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Bounded intervals on the real line have the property that every connected open set is an open ball. I wonder what other metric spaces have this property. Explicitly, let's say that $(X,d)$ has Property B if for every connected open set $U\subset X$ there exist $a\in X$ and $r\in \mathbb{R}$ such that $U=\{x\in X: d(x,a)<r\}$.

Question: is every connected metric space with Property B homeomorphic to an interval in $\mathbb{R}$?

Remarks:

  1. The converse is false: being homeomorphic to an interval of $\mathbb{R}$ does not imply Property B. For one thing, $\mathbb{R}$ itself fails the property, because $(0,\infty)$ is not a ball. For another, bounded, example, let $X = \{(x,10|x|) : |x|\le 1\} \subset \mathbb{R}^2$ with the induced Euclidean metric. Its subset $U= \{(x,10|x|) : 0 < x < 1 \}$ is not an open ball, since any ball in $X$ containing $U$ must also intersect the left half of $X$.
  2. The second example from item 1 also shows that Property B is not invariant under homeomorphisms.
  3. One might hope for a more general statement where $X$ is not assumed connected and the conclusion is that it's homeomorphic to a subset of $\mathbb{R}$. But that would be false, a counterexample being an uncountable set with the discrete metric.
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  • $\begingroup$ $(0,1)\cup (2,3)$ is a counterexmple. $\endgroup$
    – C. Ding
    Commented Jun 13, 2017 at 4:03
  • $\begingroup$ @C.Ding X is assumed to be connected. $\endgroup$ Commented Jun 13, 2017 at 9:57

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The circle (with the standard angular metric) is another example. Other than that, your question is essentially a duplicate of this MSE question.

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