# weak metric space

In the definition of a metric space, replace the triangle inequality by the weaker inequality

d (x, z) ≤ C max {d (x, y), d (y, z)},

where C is a positive constant (depending on the "metric", but not on the points x, y, z). Had structures like this ever been studied?

One can associate a more or less natural topology to this "metric", calling a set open if every point belonging to the set has a ball of positive radius, centered at this point and contained in the set. But I cannot say much on this topology. For instance, it is not obvious (and may be not true) that a ball is an open set. Neither could I prove that this topology is Hausdorff.

Any information, reference, etc. would be welcome.

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When $C=1$ I think this might hold for $p$-adic numbers... in which case the condition is stronger than the usual metric space condition, and makes $p$-adic analysis a bit easier. – Dylan Wilson Oct 16 '10 at 23:51
Yes, for C=1, this is just the so-called ultrametric en.wikipedia.org/wiki/Ultrametric_space Though why it is called (or should be called) ultrametric is not fully clear to me. – Suvrit Oct 17 '10 at 8:30

Yes, they were introduced in valuation theory by Emil Artin and remain present in many contemporary treatments, including mine: see

http://math.uga.edu/~pete/8410Chapter1.pdf

especially Section 1.2 and

http://math.uga.edu/~pete/8410Chapter2.pdf

[Added: as RW has justly pointed out, my answer here makes sense in the context of valuation theory only. The procedure that I give from passing from a "weak metric" to a metric is not going to work in general, I think, but only in the presence of some additional algebraic structure. If I were the OP, I might not choose this as the accepted answer, or at least not yet.]

The basic idea here is to consider two such guys equivalent if one can be obtained from the other by the operation of raising $d$ to some positive real number power $\alpha$. In such a way, one can make the constant $C \leq 2$ in which case one gets an actual metric. The topology one gets in this way is easily seen to be independent of the choice of $\alpha$.

As it happens, when writing up these notes for a course I taught last spring I also thought a little bit about trying to define "balls" with respect to such a weak metric (i.e., without first renormalizing). I didn't get anywhere with this either.

Finally, I should say that in valuation theory at least, these "weak metrics" (in the valuation theoretic context I called them "Artin absolute values" and then after the course was over decided to change to the terminology to just "absolute values", while retaining the word "norm" for such a guy which was actually a metric) come up as a useful tool but are not really studied on their own or in any deep way: I have yet to see a text on valuation theory where weak norms appear after page 20 or so. Whether they may have wider applicability in some other context, I couldn't say...

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The arguments you refer to (that for $C\le 2$ one gets an actual metric) heavily depend on an additional algebraic structure. I don't see how they would work in the general case. – R W Oct 17 '10 at 1:14
@RW: That may well be. I have only encountered the phenomenon in the setting of valuation theory, and I don't mean to imply that what I say works in the general case. Perhaps this should not be the accepted answer... – Pete L. Clark Oct 17 '10 at 2:56