The production of two metrics is a metric also. It's googled easy. But what's about a sum? As I can see sum is metric, as the triangle inequality of metric sum is the consequence of the inequality feature and two other axioms looks obvious. Is it correct?
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2yes, it is. <
> – mm-aops Jun 06 '14 at 21:07 -
They would need to have the same objects for one. – Christopher King Jun 06 '14 at 21:08
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Why don't you try proving it? – nomen Jun 06 '14 at 21:12
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2Indeed, google is not a theorem prover. – Ruslan Jun 06 '14 at 21:12
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1@nomen I try and described my view. – SerG Jun 06 '14 at 21:17
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1You have in essence given a proof. – André Nicolas Jun 06 '14 at 21:20
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3The product of metrics need not be a metric in general. The sum always is. – Henno Brandsma Sep 25 '18 at 04:49
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In short, yes. Your sketched argument is correct for the triangle inequality (we're just adding two valid inequalities to get a third). One does also need to remark that the sum of two numbers $\ge 0$ can only be $0$ if both summands are $0$ (and then we apply the axiom for the two constituent metrics after that).
Henno Brandsma
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For sum of metrics it's not hard exercise I think:
- Let $\rho(x,y) = \rho_1(x,y) + \rho_2(x,y)$ and $\rho_1, \rho_2$ are metrics.
- x=y: $\rho(x,y)=\rho_1(x,y)+ \rho_2(x,y)=0+0=0$
- $\rho(x,y)=\rho_1(x,y)+ \rho_2(x,y)=\rho_1(y,x)+\rho_2(y,x)=\rho(y,x)$
- We know that $\rho_1(x,y)\leq\rho_1(x,z)+ \rho_1(z,y)$ and $\rho_2(x,y)\leq\rho_2(x,z)+ \rho_2(z,y)$ It means that $\rho_1(x,y)+ \rho_2(x,y) \leq\rho_1(x,z)+\rho_1(z,y)+\rho_2(x,z)+\rho_2(z,y)= \rho_1(x,z)+\rho_2(x,z)+\rho_1(z,y)+\rho_2(z,y) = (\rho_1(x,z)+\rho_2(x,z))+(\rho_1(z,y)+\rho_2(z,y))=\rho(x,z)+\rho(z,y)$
Taraas
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