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Suppose $X$ is metric, compact, connected, and $p\in X$.

An arc is a copy of $[0,1]$.

Is it possible that every two points in $X\setminus \{p\}$ can be joined by an arc, but there is no arc in $X$ containing $p$?

Forever Mozart
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  • You may want to exclude the trivial arc at $p$: that is, $f:[0,1]\to X$ with $f(t)=p$ for all $t$. – Aweygan Jul 24 '16 at 18:18
  • That's not what I mean. Regardless of how many large the space is, the function defined above is an arc containing $p$. – Aweygan Jul 24 '16 at 18:22
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    "An arc is a copy of $[0,1]$" is somewhat imprecise. Nowhere is requested that the map $f:[0,1]\rightarrow X$ is injective. – Andrea Mori Jul 24 '16 at 18:25
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    An arc in $X$ is an injective continuous map of $[0,1]$ into $X.$ – zhw. Jul 24 '16 at 18:28
  • Aside from singletons, are there compact connected metric spaces which are path-connected but not arc-connected? Is it possible your problem is the same as asking "Is there a compact connected metric space $X$ and a point $p\in X$ so that $X-p$ is path-connected but $X$ isn't?" – William Jul 24 '16 at 19:19
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    @you Yes that is equivalent - path-connected and arc-connected are equivalent in Hausdorff spaces. See Exercise 6.3.12 in Engelking. – Forever Mozart Jul 24 '16 at 19:37
  • Ok cool. I found a space where they're different notions, but it's not Hausdorff (hence not metric). I also thought of another non-compact example: take the graph of $sin(1/x)$ for $x\in (0,1]$, along with the origin 0 (instead of a whole vertical bar). Then you can show this is not path-connected in the same way as you show the topologist's sine curve is not path-connected, and if you remove 0 it becomes path-connected, but of course it's not compact. – William Jul 24 '16 at 19:56
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    Why couldn't the arcs $A_i$ all contain points well away from $p$? – zhw. Jul 25 '16 at 07:22
  • The proposed proof strategy will not work. The remainder of a metrizable compactification of $[0, \infty)$ is connected, but need not be path-connected. On the contrary, it seems that every metrizable continuum is the remainder of a metrizable compactification of $[0, 1)$. – Niels J. Diepeveen Aug 10 '16 at 12:27
  • @NielsDiepeveen yes, that is true! (and a fairly recent result I think) – Forever Mozart Aug 10 '16 at 17:33

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The answer is "yes". See the example in my answer of Path connectedness of one point compactification. There exists a closed arcwise connected subset $X \subset \mathbb{R}^3$ such that the one-point compactifaction $X^+$ is not pathwise connected.

Paul Frost
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