Given an example of a metric space in which every sphere has two centers

This is a question from Wilansky "Topology for analysis", P.15 Prob. 103

Maybe I was thinking too Euclidean, I can't come up some other "centers" of the sphere :(

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What do you mean by the center of a sphere? – Qiaochu Yuan Jul 27 '14 at 5:56
@QiaochuYuan: Perhaps a point that's equidistant from all points in a set? Then I think $S^n$ would work. – Kyle Jul 27 '14 at 5:57
@squirrel: But would that be the ordinary center only? how about the other one? – Baffin Chu Jul 27 '14 at 5:59
@QiaochuYuan: The book is mentioned in this way: "THe following definitions apply to any semi-metric space X. For $a\in X$ and $r\in\mathbb{R}$, the cell of radius $r$ and center $a$, written $N(a,r)$, is $\{x:d(x,a)<r\}$" – Baffin Chu Jul 27 '14 at 6:01
@Sudokux: I'm not sure I understand what you mean. A (metric copy of) $S^1$ in $S^2$ separates $S^2$ into two open disks, each of which has a center. – Kyle Jul 27 '14 at 6:04

I will answer under the assumption that you refer to sets $S_{\varepsilon}(x) = \{y: \, d(x,y) = \varepsilon\}$ as spheres, and $x$ as a center of the sphere.

On $\mathbb{Q}$, given any prime $p$, you can define the $p$-adic metric $|x-y|_p$ to be $p^{-n}$, where $n$ is the unique integer such that $x-y = p^n \frac{a}{b}$ with $a,b$ integers that are coprime to $p$. (If $x=y$, then define $|x-y|_p = 0$.)

This turns out to satisfy the ultrametric inequality $|x-y|_p \le \max\{|x|_p|, |y|_p\}$, and using this, you can show that, for any $x \in \mathbb{Q}$ and $\varepsilon > 0$, $S_{\varepsilon}(x) = S_{\varepsilon}(y)$ for any $y$ such that $|x-y|_p < \varepsilon$; in particular, there are many centers of any sphere.

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+1 for using this book. It's one of my favorites.

To your question. Think about $S^1$ in $\Bbb R^2$. Then every 'sphere' is simply 2 points picked from $S^1$. This 'sphere' has 2 centers. The point that is circumferentially central to both points and the point diametrically opposed to that first point.

Think about given two points on $S^1$ and either of those 'centers' drawing a secant circle centered there and cutting through the two original points.

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How can I justify your 2nd sentence " Then every 'sphere' is simply 2 points picked from $S^1$."? – Baffin Chu Jul 27 '14 at 6:49
Two circles intersect in either one or two points, or they coincide. If they intersect in one point, they are tangent and the center of one of them lies off the boundary of the first. If they coincide, they share the same center which is not part of them. – Bryan Jul 27 '14 at 6:55

As I mentioned in my comment, the spheres $S^n$ provide a family of examples. A "sphere" in $S^{n}$ is a copy of $S^{n-1}$. If we normalize the "great circle distance" metric on $S^n$ to have $d(x,-x)=1$, then a sphere around $x$ of radius $r$ is also a sphere of radius $1-r$ around the antipodal point $-x$.

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yes, i see the point, but i think the statement "a sphere around x of radius r" is a local argument, and in the daily sense of sphere, it seems that it has to be extended to a global argument. Is it necessary here? – Baffin Chu Jul 27 '14 at 6:54
@Sudokux: The $n$-dimensional sphere $S^n$ is a (global) metric space. After I fix a particular metric, there is an unambiguous subset of points defined by $\{ y \in S^n : d(x,y) = r\}$ which we are calling a "sphere inside of $S^n$ of radius $r$". There's no concern of a local vs. global distinction. – Kyle Jul 27 '14 at 7:14
@Sudokux: To be more specific, I'm referring to the "great circle distance" on the sphere. This induces the normal metric topology on $S^n$, and has the added feature that $d(x,z)=d(x,y)+d(y,z)$ if $x,y,$ and $z$ all lie on one half of a great circle. Since every point $y \in \{y \in S^n: d(x,y)=r\}$ lies on half of a great circle between $x$ and $-x$, we have $d(x,-x)=d(x,y)+d(y,-x)$, i.e. $d(-x,y)=d(x,-x)-d(x,y)$. So if $d(x,-x)=1$ and $d(x,y)=r$, then $d(-x,y)=1-r$. – Kyle Jul 27 '14 at 7:21