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?

Resolvedafter a comment fromceltschk: 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 connectedlength spacewith 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 thebranch 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._\blacksquareI 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.

**Answer**

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 necessarily 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.

**Attribution***Source : Link , Question Author : Hagen von Eitzen , Answer Author : Martin Sleziak*