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Well, after recently answering a Hilbert's Hotel question, I've started to think: If an infinite number of people arrive, the solution is that every guest goes to the room with twice the number. However, if one imagines the doors one after the other (that is, in the order of the natural numbers), this means that each guest has to walk a distance that is proportional to the number of his room, which is unbounded. Assuming a constant walking speed, therefore also the move time is unbounded. That is not really a satisfying solution.

But this is easily solved: Just build the hotel in an infinite-dimensional Euclidean space, where room $n$ sits at the point which has coordinates $x_k=\delta_{nk}$ where $\delta$ is the Kronecker delta. That way, the distance between any two rooms is $\sqrt{2}$, and any room changing operation, no matter how complicated, can be done in constant time.

So far, so good. However, let's assume the guests in Hilbert's Hotel are conventional 3-dimensional beings, and therefore they must live on a 3-dimensional manifold; they would die in an unconstrained higher-dimensional space, let alone an infinite-dimensional one.

Therefore my question:

Does there exist a 3-dimensional Riemannian manifold which has a countably infinite number of points such that any two of them have the same finite distance on the manifold?

celtschk
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    Why do the distances have to be equal? For the sake of the original concern, travel time, why couldn't the the rooms be located at 1/n as the person in room n only needs to travel a distance of 1/2n to get to room 2n? – fleablood Oct 12 '16 at 18:20
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    @fleablood: That only works if a room is a point object. If you want a room to have some finite size, so that guests will fit into their rooms, you get in more trouble... – Micah Oct 12 '16 at 18:22
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    @fleablood: Micah has hit the nail on the head. I indeed was thinking of extended guests that cannot make themselves shrink. With points of constant distance, one would expect there to a finite space around each point that doesn't overlap with the spaces in other points (of course that would have to be proven separately; given — as HK Lee showed — that those points already don't actually exist, this question is of course moot anyway). – celtschk Oct 12 '16 at 18:28
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    I suggest putting the hotel in hyperbolic space, so that guest $n$ only has to make a trek of length $O(\log n)$, rather than of length polynomial in $n$. You might even be able to do better than this if you had a manifold whose curvature got more negative as you got farther from some central point, but I haven't worked out the details... – Micah Oct 12 '16 at 19:44
  • @Micah that $\mathcal{O}(\log n)$ distance sounds like an excellent solution, how can you prove this? – leftaroundabout Oct 12 '16 at 22:03

2 Answers2

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If $x_n$ are such points then $R=d(x_1,x_n) >0$ for all $n$ Hence $x_n\in B_R(x_1)$ Since $B_R(x_1)$ is compact so there is a convergent subsequence. Hence it does not happen.

rubik
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HK Lee
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    +1. Note that this isn't special to $\mathbb{R}^3$ - it applies to any metric space in which all closed balls of finite radius are compact. – Noah Schweber Oct 12 '16 at 18:20
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For posterity here's an implementation of Micah's suggestion.

Let $(\rho, \theta, \phi)$ denote spherical coordinates on the open unit ball, $\Omega = d\phi^{2} + \sin^{2}\phi\, d\theta^{2}$ the round metric on the unit sphere, and $f(\rho) = \rho/(1 - \rho)$ (or any smooth, monotone function defined for $0 \leq \rho < 1$ with $f(0) = 0$ and, as $\rho \to 1^{-}$, $f \to \infty$ rapidly enough that $f$ is not improperly integrable).

In the metric $$ g = d\rho^{2} + f(\rho)^{2}\, \Omega, $$ the distance from the origin to the boundary of the ball is unity, but the volume element is $$ dV = f(\rho) \sin\phi\, d\rho\, d\theta\, d\phi. $$ Geometrically, the intrinsic radii of spherical shells centered at the origin grow rapidly enough that the volume is infinite.

In this universe, there exist countably many "cells" of fixed volume (though not suitable as three-dimensional hotel rooms, as asymptotically they necessarily become "intrinsically thin in the radial direction"), but any two rooms (or points) are separated by a distance of at most $2$ because the origin is at most one unit away from each room.

Andrew D. Hwang
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