Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be the $S^1$ or a connected subset thereof, endowed with the standard metric. Then every open set $U\subseteq X$ is a disjoint union of open arcs, hence a disjoint union of open balls. Are there any other metric spaces with this property? That is: Can you give an example of a connected metric space such that every open set is the union of disjoint open balls and such that it is not homeomorphic to a subspace of $S^1$?


Edit: a few remarks, though most of them leave more questions than they give answers:

If $a,b\in X$, $a\ne b$, then the closed sets $\partial B(a,r)$, $0<r<d(a,b)$ and $\{x\in X\mid d(a,x)=\lambda d(b,x)\}$ for $\lambda>0$ are non-empty because $X$ is connected. Can anything be said about $r\to0$ or $\lambda\to \infty$? For example, does it follow that $X$ is path-connected?

I wrote "homeomorphic to a subspace of $S^1$" mostly as an abbreviation for "homeomorphic to point, open/half-open/closed bounded interval, or $S^1$-like". The same toppolocical space may however have the disjoint-ball property with one metric and not have it with another (bounded) metric. For example, in $S^1\subset\mathbb R^2$ with $d\bigl((x,y), (u,v)\bigr) =\sqrt{4(x-u)^2+(y-v)^2}$ the connected open set given by $y>0$ is not a ball: The only candidate is $B\left((0,1),\sqrt5\right)$, but it contains $(0,-1)$.

Does it follow at all that $d$ must be bounded? $X\setminus\{x\}=\bigcup_{i\in I} B(x_i,r_i)$ for some index set $I$, $x_i\in X$, $r_i>0$. For $\epsilon>0$, the ball $B(x,\epsilon)$ must intersect every $B(x_i,r_i)$ because $X$ is connected, hence $d(x,x_i)=r_i$ for all $i\in I$ and $d(x,y)\le2\sup\{r_i\mid i\in I\}$. Thus if $X\setminus\{x\}$ has only finitely many components for some $x$, then $d$ is bounded. But could $X\setminus\{x\}$ have infinitely many components for all $x$? Resolved after a comment from celtschk: $X$ itself is a ball $B(x_0,r)$, hence $d(x,y)\le d(x,x_0)+d(x_0,y)<2r$ for $x,y\in X$.

Assume a point $x_0$ looks like a finite branch, that is there is $n\in\mathbb N$, $n\ge3$ and a continuous map $h:\{1,\ldots,n\}\times[0,\epsilon)\to B(x_0,\epsilon)$ such that $d(x_0,h(i,t))=t$ and $h|_{\{1,\ldots,n\}\times(0,\epsilon)}\to B(x_0,\epsilon)\setminus\{x_0\}$ is a homeomorphism. Given $\mathbf r=(r_1, \ldots, r_n)$ with $0<r_i<\epsilon$, the set $U_{\mathbf r}=h\left(\bigcup_{i=1}^n\{i\}\times[0,r_i) \right)$ is open and connected, hence a single open ball $B(\hat x,r)$. Let $x_i=h(i,r_i)$. By continuity of $h$, we conclude $d(\hat x,x_i)=r$ and the $r_i$ can be recovered from $\hat x$ and $r$ as $r_i=\inf\{t\mid d(\hat x, h(i,t))\ge r\}=\sup\{t\mid d(\hat x,h(i,t))<r\}$. One checks that this $\phi_i\colon(\hat x, r)\mapsto r_i$ is continuous where defined. This gives a contiunuous and surjective map $(i,s,t)\mapsto (\phi_i(h(i,s),t))_i$ from a subset of the twodimensional $\{1,\ldots,n\}\times (0,\epsilon)^2$ to the $n$-dimensional $(0,\epsilon)^n$. Unless this is some space-filling monster (can it be?), we conclude that $X$ cannot have branch-points. (This is ugly - can it be made prettier? Or branch-point be defined friendlier?)


Edit: Meanwhile I am confident that everey connected length space with the disjoint open ball property is one of the known spaces (i.e. homeomorphic to a connected subspace of $S^1$). So, how far is a connected metric space from being a length space? Could the ideas be transferred or do they give hints for counterexamples?

In what follows, let $(X,d)$ be a connected length space with the disjoint open ball property. We can define the branch degree (or is there a standard name for this?) $\beta(x)$ for $x\in X$ as the (possibly infinite) number of connected components of $X\setminus\{x\}$.

Lemma 1: For $x\in X$, we have $\beta(x)\le2$.

Proof: Assume $\beta(x)\ge 3$, i.e. $X\setminus \{x\}$ has connected components $U_i$, $i\in I$ and wlog. $\{1,2,3\}\subseteq I$. For $i\in\{1,2,3\}$ select a point $x_i\in U_i$ and let $\rho=\min\{d(x,x_1),d(x,x_2),d(x,x_3)\}$. Then for $r<\rho$ and $i\in\{1,2,3\}$ we have that $U_i\cap B(x,r)$ is connected because an (approximately) geodesic path to $x$ cannot leave the conected component and stays within the ball. Also, we can find a point $\in U_i$ at distance $r$ from $x$. The set $$U:=(U_1 \cap B(x,\rho))\cup(U_2 \cap B(x,\frac12\rho))\cup B(x,\frac13\rho)\cup\bigcup_{i\in I\setminus\{1,2,3\}}U_i$$ is open and connected, hence $U=B(y,R)$ for some $y\in X$, $R>0$. As paths between points in different $U_i$ must pass through $x$, we conclude that $R=d(x,y)+\rho$ if $y\notin U_1$, $R=d(x,y)+\frac12\rho$ if $y\notin U_2$ and $R=d(x,y)+\frac13\rho$ if $y\notin U_3$. Since $y$ is in at most one of $U_1, U_2, U_3$, we arrive at a contradiction.$_\blacksquare$

I think the following should be possible to prove:

$Lemma 2:$ If $a,b,c$ are three distinct points $\in X$, then $X\setminus\{a,b,c\}$ is not connected.

Proof: ???

Since I'm not yet sure about a proof of lemma 2, the rest is left in handwaving stage:

Assume there exists $x\in X$ with $\beta(x)=0$. Then $X$ is just a point and we are done.

Assume there exists $x\in X$ with $\beta(x)=2$. Write $X\setminus\{x\}=U_1\cup U_2$ and definie $f:X\to\mathbb R$ by $$f(y)= \begin{cases}d(x,y)&y\in U_1\\-d(x,y)& y\in U_2\end{cases}$$ I claim that $f$ is injective and in fact it should be possible to show this with Lemma 2 or some similar result.

Then we are left with the case that $\beta(x)=1$ for all $x$. Then for any such point either $X\setminus\{x\}$ should have a point $y$ with $\beta(y)=2$ hence $X\setminus\{x\}$ "is" an interval and we conclude that $x$ is one of its endpoints (or possibly "is both" endpoitns, making an $S^1$). Or otherwise at least $X\setminus\{x,y\}$ must have a point $z$ with $\beta(z)=2$ (by lemma 2) and hence $X\setminus\{x,y\}$ "is" an interval and $x,y$ are its endpoints.

share|improve this question
6  
@GiuseppeNegro: Really? Every open set in the plane is the union of open balls, but not, as far as I can see, of disjoint open balls. For example, take the open unit square. As soon as you select one open ball to be in the union, none of the points on its boundary can ever be covered by an open ball disjoint from the first one. But most of the boundary will have to be inside the open unit square. –  Henning Makholm Sep 13 '12 at 22:27
6  
The key observation is that a disjoint union of several open balls is not connected, hence such spaces "tend to be" totally disconnected (like $p$-adics). –  Hagen von Eitzen Sep 13 '12 at 22:34
1  
@HenningMakholm: Of course you are right. I have checked the property I was referring to and I found out that I stated it incorrectly. The correct property says: "every nonempty open set in $\mathbb{R}^k$ is a countable union of disjoint boxes" (Rudin R&CA, §2.19(d)). Here a "box" is a set like this: $$\{x\in \mathbb{R}^k\ :\ \alpha_i \le x_i<\alpha_i+\delta\}, $$ Of course this is not a ball. –  Giuseppe Negro Sep 14 '12 at 1:52
1  
This is not directly related but when thinking about possible examples I was reminded of a nice characterization of metric trees given by Stefan Wenger as Corollary 1.2 here: metric trees are precisely the geodesic metric spaces with the property that the intersection of closed balls is either empty or again a closed ball. –  t.b. Sep 14 '12 at 8:52
2  
The distance must be bounded from above, because the complete metric space is an open set, thus a disjoint union of open balls. If it were the disjoint union of more than one open balls, it would not be connected, thus it is a single open ball. But every open ball consists of points which are less than some radius $r$ from some given point $P$. Now take some arbitrary points $A,B$. Then because they are in a single open ball, by the triangle inequality we get $d(A,B)\le d(A,P)+d(P,B)<2r$. –  celtschk Sep 15 '12 at 14:33
show 12 more comments

2 Answers 2

I had some new thoughts on this that are too long for a comment. I have a proof sketch if the space is path-connected, but it would only work if you can show that the space can't contain a Y-shaped graph.

I will use the fact that any injective map of the interval into a metric space is an embedding.

Assuming the space is path-connected, choose $x,y$ in the space, and let $\alpha$ be an arc connecting them. Let $\beta$ be a path (not neccessarily an arc) from $x$ to $y$ that only intersects each endpoint once and that does not lie entirely in the image of $\alpha$ (if such a path exists). I claim that $\beta$ and $\alpha$ intersect only in their end points $x$ and $y$. Otherwise, there exists a point $p$ in the images of both $\alpha$ and $\beta$ such that there is an $\epsilon>0$ with $\beta((\beta^{-1}(p),\beta^{-1}(p)+\epsilon)$ is disjoint from $\alpha$. You choose $p$ to be the first point after $x$ where $\beta$ leaves $\alpha$.

Let $\gamma=\beta([\beta^{-1}(p),\beta^{-1}(p)+\frac{1}{2}\epsilon]$. Then $\gamma\cup\alpha$ is an embedded space with a branch point, which is not possible as discussed above.

Thus, any two paths from $x$ to $y$, each intersecting the endpoints only once, must be disjoint except in their endpoints. We can choose $\beta$ to be an arc by shrinking it; it will still be disjoint from $A$. There cannot be three such arcs, since we would get another branch point. Thus, if there are two points where two such arcs exist, the arcs must be surjective and we have a circle.

If there is only one path between each pair of points, then the space is the union of closed intervals pasted along closed intervals, and is an interval.

Edit: This is just a proof sketch; some of the details may need hammering out.

share|improve this answer
add comment

$\mathbb{R}$ with trivial topology has the open ball property but its not homeomorphic to any subset of $S^1$

So I guess you want a manifold with open ball property.

Claim: any such manifold with open ball property will have topological dimension 1.

share|improve this answer
    
In order to speak of open balls, you need not only a topology but a metric space. So you'd need the trivial metric rather than the trivial topology. In any case, that doesn't satisfy the question's criterion of being connected. So what's left in this answer is essentially just a weaker version of the question. –  Henning Makholm Apr 12 '13 at 14:47
    
the trivial topology on $\mathbb{R}$ is given by the metric $d(x,y)=0 \forall x,y \in \mathbb{R}$ And it is connected since the only open and also closed sets are the whole space and the empty set. –  baharampuri Apr 12 '13 at 18:12
4  
Usually the definition of a metric space doesn't allow $d(x,y)=0$ when $x\ne y$. –  Henning Makholm Apr 15 '13 at 11:32
2  
Indeed, a space that has such pairs and is otherwise metric-like is called a pseudo-metric (or pseudometric) space. –  dfeuer Apr 16 '13 at 20:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.