Suppose $X$ is a connected separable metric space, and $X\setminus \{x\}$ has exactly two connected components for every $x\in X$.

If $X$ is locally connected, then $X\simeq \mathbb R$. This was noted in previous answers to




My questions is, is the conclusion the same ($X$ homeomorphic to the reals $\mathbb R$) if $X$ is locally compact?


$X=\{(x, \sin(\frac{1}{x}): x >0\} \cup \{0\} \times \Bbb R$ looks like a potential counterexample.

edit Removing the non-cutpoints so

$X=\{(x, \sin(\frac{1}{x}): x >0\} \cup \left(\{0\} \times (\Bbb R \setminus (-1,1))\right)$ seems to work better (thanks to David Hartley's comment).

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    $\begingroup$ No, points in $\{0\} \times (-1,1)$ aren't cut points. $\endgroup$ – David Hartley Sep 6 '19 at 23:33
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    $\begingroup$ @DavidHartley. Agreed............. $\endgroup$ – DanielWainfleet Sep 7 '19 at 1:05
  • $\begingroup$ @DavidHartley And if we remove that segment from $X$? Or do we lose local compactness at $(0,1)$ and $(0,-1)$? $\endgroup$ – Henno Brandsma Sep 7 '19 at 5:28
  • $\begingroup$ Your comment is correct, you lose local compactness when you remove that segment. $\endgroup$ – David Hartley Sep 7 '19 at 20:06
  • $\begingroup$ @DavidHartley Can you explain why? $\endgroup$ – Henno Brandsma Sep 7 '19 at 20:07

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