# Open balls in path connected metric spaces with more than one point

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?

## 1 Answer

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.

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