Connected metric spaces with disjoint open balls

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