Is there any arc-connected set $X\subset\mathbb{R}^n$ such that $\overline{X}$ is not arc-connected? Could someone give me an example of an arc-connected set $X\subset\mathbb{R}^n$ such that $\overline{X}$ is not arc-connected?
Thanks.
 A: The topologist's sine curve $S:=\left\{\left(x,\sin\left(\frac1x\right)\right)\mid x>0\right\}$ is path-connected and thus arc-connected since every path-connected Hausdorff space is arc-connected (although in this case it is trivial to show arc-connectedness directly). Its closure $\overline S=S\cup(\{0\}\times[-1,1])$ is not path-connected, but still connected, as is any set between a connected set and its closure.
Remark: The closure $\overline S$ is also not locally connected. The proof looks somewhat similar to the proof of path-disconnectedness (in the comments). Indeed, path-connected and local connectedness are related in a certain way:


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*Once you know that $\overline S$ is not path-connected, you can conclude that $S\cup A$, where $\emptyset\neq A\subset\{0\}\times[-1,1]$, is not path-connected and thus not locally path-connected. This is because a connected, locally path-connected space must be path-connected. This follows from the fact that path-components in locally path-connected spaces are open and at the same time closed.

*There is sort of a converse: Assume that you know already that $S\cup A$, where$\emptyset\neq A\subseteq\{0\}\times[-1,1]$, is not locally connected at the points of $A$. This can be used to show that $S\cup A$ is not path-connected. For if $p:I\to S\cup A$ connects a point $s\in S$ to a point $t\in A$, then $p$ is a closed map. Now quotient maps preserve local (path-)connectedness, so the path $p[I]$ had to be locally-connected. But $p[I]$ covers everything of $S\cup A$ which is on the left of $s\times\Bbb R$, and $t$ does not have arbitrarily small connected neighborhoods, contradicting the local-connectedness of $p[I]$.

