- #1

chwala

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- Homework Statement
- See below

- Relevant Equations
- Metric spaces

Am refreshing on Metric spaces been a while...

Consider the axioms below;

1. ##d(x,y)≥0## ∀ ##x, y ∈ X## - distance between two points

2. ## d(x,y) =0## iff ##x=y##, ∀ ##x,y ∈ X##

3.##d(x,y)=d(y,x)## ∀##x, y ∈ X## - symmetry

3. ##d(x,y)≤d(x,z)+d(z,y)## ∀##x, y,z ∈ X## - triangle inequality

The proofs are clear to me, i just read on that. I wanted to check how to show that axiom ##2## holds...

My take is given set ##R## with usual metric si defined by,

##d_1(x,y)##=##|x-y|##, ∀ ##x, y ∈ X##, then ##d_1(x,y)##= ##\sqrt {(x-x)^2+(x-x)^2}## since ##x=y##

Consider the axioms below;

1. ##d(x,y)≥0## ∀ ##x, y ∈ X## - distance between two points

2. ## d(x,y) =0## iff ##x=y##, ∀ ##x,y ∈ X##

3.##d(x,y)=d(y,x)## ∀##x, y ∈ X## - symmetry

3. ##d(x,y)≤d(x,z)+d(z,y)## ∀##x, y,z ∈ X## - triangle inequality

The proofs are clear to me, i just read on that. I wanted to check how to show that axiom ##2## holds...

My take is given set ##R## with usual metric si defined by,

##d_1(x,y)##=##|x-y|##, ∀ ##x, y ∈ X##, then ##d_1(x,y)##= ##\sqrt {(x-x)^2+(x-x)^2}## since ##x=y##

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