Inverse of Metric in Topology: Schwarz Inequality

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Discussion Overview

The discussion revolves around the concept of whether the multiplicative inverse of a metric can itself be considered a metric, particularly in the context of the triangle inequality and the Schwarz inequality in topology.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions if the multiplicative inverse of a metric, defined as ##D(x,y) = \frac{1}{d(x,y)}##, respects the triangle inequality.
  • Another participant suggests taking the reciprocal of the inequality and flipping it, but does not clarify the implications for the triangle inequality.
  • A participant expresses uncertainty about whether the new metric ##D## can be a valid metric if it does not satisfy the triangle inequality.
  • Some participants argue that the proposed metric ##D## may not be a metric at all, citing potential violations of the triangle inequality.
  • A counterexample is provided using the standard Euclidean metric, demonstrating that the triangle inequality does not hold for the proposed inverse metric in specific cases.

Areas of Agreement / Disagreement

Participants generally disagree on whether the inverse of a metric can be considered a valid metric, with multiple competing views and no consensus reached on the validity of the proposed metric.

Contextual Notes

There are limitations regarding the definition of the inverse metric, particularly in cases where the original metric is zero, which leads to undefined behavior. Additionally, the discussion highlights the need for further exploration of the properties of the proposed metric.

Bachelier
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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?
 
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Hey Bachelier.

Try taking the reciprocal (also flip the inequality) and collect terms.
 
chiro said:
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.
 
I don't think your D metric will be a metric at all.
 
chiro said:
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.
 
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.
 

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