It is well known and easy to see (modulo standard basic facts) that any compact 1-dimensional Alexandrov space with curvature bounded from below is isometric either to a circle or to a segment.
I am looking for a precise reference.
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$\begingroup$ There are several different notions called "Aleksandrov space". Which one do you have in mind? $\endgroup$– Alexandre EremenkoCommented May 24, 2022 at 12:59
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$\begingroup$ @AlexandreEremenko: The one introduced in Burago-Gromov-Perelman paper chrome-extension://efaidnbmnnnibpcajpcglclefindmkaj/mathnet.ru/links/a40e90f6161e3bd935c7a2da6397259b/rm4489.pdf $\endgroup$– asvCommented May 24, 2022 at 13:02
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2 Answers
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See 15.18 in our Alexandrov geometry: foundations.
answered Jul 1, 2022 at 13:31
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I don't think anyone bothered to write a formal proof down but this is indeed quite easy. Tangent spaces (which are always metric cones) can only be lines or half lines. From this it's immediate that the set of regular point (with tangent space $\mathbb R$ ) is open and connected. By the BGP paper you mention it follows that that set of regular points is bilipschitz to a 1-manifold, so isometric to an open interval and the whole space it the metric completion of that interval.
answered May 24, 2022 at 15:28