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As an exercise, I am considering a real-world example and I am thinking if I can model it through metric spaces, but apparently it is not possible (or I may be very wrong in my reasoning).

Say that I have a bunch of cities and I measure the CO2 levels at some point of the day in all the cities. This leads me to a set of numbers which I can use for my metric space.

Then, I also want know how "far" are two cities in terms of measured CO2 levels. I can take the absolute value of the difference of any pair of cities to see how "far" they are. By that, I think I should have defined a metric space.

However, I cannot figure if one city is more or less polluted than another, which is something desirable to know. I could remove the absolute value from my distance function, but then the function I am using is no longer a distance.

Am I using a wrong mathematical structure? Or what shall I do to capture that? What if I am interested in both CO2 and NOx for which my measurements would be 2D vectors?

EDIT: I tried to pick $d(x,y)=exp(x-y)$ but then I would have $d(x,x)=1$ and therefore that is not a good distance function.

Barzi2001
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    You probably want a norm instead of metric. – freakish Nov 08 '23 at 16:46
  • The problem here is that you are trying to use one function for two different purposes, and while those purposes are closely related, they are not entirely consistent. The concept of distance is symmetric. $x$ is as far from $y$ as $y$ is from $x$. The concept of order is anti-symmetric: If $x$ is less than $y$, then $y$ is not less than $x$. Accept that these are two different things that cannot be completely combined into one measurement, and you will be fine. – Paul Sinclair Nov 09 '23 at 15:13
  • @freakish point! But aren’t norms defined on vector spaces? In my case I don’t have a vector space because the “candidate” vector space of the measurements is on R+. I may restrict the field to R+ but then I don’t have a field… how to make everything to stand up? – Barzi2001 Nov 10 '23 at 03:24
  • @PaulSinclair thanks! But I am thinking: what if we relax the symmetry property of the distance? That is going from x to y is not the same to go from y to x? Would that fix my issue? I am thinking to fix a point and then measure the distance of each element from such a point. The non-symmetry could tell me if the pollution is higher or smaller… just guessing… – Barzi2001 Nov 10 '23 at 03:27
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    Yes, you can drop that symmetry property! But then it isn't a distance, it's a difference. That means it isn't a metric, as it has different properties. As for vector spaces, $\Bbb R^+ \subset \Bbb R$, so yes, you do have a vector space, even if you are not using all of it. – Paul Sinclair Nov 10 '23 at 05:03

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