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Is there a chance to make a sound argument for the triangle inequality - characterizing distances - from general considerations only, e.g. like this:

Given arbitrary distances $d(x,y)$ and $d(y,z)$ the distance $d(x,z)$ cannot be greater than a specific value which depends on $d(x,y)$ and $d(y,z)$. That is

$$d(x,z) \leq F(d(x,y),d(y,z))$$

Because $F$ must be symmetric, scale-invariant, and - by Occam's razor - as simple as possible, $F(a,b)$ must be $a + b$.

Is the last step (a) valid, and if not so (b) why?

I ask this question thinking of closeness $c(x,y) \in [0,\infty)$ as a natural counter-concept to distance. You can measure closeness just as you can measure distance (by applying yardsticks correspondingly). Closeness is symmetric, and closeness is maximal (as opposed to minimal) iff $x=y$. All in all: closeness is obviously nothing but and definable as the reciprocal of distance. BUT: one might want to establish it on its own feet. And this would demand of a reciprocal triangle inequality:

Given arbitrary closenesses $c(x,y)$ and $c(y,z)$ the closeness $c(x,z)$ cannot be smaller than a specific value which depends on $c(x,y)$ and $c(y,z)$. That is

$$c(x,z) \geq G(c(x,y),c(y,z))$$

Because $G$ must be symmetric, scale-invariant, and - by Occam's razor - as simple as possible, $G(a,b)$ must be $\frac{ab}{a+b}$.

Questions:

  1. Does the last step here disturb you more than in the first case?

  2. If so: why?

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The form of the triangular inequality is justified by the fact that it is useful in practice—surely, the fact that the euclidean distance satisfies is and that it comes naturally in that context is nice, but if in practice it were not be useful, we would not define metric spaces using it. What are you trying to solve with your definition? Knowing that will allow you to answer your question. –  Mariano Suárez-Alvarez May 4 '12 at 2:30
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Can't we argue for associativity? That cuts down the options by a lot. –  Will Sawin May 4 '12 at 2:32
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Why are algebraic operations (addition, multiplication, and division) simpler than order operations (like taking mins and maxes)? Occam's Razor as a meta-scientific principle is far from a sound inference method, nor does it always give the correct answer i.e. Newtonian mechanics versus general relativity. –  BSteinhurst May 4 '12 at 3:20
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When defining the range of the closeness function, do you mean $c(x, y)\in (0, \infty]$? That would seem to make more sense, if you're inverting the notion of distance. –  Noah S May 4 '12 at 3:27
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Changing the triangle inequality to the max inequality gives you the condition for a non-archimedean valuation, which are rather important in number theory, given that they're also called "primes". Basic concepts like these are rather dense with meaning - it's hard to stumble anywhere without hitting math. –  Will Sawin May 4 '12 at 4:31
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2 Answers

What is it about distance which lets us know that $d(x,z) \leq F(d(x,y),d(y,z))$ for some function $F$ but does not let us know that $F$ is addition? Suppose each integer lattice point $(x,y)$ is the center of a disk of radius $\frac{1}{5}$ We could have a space consisting of the disks and define the distance between two disks to be the length of the shortest line segment with one endpoint on the boundary of each disk. Then $d(x,z) \leq d(x,y)+d(y,z)+\frac{1}{5}$ with equality only when the centers are collinear and in the right order. Imagine these as islands where walking on an island is free but otherwise we fly and it costs (let's be birds). Then we can define a new and well behaved distance as the least amount of flying to get from one island to another with as many stops as desired on the way. It might be interesting to characterize that function (or maybe it is trivial, but that is another question.)

OIn an an ultrametric space one has $d(x,z) \leq \max(d(x,y),d(y,z)).$ For example a space where $d(x,y)$ is $0$ or $1$ according as $x=y$ or $x \ne y.$ Or a space of polynomials with distance defined to be $c^j$ for two unequal polynomials which differ for the first (or last) time in the coefficient of $x^j.$

Since there are these other distance notions, closeness need not be limited to $G=\frac{ab}{a+b}$

Now you might say "those are not what I intend by distance!" In which case I'd say that your idea of distance includes the standard triangle inequality.

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@Aaron: Cross-checking your birds-living-on-islands example just now I stumble over the $\Delta$ inequality: shouldn't the "correction" term be $\frac{2}{5}$ or shouldn't be $\frac{1}{5}$ the diameter of the disks? –  Hans Stricker May 7 '12 at 20:33
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To a(n otherwise nice) distance function $d$ that fails to satisfy the triangle inequality, one may canonically, and in the proper context, even functorially associate another, namely $ d'(x,y)= \inf_{\{a_1,\ldots,a_n\}} d(x,a_1) +(\sum_{i=1}^{n-1} d(a_i,a_{i+1}))+d(a_n,y)$.

If, in the context of a given application, $d'$ still contains all the essential information provided by the original $d$, then one may assume the triangle inequality as a matter of convenience. In this optic, the triangle inequality emerges less as self-evident truth and more as a useful convention.

In particular, when metrics arise in topology, at least in a certain sense, only the small values matter: $\min(d(x,y),B)$ defines the same topology as $d(x,y)$ no matter how small the $B$. In any case $d$ and $d'$ will induce the same topology, so the triangle inequality does not limit generality.

Outside of topology distance functions may carry less robust information. Viewing $d(x,y)$ as the cost of getting from $x$ to $y$, the functorial construction above depends on the assumption $\forall x\ d(x,x)=0$. Of course in the real world adding an extra leg to a journey may entail a fixed cost: anyone who thinks there's no triangulation tax has never tried to get a decent meal while waiting for a connection at the airport.

One may modify construction of $d'$ from $d$ to account for a fixed or variable triangulation tax. If, say, the tax is bounded by $B$ one gets $d'(x,z)\leq d'(x,y)+d'(y,z)+B$.

Final note to OP: don't you want to impose $F(D,0)=F(0,D)=D$?

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