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Is it true that in every path connected metric space with more than one point, every open ball contains a subset with more than one point which is connected (or possibly more, path connected)? Suppose $B$ is an open ball with center $x$ and let $x \ne y \in B$ (such $y$ exists by Does there exist a connected metric space, with more than one point and without any isolated point, in which at least one open ball is countable?) and consider $\hat B:=\{y \in B: x,y$ are connected by a path lying in $B\}$ ; then does $\hat B$ has more than one point? Is $B$ connected?

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The answer is positive: in every path connected metric space with more than one point, every open ball contains a path-connected subset with more than one point.

Proof. If $B$ is an open ball and $p\in B$, by assumption there is a nonconstant continuous map $\gamma:[0,1]\to X$ such that $\gamma(0)=p$. If $\gamma([0,1])\subset B$, it is the desired set. Otherwise let $b =\inf\{t: \gamma(t)\notin B\}$. The continuity of $\gamma$ implies $b>0$. The set $\gamma([0,b))$ is what you want.


By the way: for general connected spaces, the answer is negative. For example, the Knaster–Kuratowski fan is a connected metric space $X$ that contains a point $p$ such that $X\setminus \{p\}$ is totally disconnected (has no connected subsets with more than one point). In particular, an open ball that is disjoint from $p$ does not contain any connected subsets with more than one point.

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  • $\begingroup$ But does the set $\gamma ([0,b))$ has more than one point ? $\endgroup$
    – user228168
    Commented Sep 27, 2015 at 16:06
  • $\begingroup$ Yes. Otherwise $\gamma$ would be constant on $[0,b)$, and then $\gamma(b)=\gamma(0)\in B$, a contradiction to the choice of $b$. $\endgroup$
    – user147263
    Commented Sep 27, 2015 at 16:07

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