov 29 '16 at 0:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.