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Apr 15, 2021 at 7:43 comment added erz Again, your argument seems to rest on the choice of the Euclidean metric on $H$. In particular, it is not clear why the length of horizontal segments converge to $0$
Apr 15, 2021 at 6:28 comment added John Samples @erz IMO it's already obvious from the wording of the post, e.g. "if this argument is wrong" and "I think" etc. Also I feel it shouldn't be very far off (assuming it's correct); the argument I gave in the comment about 'midpoints' of the horizontal arcs not converging to $x$ while both sets of endpoints do seems on the verge of a complete proof, I am just waiting for a length space expert to give the final OK and 'precise statement'.
Apr 15, 2021 at 5:21 comment added erz ok, but then please indicate clearly that this is just a suggestion of a counterexample (i agree, it seems plausible, but i am afraid the proof is pretty far at this point)
Apr 15, 2021 at 4:35 comment added John Samples @erz But this configuration is along neighborhood boundaries, so should be topological. It's not supposed to be a complete proof, someone with more experience in length spaces can either finish the details or show that it's not a counterexample, and maybe the same for extending along $\mathbb{Q}$ instead of a point. But I think this space should work. No other relevant space in Steen & Seebach fails to have an intrinsic metric, so if these spaces do then maybe it's true after all. Maybe an analogue where $\mathbb{H}^2$ is replaced by some wild manifold could be interesting as well.
Apr 15, 2021 at 3:33 comment added erz but again this way you only prove that the extended Euclidean metric is not intrinsic. Also, please use @username when responding, because i wasn't aware of your reply (you are aware of mine, because it is a comment under your post)
Apr 12, 2021 at 2:22 comment added John Samples For example, take a straight line emanating from $x$ to the left, but tangent to, $C$. Then points on this line converge to $x$, as do the points on $V$, and they are getting arbitrarily close to each other, as seen just by taking horizontal segments. But by the topological agreement criterion, points along $C$, which are contained in these horizontal segments, will have to have horizontal distance from $V$ bounded away from $0$. This violates the triangle inequality on the horizontal segments.
Apr 12, 2021 at 1:49 comment added John Samples The Euclidean metric on $\mathbb{H}^2$ is already a length metric (it's convex).
Apr 12, 2021 at 1:40 comment added erz I think what you have proved is that the metric induced by the Euclidean metric on $H$ (how do you extend it on $H\cup \{x\}$ ?) is not intrinsic. Why does this mean that there is no equivalent intrinsic one?
Apr 11, 2021 at 20:25 history edited John Samples CC BY-SA 4.0
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Apr 4, 2021 at 8:11 comment added მამუკა ჯიბლაძე Hmmm don't know. There might be different topologies on that space. Maybe some of them are connected and not path-connected? I vaguely remember (I think from Ken Brown's book) that at least one of these is a CW-complex topology, but I believe that, say, the modular forms people use a different one, with $(ds)^2=((dx)^2+(dy)^2)/y^2$
Apr 4, 2021 at 8:01 comment added John Samples Also, a possibly relevant fact: The path components of a length space (in this case, the whole space) coincide with the 'finite distance components', i.e. maximal sets each of whose pairs of points can be joined by a finite-length path.
Apr 4, 2021 at 7:49 comment added John Samples I don't think so; otherwise the space would be disconnected. When I used "$\mathbb{H}$" I mean with the metric induced from the usual Euclidean one (though I suppose that theoretically it shouldn't matter, but then the arguments will be different). Sadly I'm not an expert on this space, either xD Someone will probably point out some technical issue(s) to deal with, then maybe we will have a clearer picture after sorting through them.
Apr 4, 2021 at 7:23 comment added მამუკა ჯიბლაძე In fact I am confused now. In the hyperbolic metric, is not $x$ an ideal point? I mean, at infinite distance from any internal point of the open upper halfplane?
Apr 4, 2021 at 7:04 comment added John Samples Oops, that probably needs to happen for every subsequence of the $c_n$, or at least in some limiting sense.
Apr 4, 2021 at 6:48 comment added John Samples I don't think so; I think if you draw consecutive segments of points approaching $x$ from the left (WLOG) with increasing $x$-coordinates and decreasing $y$-coordinates in the plane, then their angles should have to be bounded away from zero. Here the topological invariance should be due to $c_n$ being on the boundary of a neighborhood (a 'tangent disc topology circle'), and this boundary also belonging to a 'Euclidean topology circle', and they agree except at $x$.
Apr 4, 2021 at 4:58 comment added მამუკა ჯიბლაძე Is there a rectifiable path to $x$ passing through all $c_n$?
Apr 4, 2021 at 4:17 history edited John Samples CC BY-SA 4.0
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Apr 4, 2021 at 4:10 history answered John Samples CC BY-SA 4.0