Obviousness – norm implies metric
We have a norm, which satisfies positivity, linearity and the triangle inequality. We have a metric, which satisfies positivity, symmetricity and the triangle inequality. We wish to prove that every norm is a metric. "Obviously", the only property we need to prove is... the triangle inequality.
At the time, I couldn't see how linearity implied symmetry was obvious. Now, I think it's about as obvious as the triangle inequality was - there's a step to take, even though it's simple.
Linearity requires that . Symmetricity requires that d(x,y)=d(y,x).
To prove it, we let d(x,y)= ||x-y|| = |-1|||x-y|| = ||-x-(-y)|| = ||y-x|| = d(y,x).
Christopher Midgley
12 Jan 2012 10:39
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Tags: Maths
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Nick
I agree entirely: finding that step is no easier than deducing the triangle inequality from the norm.
On this topic, I think the requirement that d(x,y)>=0 for all x & y in a metric space is actually unnecessary: we can prove it from the other axioms as follows:
2d(x,y)=d(x,y)+d(x,y)=d(x,y)+d(y,x) as d is symmetric
and d(x,y)+d(y,x)>=d(x,x) by the triangle inequality
and d(x,x)=0 so 2d(x,y)>=0 and hence d(x,y)>=0
12 Jan 2012, 12:19
Christopher Midgley
My first problem sheet goes slightly further: by modifying the triangle inequality requirement to be d(x,z)<=d(x,y)+d(z,y), we can remove the requirement for symmetry as well (it can be derived).
I’ll suppose it’s listed as required because it’s easier to check a metric satisfies symmetry than to check it satisfies the modified triangle inequality – if it fails, you can stop there.
14 Jan 2012, 11:30
Nick
Oh, I see: you can replace y with x to get d(x,z)<=d(z,x) (since d(x,x)=0)
and then by symmetry d(z,x)<=d(x,z)
so d(x,z)=d(z,x).
Thanks for pointing that out – it’s neat!
15 Jan 2012, 10:56
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