### 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).

## 3 comments by 1 or more people

## 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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