Timeline for Metric "in the limit"?
Current License: CC BY-SA 4.0
22 events
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| Nov 13, 2020 at 17:00 | answer | added | Gabe K | timeline score: 1 | |
| Nov 13, 2020 at 16:32 | comment | added | Gabe K | Not exactly. Consider the space $[0,2]$ with a "distance at infinity". Under your definition, the sequence $z_n = 1 + \frac{1}{n}$ and this would satisfy the assumption of "going to infinity." In short, what would happen is that if the "distance at infinity" were continuous (with respect to the usual topology), it would force the "distance at infinity" to be a metric in the usual sense, which is what you are trying to generalize. | |
| Nov 13, 2020 at 15:43 | comment | added | Bjørn Kjos-Hanssen | @Gabe I could add boundedness $d(x,y)\le 2$ for all $x,y$, does that take care of your first concern? | |
| Nov 13, 2020 at 13:24 | comment | added | Gabe K | Second, your first assumption seems like it should also be stronger. In particular, if you fix $x$ and $y$, then this inequality is satisfied whenever either $d(x,z_n)$ or $d(z_n,y)$ go to infinity. However, that's not really strong enough to imply "metric-like" behavior. Another idea might be that if $x_n, y_n$ and $z_n$ all go to infinity, then $$ \liminf_{n \to \infty} d(x_n,z_n)+d(z_n,y_n)-d(x_n,y_n) \geq 0.$$ In other words, the asymptotic triangle inequality holds. | |
| Nov 13, 2020 at 13:16 | comment | added | Gabe K | 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). | |
| Nov 13, 2020 at 8:21 | comment | added | Ville Salo | 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.) | |
| Nov 13, 2020 at 8:06 | comment | added | Bjørn Kjos-Hanssen | @VilleSalo That's interesting. Would be nice if a "metric in the limit" space is associated with an actual metric space, somehow | |
| Nov 13, 2020 at 6:29 | comment | added | Ville Salo | It might be interesting how this plays together with the asymptotic cone. | |
| Nov 13, 2020 at 6:15 | comment | added | Bjørn Kjos-Hanssen | @PietroMajer constant sequences don't go to infinity :) | |
| Nov 13, 2020 at 5:24 | comment | added | Pietro Majer | I don't get the point, because if you consider constant sequences you recover the usual axioms of a distance, and on the other hand the usual axioms imply the inequalities with limits. | |
| Nov 12, 2020 at 23:53 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |
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| Nov 12, 2020 at 23:51 | comment | added | Bjørn Kjos-Hanssen | @mathworker21 thanks, yes, that's what I mean by "eventually not equal to". I have edited it in now. | |
| Nov 12, 2020 at 23:45 | comment | added | mathworker21 | The $\lim_{n \to \infty} d(z_n,z_n) = 0$ condition seems silly. If $S$ contains at least two points, say $x$ and $y$, then we may let $z_1,z_2,z_3,z_4,\dots = x,y,x,y,\dots$ to see $d(x,x) = 0$ and $d(y,y) = 0$. Do you mean to say that $z_n$ goes to infinity if it's not the case that some $x \in S$ has $z_n = x$ for infinitely many $n$? | |
| Nov 12, 2020 at 23:04 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |
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| Nov 12, 2020 at 22:55 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |
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| Nov 12, 2020 at 22:46 | history | edited | Bjørn Kjos-Hanssen |
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| Nov 12, 2020 at 22:40 | comment | added | Bjørn Kjos-Hanssen | @JackL. thanks, I edited the question | |
| Nov 12, 2020 at 22:39 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |
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| Nov 12, 2020 at 22:38 | history | undeleted | Bjørn Kjos-Hanssen | ||
| Nov 12, 2020 at 22:19 | history | deleted | Bjørn Kjos-Hanssen | via Vote | |
| Nov 12, 2020 at 22:17 | comment | added | Jack L. | But $d(x,z_n)$ may not converge as $n\to\infty$. | |
| Nov 12, 2020 at 22:09 | history | asked | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |