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EDIT: I have asked a better version of this question here.

Why does the distance from a point to itself need to be $0$? Doesn't it only need to be the smallest distance possible in that space? Do we not obtain an equivalent theory of metric spaces if we declare that $\forall a,b,c,d(a,a)≤d(b,c)$ and $\forall x,y, d(x,x) =d(y,y)$? What "goes wrong" if we substitute these two axioms for the usual $d(x,x) = 0$?

Andrés E. Caicedo
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Perry Bleiberg
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    What do you do for your triangle inequality? – Angina Seng Jun 13 '17 at 04:32
  • Keep it the same. I was thinking that's where things ought to go wrong, but I couldn't quite make it work. – Perry Bleiberg Jun 13 '17 at 04:32
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    @user52969 Besides the triangle inequality, you'd also lose $\operatorname{d}(x,y)=0 \iff x=y$. – dxiv Jun 13 '17 at 04:41
  • @dxiv I guess I was implicitly thinking of replacing $d(x,y)=0⟺x=y$ with $d(x,y)=\epsilon⟺x=y$, where $\epsilon$ is always smaller than any other possible distance, and just moving on with life. What I'm trying to ask is: "Is there anything about 0, other than the fact that it is the smallest nonnegative number, that makes it a good value for the distance we say things are from themselves?" Can you articulate why I've lost the triangle inequality? – Perry Bleiberg Jun 13 '17 at 04:46
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    We are familiar with metric spaces, we live in one. It is nice to have a correspondence between experience and the mathematics that we use to describe it. There is no distance between a point and itself. – Tucker Jun 13 '17 at 04:49
  • @Tucker I would agree if I were trying to study the real world, but, alas, I am trying to find an interpretation of metric spaces as 2-categories. (Btw, can anyone link me to anywhere this has been done before?) – Perry Bleiberg Jun 13 '17 at 04:52
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    You'd also need to say how to translate this into a topology or similar structure. For example, many concepts in topology are defined by considering balls of arbitrarily small radius. Would those be replaced here by arbitrarily small balls of radius $> \epsilon$? If so, that would ascribe a special status to the number $\epsilon$, which would be different for different spaces. – user49640 Jun 13 '17 at 04:54
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    It seems like a pseudometric where all distances have been shifted from $0$ by some fixed constant. So, in my opinion, it is nothing more interesting than a pseudometric. – Rocket Man Jun 13 '17 at 04:56
  • @user49640 When you say "need to," do you mean "need to so your theory could actually be useful" or "need to, or else you haven't actually come up with an equivalent theory" ? I am new to the notion of interpretation functions so excuse me if I am missing something basic. – Perry Bleiberg Jun 13 '17 at 04:57
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    Yes, "need to" so that the theory would be useful. Metric spaces are useful because they allow us to talk about continuity, for example. – user49640 Jun 13 '17 at 04:58
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    @user52969 d(x,y)=ϵ ⟺ x=y This doesn't follow from the conditions stated in the posted question. Maybe you should articulate the complete definition of $\operatorname{d}(x,y)$ that you have in mind. why I've lost the triangle inequality What would be $\operatorname{d}(\epsilon\cdot x, \epsilon \cdot x),$? – dxiv Jun 13 '17 at 05:04
  • @dxiv Why there should be a multiplication? –  Jun 13 '17 at 05:23
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    It's equivalent to a metric space with all distances increased by a constant, as long as you replace the triangle inequality with $d(x,y)+d(y,z)\le d(x,z)+\epsilon$. –  Jun 13 '17 at 05:25
  • @PaulK Fair point, I was thinking at metrics on vector spaces. That said, I'll leave my previous comment in for now, until the OP clarifies "the complete definition of $\operatorname{d}(x,y)$ that you have in mind" which I asked for. If that turns out to be on something other than a vector space, then I'll remove said comment. – dxiv Jun 13 '17 at 05:34
  • Thanks for your answers everybody. I have rethought this question a bit and asked it more articulately here: https://math.stackexchange.com/questions/2323058/alternative-categorification-of-metric-spaces – Perry Bleiberg Jun 15 '17 at 00:29

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If x_n = x for every n +1,2,3,... you do want the sequence to converge to x which means d(x,x) = d(x_n ,x ) ---> 0 and this implies d(x,x) = 0 . On the other hand leaving out the axiom , : d(x,y)=0 ==> x=y and you get what is referred to as a pseudo metric space and these have importance for say spaces of integrable functions which leads to an identification of functions which have 0 distance between them ,in order to get uniqueness of limits .

user439545
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  • According to wikipedia a pseudometric still fulfils the requirement that $d(x,x)=0$ while it allows for $d(x,y)=0$ for $x\ne y$. – skyking Jun 13 '17 at 05:07
  • Also note that that definition of convergence is not necessarily the one to be used. If we define open sets in the same way as earlier you still can have convergence despite the distance don't approach zero. – skyking Jun 13 '17 at 05:10
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Maybe you could drop that requirement, but of course then it would not fulfil the standard definition for a metric space.

Dropping that requirement would alter the topological properties of the space. If you have for some $a$ that $d(a,a)>0$ then that would mean that we have open points. For example this would make the space non-connected.

Note that if you instead require that $d(a,a)$ be the smallest distance available you can define a proper metric by setting $\delta(x,y) = d(x,y)-d(a,a)$.

skyking
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Suppose you make the axiom changes you're proposing. Then $d(x,x) + d(x,x) = 2\epsilon > \epsilon = d(x,x)$ since $\epsilon$ for you is an infinitesimal but non-zero number. This gives you a problem: pick any other point $y$ in a small enough neighbourhood of $x$ and you can find a number $N$ so that $Nd(x,x) = d(x,y)$ which is surely not what you're intending.

You mention in the comments (I think you should write it explicitly in the question) that you're trying to treat a metric space as a 2-category. You could try here: https://ncatlab.org/nlab/show/metric+space as a starting point, and look at the Lawvere metric spaces. I know too little about category theory to know how helpful it might be.

postmortes
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  • Maybe if distances come from a non-Archimedean field? – G Tony Jacobs Jun 13 '17 at 05:22
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    @GTonyJacobs for the benefit of others, you mean that we might be in a field where it's not true that all numbers are within an integer multiple of each other? In this case I think you're right: there's no real problem with making this axiom change (I think we still lose the notion of separation, but I'm not sure that's important if we're working in non-Archimedean fields!) – postmortes Jun 13 '17 at 05:29
  • Yeah, at that point, we would be defining a very peculiar structure, and I'm not sure we'd want to call the function $d(\cdot,\cdot)$ a "distance" function at all! – G Tony Jacobs Jun 13 '17 at 05:32