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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

https://mathoverflow.net/a/76139/95718.

In general, $X$ need not be homeomorphic to $\mathbb R$. Consider the plane set $$\{\langle x,\sin(1/x)\rangle:x\in (-\infty,0)\}\cup ([0,\infty) \times \{0\}).$$ This example fails to be locally connected (of course). Also it is not locally compact.

Question: If $X$ is locally compact, then is $X$ necessarily homeomorphic to the real line?

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    $\begingroup$ (I voted for reopening the first linked question, currently closed as duplicate.) $\endgroup$ – YCor Sep 6 at 19:07
  • $\begingroup$ I now have a proof. Surely this result is not new, and there should be a reference for it... $\endgroup$ – D.S. Lipham Sep 7 at 18:50
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Franklin and Krishnarao showed every connected separable regular locally connected space in which every point splits the space into exactly two connected components is homeomorphic to the real line in this paper.

In a one-page addendum, they showed locally connected can be replaced with either locally compact or rim-compact.

Franklin, S. P.; Krishnarao, G. V., On the topological characterization of the real line: an addendum, J. Lond. Math. Soc., II. Ser. 3, 392 (1971). ZBL0209.54302.

EDIT: There is an error in the proof in the addendum paper, at this point:

Indeed, Kok shows [1; Theorem 1] in a connected Hausdorff space, each point being a strong cut point is equivalent to (S") given three distinct points, some one separates the other two.

I explain the error and correct the proof here: https://www.researchgate.net/publication/335831043_Compactification_of_cut-point_spaces.

Kok, H., On conditions equivalent to the orderability of a connected space, Nieuw Arch. Wiskd., III. Ser. 18, 250-270 (1970). ZBL0202.21803.

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