TITLE: In a metric space, why must $d(x,x) = 0$?
QUESTION [6 upvotes]: 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 $d(x,x) = 0$?

REPLY [2 votes]: 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.