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 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 $0<r_i<\epsilon$, 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 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.