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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.

Stefan Hamcke
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Pedro
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  • A similar question: http://math.stackexchange.com/questions/391595/path-connectedness-and-compactifications – Seirios Jul 18 '13 at 15:02

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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:

  • 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 $p[I]$.
Stefan Hamcke
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  • How to prove that $S\cup({0}\times[-1,1])$ is not path-connected? I've tried to do this by supposing that there is a path from $(0,0)$ to $(2/\pi,1)$, but I'm not able to finish. Could you give any suggestion? – Pedro Jul 18 '13 at 00:06
  • Assume that $p:I\to\overline S$ is a path from $p(0)=(0,z)$ to $p(1)=(x,\sin(1/x))$. Since ${0}\times[-1,1]$ is closed, its preimage under $p$ is a closed set in $I$, so it has a maximum $m$. Assume that $p(m)=(0,y)$. Then $p[(m,1]]\subseteq S$. Now take a small neighborhood about $(0,y)$, say $U=[0,\epsilon)\times(y-\epsilon,y+\epsilon)$ where $\epsilon<1-y$ or $\epsilon<y+1$. Show that no neighborhood of $m$ is mapped entirely into this $U$, contradicting the continuity of $p$. – Stefan Hamcke Jul 18 '13 at 00:23