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If $X$ is a locally compact, locally connected, connected metrizable space, does that imply that there must be a metric $d$ on $X$ such that $B(x, r) = \{y\in X : d(x, y) < r\}$ is connected for all $x\in X, r > 0$?

Note that if $X$ admits such metric then it's necessarily locally connected and connected.

If $X$ has at least two points, then we can always find a metric on $X$ with a disconnected open ball by choosing non-empty disjoint open $U, V\subseteq X$, $x\in U\cup V$ and a metric $d$ on $X$ such that $U\cup V$ is open unit ball around $x$. See here.

Edit: Added the assumption of local compactness.

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    $\begingroup$ The answer is positive if you assume also that $X$ is locally compact. $\endgroup$ Aug 25, 2023 at 15:59
  • $\begingroup$ It is up to you. $\endgroup$ Aug 25, 2023 at 16:05
  • $\begingroup$ If we can obtain a positive result by just adding local compactness, that's totally fine for me! $\endgroup$
    – Jakobian
    Aug 25, 2023 at 16:12

2 Answers 2

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Here is a direct proof that only assumes $X$ is connected, locally connected, and metrizable. $\newcommand{\diam}{\operatorname{diam}}$

Let $X$ be connected and locally connected metrizable, with topology induced by bounded metric $d$ (note we always have such a bounded metric, as we can replace $d$ with $\min(d,1)$). For $x,y\in X$, let $d'(x,y)=\inf_\Gamma \diam(\Gamma)$, where the infimum is taken over connected subsets $\Gamma\subseteq X$ containing $x$ and $y$, and the diameter is taken with respect to $d$.

Then certainly $d'(x,y)\geq d(x,y)$. Moreover, $d'$ is a metric - finiteness comes from connectivity of $X$ and boundedness of $d$, and the triangle inequality follows from the fact that if the connected set $\Gamma_1$ joins $x$ to $y$ and connected $\Gamma_2$ joins $y$ to $z$, then $\Gamma_1\cup \Gamma_2$ joins $x$ to $z$, and has diameter at most $\diam(\Gamma_1)+\diam(\Gamma_2)$.

Finally, if $d(x_n,x)\to 0$, then by local connectivity $d'(x_n,x)\to 0$, so $d$ and $d'$ are equivalent metrics.

But if $y\in B(x,R)$, where the (open) ball is taken with respect to $d'$, then there is a connected set $\Gamma$ joining $x$ to $y$ with $\diam(\Gamma)< R$, and then every $z\in \Gamma$ is also joined to $x$ by a connected set with diameter less than $R$ (namely $\Gamma$), so $d'(z,x)<R$ as well. Hence $\Gamma\subseteq B(x,R)$, and since this applies to all $y\in B(x,R)$, the ball is connected.

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    $\begingroup$ I really like this argument, you can easily generalize it to other settings. $\endgroup$
    – Jakobian
    Feb 10 at 15:11
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Definition. A metric $d$ on a space $X$ is called geodesic if every two points $p, q\in X$ are connected by a geodesic, i.e. a path whose length equals the distance between $p$ and $q$. Equivalently, there exists an isometric map $c:[0, d(p,q)]\to (X,d)$ such that $c(0)=p, c(d(p,q))=q$.

Clearly, geodesic metric spaces have path-connected metric balls (both open and closed).

Theorem. Let $X$ be a separable metrizable connected locally connected locally compact topological space. Then $X$ admits a complete geodesic metric.

See

A. Tominaga and T. Tanaka, Convexification of locally connected generalized continua, J. Sci. Hiroshima Univ. Ser A 19 (1955), 301-306.

Their proof builds upon the earlier work by Moise and Bing.

It is not impossible that one can drop the assumption of local compactness and separability and still get existence of a metric where open metric balls are connected.

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