# Connected metric spaces with disjoint open balls

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.

• 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. Sep 13, 2012 at 22:07
• @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. Sep 13, 2012 at 22:27
• 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). Sep 13, 2012 at 22:34
• @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. Sep 14, 2012 at 1:52
• 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 $d(A,B)\le d(A,P)+d(P,B)<2r$. Sep 15, 2012 at 14:33

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.

I do not know what happens for general metric spaces, here is a proof which covers the case of length metric spaces.

Definition. Call a metric space $X=(X,d)$ a B-space if every open subset of $X$ equals the disjoint union of some open metric balls in $X$.

Note that each length metric space $(X,d)$ satisfies the following two properties:

a. $X$ is locally path-connected. (Since open balls in $X$ are path-connected.)

b. Every closed metric ball $\bar B(a,r)=\{x\in X: d(a,x)\le r\}$ equals the closure of the corresponding open ball $B(a,r)=\{x\in X: d(a,x)< r\}$.

I will use the notation $S(a,r)$ for the metric sphere $S(a,r)=\{x\in X: d(a,x)=r\}$.

Theorem 1. Suppose that $X$ is a $B$-space which consists of more than 1 point and which satisfies (a) and (b). Then $X$, is a topological 1-dimensional manifold (possibly with boundary). In other words, $X$ is homeomorphic to a connected subset of $S^1$.

Proof. It suffices to consider the case when $X$ is connected. Since $X$ is a Hausdorff path-connected space, it is also arc-connected, i.e. any two points in $X$ belong to an arc, i.e. a subset homeomorphic to a closed interval, see

S. Willard, General Topology, Addison-Wesley 1970. Theorem 31.2. (Compare the discussion here.)

Given an arc $\alpha\subset X$ which is the image of a homeomorphism $f: [0,1]\to \alpha$, let $\alpha^\circ$ denote $f((0,1))$, the corresponding open arc.

Since $X$ is locally connected, each $B(a,r)$ contains a unique maximal open connected subset containing $a$, I will denote this subset $B_c(a,r)$.

Lemma 1. Each open arc $\alpha^\circ$ is an open subset of $X$.

Proof. To prove this, consider for each $r>0$ the open $r$-neighborhood $U_r(\alpha)$ in $X$: $$U(\alpha,r)=\bigcup_{a\in \alpha} B(a,r).$$ This neighborhood will contain an open connected sub-neighborhood of $\alpha$: $$U_c(\alpha,r)= \bigcup_{a\in \alpha} B_c(a,r).$$ Since $X$ is a B-space, there are points $a_r\in X$ and radii $R=R(r)$ such that $$U_c(\alpha,r)= B(a_r, R(r)).$$ For each $r$ there exists $x_r\in \alpha$ such that $d(x_r, a_r)<r$. By compactness of $\alpha$, there is a sequence $r_i\to 0+$ such that $$x_{r_i}\to a\in \alpha$$ and $R(r_i)\to R(0)$. Since $\alpha$ is not a singleton, $R(0) >0$.

Clearly, $a_{r_i}\to a$, $$\bigcap_{r_i} U_c(\alpha,r)= \alpha$$ and $$\alpha= \bar{B}(a, R(0)).$$

We conclude from this that $\alpha\cap B(a,R(0))$ is an open subset of $X$. From this, we see that the subset of $X$ where $X$ is a 1-dimensional manifold is dense (and clearly open) in $X$. So far, we did not use the Property (b).

Claim. I claim that $\alpha^\circ \cap S(c, R(0))=\emptyset$.

Proof. If $\alpha^\circ \cap S(a, R(0))\ne \emptyset$, the arc $\alpha$ contains a subarc $\beta$ connecting points of $S(a, R(0))$ and not containing the center $a$. Let $b\in \beta$ be a point with minimal distance from $a$; by our assumption that $a\notin \beta$, $\rho=d(b,a) > 0$. Since $\alpha\cap B(a, R(0))$ is a 1-dimensional manifold of $B(a, R(0))$, the point $b$ cannot be the limit of any sequence $b_i\in B(a, R(0))$ such that $(d(a,b_i))_{i\in {\mathbb N}}$ converges to $d(a,b)$ from the left. This contradicts the Property (b) (the point $b\in S(a,\rho)$ does not belong to the closure of $B(a, \rho)$.) qed

The claim implies that $\alpha^\circ = \alpha \cap B(a, R(0))$, i.e. $\alpha^\circ$ is open in $X$. This concludes the proof of Lemma 1. qed

Lemma 1 implies that $X$ is a 1-dimensional manifold near each point contained in an open arc. Suppose, therefore, that $x\in X$ is a point which does not belong to any open arc. Since $X$ is not a singleton, $x$ is the end-point of some nondegenerate arc $\alpha\subset X$ connecting points $x$ and $y$ in $X$.

Lemma 2. I claim that $\alpha-\{y\}$ is a neighborhood of $x$.

Proof. If not, there is a sequence $x_n\to x$, $x_n\notin \alpha$. Since $X$ is locally arc-connected, there exist arcs $\beta_n$ connecting $x_n$ to $x$ and disjoint from $y$ for all sufficiently large $n\ge n_0$. Lemma 1 implies that $\beta_n\cap \alpha= \{x\}$ for all $n\ge n_0$. Therefore, the concatenation $\gamma_n$ of $\beta_n$ and $\alpha$ is an arc in $X$ (for $n\ge n_0$) such that $x\in \gamma_n^\circ$. A contradiction. qed

Combining the Lemmata 1 and 2 we see that $X$ is a 1-dimensional manifold (possibly with boundary). qed