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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 $\partial B(a,r)$, $00$ 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 toppological 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


Edit: Meanwhile I am confident that every 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 connected 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" endpoints, 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.

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    $\mathbb{R}^2$.2012-09-13
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    The Euclidean plane has this property. More generally, the Euclidean space of whatever dimension has this property. It is something that is used in constructing Lebesgue measure, for example.2012-09-13
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    @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.2012-09-13
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    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).2012-09-13
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    @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.2012-09-14
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    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](http://arxiv.org/abs/math/0603539): 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.2012-09-14
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    @t.b.: Still sounds interesting. Can we get anywhere near such spaces? Given $x\in X$ can we find a space $Y$, $\epsilon>0$ and continuous $h:Y\times[0,\epsilon)\to X$ such that $d(x,h(y,t))=t$ and $h|_{Y\times (0,\epsilon)}\to B(x,\epsilon)\setminus\{x\}$ homeomorphic? From this I think one should be able to decude $|Y|\le 2$: For every $r:Y\to(0,\epsilon)$, the image of $\bigcup \{y\}\times [0,r(y))$ is connected and open, hence an open ball around one of its points. It looks like there are "more" choices for functions $r$ than for balls ... Just an idea.2012-09-15
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    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 $d(A,B)\le d(A,P)+d(P,B)<2r$.2012-09-15
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    Are you aware of any other examples where the union of any two non-disjoint balls is a ball?2012-09-15
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    @celtschk: don't you mean the *entire* metric space, rather than *complete*? :) (I hope there is no such thing as "entire metric space"; either way, I'm pretty sure it's not quite as common as "complete metric space").2012-09-15
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    I don't think space-filling monsters can be considered branches. ;) It only makes sense if the branches are locally non-intersecting (themselves or one another) near $x_0$.2012-09-15
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    @tomasz: I mean, of course, the set of all points of the metric space. I think this formulation should be completely unambiguous. :-)2012-09-15
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    @celtschk: to be honest, I was a little confused at first, and was wondering where does completeness come into play. But, admittedly, that was before I took care to actually read anything properly.2012-09-15
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    One idea for a metric space which *might* have that property: Take the [Vicsek fractal](https://en.wikipedia.org/wiki/Vicsek_fractal) (which if I understand it correctly is path connected), and as metric take the length of the shortest path connecting the two points inside the fractal (where the path length is calculated using the Euclidean metric of the 2D space it is embedded in). I didn't, however, do any calculation; it's only an intuition which might be wrong.2012-09-15
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    @celtschk: If I'm not mistaken, no: One of the connected components of the open set $U$ given by $|y|<\epsilon$ contains the end points $(-1,0)$ and $(1,0)$, hence the open ball coverng both ends must cover some points of type $(0,y)$ with $|y|>\epsilon$, hence outside $U$.2012-09-15
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    @HagenvonEitzen: Yes, you're right.2012-09-15
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    Has this been posted on mathoverflow? Seems like a good spot for it.2013-01-01
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    I know, it is an old question, but do you want to see a proof of your conjecture for geodesic metric spaces? (With some messy modifications it will also work for general length spaces.) As for general path-connected metric spaces, they are _very_ far from length spaces; they tend to be length-metrizable, but this changes the metric quite a bit.2017-06-17
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    @MoisheCohen Yes please!2017-06-18

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