Inverse of Metric in Topology: Schwarz Inequality

[Bachelier]

In Topology:

is the multiplicative inverse of a metric, a metric?

How do we define the Schwarz inequality then?

if $d(x,z) ≤ d(x,y) + d(y,z)$ the inverse $1/d(x,z)$ would give the opposite?

 

[chiro]

Hey Bachelier.

Try taking the reciprocal (also flip the inequality) and collect terms.

 

[Bachelier]

Hey Bachelier.

Try taking the reciprocal (also flip the inequality) and collect terms.

Hi Chiro:

I don't know if what you mean is taking the reciprocal of $d(x,y)$

indeed if I start with $d(x,z) ≤ d(x,y) + d(y,z)$

then $\frac{1}{d(x,z)} ≥ \frac{1}{d(x,y) + d(y,z)}$

would lead to $d(x,z) ≤ d(x,y) + d(y,z)$ again.

But what I feel like is the need to show that: my new metric defined as $D(x,y) = \frac{1}{d(x,y)}$ respects the triangle inequality.

$i.e. D(x,z) ≤ D(x,y) + D(y,z)$ (result A)

but the way the metric $D$ is defined would give me

$D(x,z) ≥ D(x,y) + D(y,z)$ not result A

Failing this will not make $D$ a metric.

 

[chiro]

I don't think your D metric will be a metric at all.

 

[Bachelier]

I don't think your D metric will be a metric at all.

I thought so. I heard someone talking about the inverse of a metric as being a metric on the same space.

 

[jbunniii]

Consider the standard euclidean metric on the real numbers d(x,y) = |x-y|, and let D(x,y) = 1/d(x,y). Even overlooking the problem that this is not defined when x=y, we can find a simple counterexample. Let x = 0, y = 3, z = 1. Then D(x,y) = 1/3, D(y,z) = 1/2, and D(x,z) = 1, so the triangle inequality D(x,z) <= D(x,y) + D(y,z) is not satisfied.