Prove that d is a METRIC

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Does that seem like it covers the case?In summary, a metric d on X x Y is defined by d((x1,y1),(x2,y2))=max(ρ(x1,x2),σ(y1,y2)). To verify that d is a metric, we need to prove the "triangle inequality" property, which states that for any (x1,y1),(x2,y2),(x3,y3) in X x Y, max(ρ(x1,x2),σ(y1,y2))≤ max(ρ(x1,x3),σ(y1,y3)) + max(ρ(x3,x2),σ(y3,y2)). To count the cases, we can consider two cases
  • #1
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Homework Statement


Let (X,ρ) and (Y,σ) be metric spaces.
Define a metric d on X x Y by d((x1,y1),(x2,y2))=max(ρ(x1,x2),σ(y1,y2)).
Verify that d is a metric.

Homework Equations



The Attempt at a Solution


I proved positive definiteness and symmetry, but I am not sure how to prove the "triangle inequality" property of a metric. How many cases do we need in total, and how can we prove it?

Any help is appreciated!
 
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  • #2
So to verify the triangle inequality, we need to prove that
max(ρ(x1,x2),σ(y1,y2))≤ max(ρ(x1,x3),σ(y1,y3)) + max(ρ(x3,x2),σ(y3,y2)) for ANY (x1,y1),(x2,y2),(x3,y3) in X x Y.

How many separate cases do we need? I have trouble counting them without missing any...Is there a systematic way to count?

Case 1: max(ρ(x1,x2),σ(y1,y2))=ρ(x1,x2), max(ρ(x1,x3),σ(y1,y3))=ρ(x1,x3), max(ρ(x3,x2),σ(y3,y2)) =ρ(x3,x2)

This case is simple, the above inequality is true since ρ is a metric.


Case 2: max(ρ(x1,x2),σ(y1,y2))=ρ(x1,x2), max(ρ(x1,x3),σ(y1,y3))=σ(y1,y3), max(ρ(x3,x2),σ(y3,y2)) =ρ(x3,x2)

For example, how can we prove case 2?


Any help is appreciated!
 
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  • #3
Suppose that [tex]\rho(x_1,x_2)\ge\sigma(y_1,y_2)[/tex]. What do you know about [tex]\rho(x_1,x_3)+\rho(x_3,x_2)[/tex]? Can you infer anything about the right-hand side of your inequality based on that?
 
  • #4
Case 2: max(ρ(x1,x2),σ(y1,y2))=ρ(x1,x2), max(ρ(x1,x3),σ(y1,y3))=σ(y1,y3), max(ρ(x3,x2),σ(y3,y2)) =ρ(x3,x2)

Tinyboss said:
Suppose that [tex]\rho(x_1,x_2)\ge\sigma(y_1,y_2)[/tex]. What do you know about [tex]\rho(x_1,x_3)+\rho(x_3,x_2)[/tex]? Can you infer anything about the right-hand side of your inequality based on that?
We'll have ρ(x1,x3)+ρ(x3,x2) ≥ σ(y1,y2)).

But I think for case 2, we need to prove that ρ(x1,x2)≤σ(y1,y3)+ρ(x3,x2) instead?? How can we prove it?

Thanks!
 
  • #5
Still confused...please help...
 
  • #6
Okay, you need to show, for all points, that

[tex]\max(\rho(x_1,x_2),\sigma(y_1,y_2))\le\max(\rho(x_1,x_3),\sigma(y_1,y_3))+\max(\rho(x_3,x_2),\sigma(y_3,y_2))[/tex].

Suppose [tex]\rho(x_1,x_2)\ge\sigma(y_1,y_2)[/tex]. Since [tex]\rho[/tex] is a metric, you know that [tex]\rho(x_1,x_3)+\rho(x_3,x_2)\ge\rho(x_1,x_2)[/tex].

So what do you know about [tex]\max(\rho(x_1,x_3),\square)+\max(\rho(x_3,x_2),\square)[/tex], regardless of what's in the squares? You know it's at least as big as [tex]\rho(x_1,x_2)[/tex].
 
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