# Locally compact cut-point space must be $\mathbb R$?

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

https://mathoverflow.net/a/319872/95718

and

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

## 1 Answer

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

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