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Let's say that a function $d:S\times S\to [0,\infty)$ for a countable set $S$ is a metric in the limit if $$d(x,y)\le \liminf_{n\to\infty} d(x,z_n)+d(z_n,y),$$ $$\lim_{n\to\infty} d(z_n,z_n)=0, \quad\text{and}\quad \lim_{n\to\infty} d(x,z_n)- d(z_n,x)=0$$ for any sequence $z_n$ that goes to infinity in the sense that $(\forall x)(\exists n_0)(\forall n\ge n_0)(z_n\ne x)$.

We could require that $d(x,y)=0\implies x=y$ as well.

I'm looking for an answer to any one of these:

  1. Is this a familiar idea?
  2. Does some of the theory of metric spaces still hold for such functions?
  3. Perhaps there exists a "better" variant of this idea?
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  • $\begingroup$ But $d(x,z_n)$ may not converge as $n\to\infty$. $\endgroup$ – Jack L. Nov 12 '20 at 22:17
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    $\begingroup$ @PietroMajer constant sequences don't go to infinity :) $\endgroup$ – Bjørn Kjos-Hanssen Nov 13 '20 at 6:15
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    $\begingroup$ It might be interesting how this plays together with the asymptotic cone. $\endgroup$ – Ville Salo Nov 13 '20 at 6:29
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    $\begingroup$ Yes, that's exactly the sort of thing I had in mind, shadowing a weakened structure by a rigid one is always interesting. Another comment, just like with topology/coarse geometry, one could separately study small-scale behavior and large-scale behavior for countable spaces with a "metric in the limit". There's also a zooming-in version of asymptotic cones, IIRC called tangents. (I'm not an expert on any of this, and I didn't play with your specific axioms.) $\endgroup$ – Ville Salo Nov 13 '20 at 8:21
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    $\begingroup$ I'm not sure of the answer but here are two comments. I suspect that you might want to make your notion of going to infinity stronger. Your current assumption would allow for Cauchy sequences that converge to $x$ but don't actually hit $x$. Going to infinity is more that $z_n$ eventually leaves any compact set (perhaps any $d$-neighborhood of x). $\endgroup$ – Gabe K Nov 13 '20 at 13:16
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This is not a complete answer, but was too long for a comment. I don't know of any place where this exact idea has been studied. However, the notion of metrics at infinity play an important role in the analysis of $\delta$-hyperbolic spaces, so this might be a good model for the idea. In particular, the Gromov boundary [1] of a $\delta$-hyperbolic space inherits a metric structure which is in some sense the "metric in the limit."

A $\delta$-hyperbolic $(M,d)$ space is a metric space, but I suspect that many of the properties of the Gromov boundary can be recovered if you weaken the assumptions on $d$ to only involve metricity "in the limit."

In order for this to work, the definition for a "metric in the limit" is really going to have to reflect the fact that the sequences are "going to infinity" and don't converge to anything. You also want to gain metricity in the limit, and I'm not sure your first definition is strong enough to do that.

As such, my proposal would be to add the assumption that $$ \liminf_{n\to\infty} d(x_n,z_n)+d(z_n,y_n)-d(x_n,y_n) \geq 0,$$ for any sequences $x_n,y_n, z_n$ which go to infinity. I would also suggest changing the definition of "going to infinity" to be the following:

There exists an $\epsilon$ so that for all $x \in S$, there exists an $n_0$ so that for all $n\ge n_0$, $d(z_n, x)>\epsilon$.

On their own, these definitions are not enough to recover any of the properties of the Gromov boundary. However, with a rough version of $\delta$-hyperbolicity and an extra assumption so that "going to infinity" means that the $d(x,z_n) \to \infty$, I suspect that you might be in business.

[1] Väisälä, Jussi, Gromov hyperbolic spaces, Expo. Math. 23, No. 3, 187-231 (2005). ZBL1087.53039.

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  • $\begingroup$ Your displayed equation is not formally a strengthening, but rather incomparable to mine I guess? I didn't say it but I was thinking of a bounded space, $\exists c\forall x,y\, d(x,y)\le c$, which is maybe a different direction than you're going in. Still, thanks for the food for thought. $\endgroup$ – Bjørn Kjos-Hanssen Nov 13 '20 at 18:02
  • $\begingroup$ That's a good point. I'll edit the answer to acknowledge that. $\endgroup$ – Gabe K Nov 13 '20 at 19:21
  • $\begingroup$ In order to be amenable to discrete topologies, you might want to switch the order of $x$ and $\epsilon $ in that definition of "going to infinity" so that the $\epsilon $ may depend on $x$. That will be equivalent to your definition if the topology induced by $d(x, \cdot)$ is discrete. $\endgroup$ – Gabe K Nov 13 '20 at 19:29

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