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Suppose $d$ is a bounded metric on $X$, i.e. $d(x,y)< K<\infty$ for all $x,y\in X$. Is there a standard way to convert $d$ into another metric $\widetilde{d}$ on $X$ with the property that $\widetilde{d}(x,y)\to\infty$ if and only if $d(x,y)\to K$? One way would be to find some function $f$ such that $\widetilde{d}(x,y)=f(d(x,y))$ satisfies the given conditions, but it is not obvious that this is always possible.

The following properties are additionally useful, but not necessary:

  • $\widetilde{d}$ preserves the topology of $(X,d)$
  • $d(x,y)>d(x',y')\implies \widetilde{d}(x,y)>\widetilde{d}(x',y')$

Also, if this is possible when $(X,d)$ satisfies certain extra assumptions but not in general, answers in this direction are welcome.

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    $\begingroup$ I'll assume that $K=1$. Then set $\tilde d:=\frac{d}{1-d}$. I voted to close the question as off-topic. $\endgroup$ Jan 7, 2018 at 22:11
  • $\begingroup$ That's a typo (fixed). Does your answer preserve either of the additional properties I mentioned? (Will happily accept this answer if you can add a brief discussion.) $\endgroup$
    – JohnA
    Jan 7, 2018 at 22:16
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    $\begingroup$ @Loïc Teyssier: that's not a metric (consider three points satisfying $d(a,b)=d(b,c)=1/2$ and $d(a,c)=1$). In fact this example shows that nothing of the form $f\circ d$ can work. $\endgroup$
    – Nik Weaver
    Jan 7, 2018 at 23:32
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    $\begingroup$ Obviously this is not possible if there are points $x,y$ with distance $1$ in the old metric, so you probably meant to assume $d(x,y)<K$ for all $x,y$, with $K=\sup d(x,y)$ and are then asking about a new $d'$ such that $d'\to \infty$ if $d\to K$ (where is that $1$ coming from anyway?). $\endgroup$ Jan 8, 2018 at 0:50
  • $\begingroup$ @ChristianRemling Yes, you are correct. I fixed my question to reflect these edits. $\endgroup$
    – JohnA
    Jan 8, 2018 at 3:04

3 Answers 3

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I do not know of any procedure for constructing such metric space in the general setting. On the contrary, if you restrict your class of metric spaces to the smaller class of complete locally compact path-metric spaces then the Hopf-Rinow theorem, see the First Chapter of [Gr], will prevent the construction of your $(X,\tilde{d})$.

If $(X,d)$ is in the intermediate class of (non-complete and non-compact) locally compact path-metric spaces, then I recall having seen the construction of a $(X,\tilde{d})$ satisfying

  1. $(X,\tilde{d})$ is a complete locally compact path-metric space
  2. The topology of $X$ is preserved.
  3. $d(x,y) \leq \tilde{d}(x,y)$.

The procedure will be something like completing $(X,d)$, embedding $X$ inside its metric completion $\bar{X}$ and then enlarging the metric of $X$ around the points in $\bar{X} \setminus X$ (think of this as sending the extra points "to infinity"). I ignore whether your second condition can be preserved under this construction

[Gr] Gromov, Misha, Metric structures for Riemannian and non-Riemannian spaces. Progress in Mathematics (Boston, Mass.). 152. Boston, MA: Birkhäuser. xix, 585 p. (1999). ZBL0953.53002.

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I don't think there's any standard procedure, but here is one way to do this if there is a "base point" $e\in X$ satisfying $d(e,p) < K/2$ for all $p\in X$. I am not sure whether such a point can always be added. Wlog take $K=2$ and fix $e$. For any $p,q \in X$ find $p', q'$ in the open unit disc equipped with the euclidean metric such that the distances between $p'$, $q'$, and $0$ equal the distances between $p$, $q$, and $e$. There should be only one way to do this up to rotations and reflections. Then define $\tilde{d}(p,q)$ to be the distance between $p'$ and $q'$ for the hyperbolic metric on the disc.

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Not unless you're willing to lose the triangle inequality.

Nik Weaver's counterexample in the comments doesn't require $\tilde{d}$ to be of the form $f \circ d$: If $(X,d) = [0,1]$ with the usual metric, then you'd be requiring $\tilde{d}(0,1) = \infty$ while $\tilde{d}(0,0.5)$ and $\tilde{d}(0.5,1)$ are finite.

For the specific case of $f \circ d$ on $[0,1]$, the triangle inequality appears to be related to concavity of $f$.

The closest thing I can think of that respects the triangle inequality is dividing a Riemannian metric by the distance* to some set that you're about to delete.

*If you want the result to be Riemannian too, you still have to worry about smoothness, which probably requires constructing a different function to divide by; e.g. assume the deleted set is a submanifold and modify the metric only on a tubular neighborhood.

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  • $\begingroup$ I don't think my example shows anything if you drop the assumption that $\tilde{d}$ is of the form $f\circ d$. $\endgroup$
    – Nik Weaver
    Jan 8, 2018 at 0:31
  • $\begingroup$ Good point about concavity. This is necessary but not sufficient (see mathoverflow.net/q/35669/99132), so its not clear what happens in the general case. Also, if there are points with $d(x,y)=K$, then the answer must be no for trivial reasons so I modified the question to require $d(x,y)<K$, which seems to invalidate the counterexample. $\endgroup$
    – JohnA
    Jan 8, 2018 at 14:06
  • $\begingroup$ That should say "sufficient but not necessary"... $\endgroup$
    – JohnA
    Jan 8, 2018 at 14:11
  • $\begingroup$ It doesn't invalidate the counterexample. You just take points whose distance is close to $K$. $\endgroup$
    – Nik Weaver
    Jan 8, 2018 at 15:36
  • $\begingroup$ If $f$ is subadditive and $d(a,c)=K-\delta$ for some $\delta>0$, it seems to work, but maybe you have a different argument in mind from what @pteromys suggested. Feel free to clarify if I am missing something. $\endgroup$
    – JohnA
    Jan 8, 2018 at 17:08

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