0
$\begingroup$

Given $X$ a set, and $\mid\mid\cdot\mid\mid:X\rightarrow \mathbb{R}$ a function, we define $\mid\mid\cdot\mid\mid$ to be a norm on $X$ following the normed space axioms.

My textbook states that if we let $d(x,y)=\mid\mid x-y \mid\mid$, we can show that this distance is indeed a metric space following the usual metric space axioms.

This is fairly straightforward, but got me thinking - surely I could use $d(x,y)=\mid\mid x+y \mid\mid$ and still be able to prove it is a metric using the usual metric space axioms?

Is there a reason why the textbook suggests to use $d(x,y)=\mid\mid x-y \mid\mid$ and would both be valid metric spaces?

$\endgroup$
  • 2
    $\begingroup$ With the sum you don't get the property that $d(x,y)=0$ iff $x=y$. $\endgroup$ – orole Jan 16 '18 at 12:59
3
$\begingroup$

The mapping $$d: (x,y)\mapsto \|x+y\|$$

is not a metric, because it does not satisfy the axiom

$$\forall x\in X:d(x,x)=0$$

(except if $X\neq \{0\}$, but then you only have one metrix on $X$ anyway).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see. How would I go about proving that this axiom is satisfied for $\mid\mid x-y \mid\mid$ as I must have done this wrong then $\endgroup$ – Thomas Smith Jan 16 '18 at 13:03
  • 1
    $\begingroup$ @BenHarvey Norms have the property that $\|x\|=0$ if and only if $x=0$. Therefore, $\|x-y\|=0$ if and only if $x-y=0$, and therefore $x=y$. You cannot do the same with $\|x+y\|=0$, since you can, from $\|x+y\|=0$, only conclude that $x=-y$. $\endgroup$ – 5xum Jan 16 '18 at 13:05
1
$\begingroup$

Other answers are pointing to the formal reasons why $d(x,y)=\|x+y\|$ is not a metric. However, I would also suggest that you build your geometrical intuition here.

In geometry, when you have points $A$ and $B$ and the origin $O$ in, say, a plane, you would calculate the vector $\vec{AB}=\vec{OB}-\vec{OA}$ and then you would use the ordinary vector norm to calculate the distance of $A$ to $B$:

$$d(A, B)=\|\vec{AB}\|=\|\vec{OB}-\vec{OA}\|=\|\vec{B}-\vec{A}\|$$

(if we agree to not write $O$ and just identify any point $X$ with the vector $\vec{X}=\vec{OX}$).

That is where the minus sign comes from. It would not make much sense to use the plus sign ($\|\vec{OB}+\vec{OA}\|$). You can calculate with it, but you certainly don't have a clear geometrical intuition behind it.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, this was really useful. $\endgroup$ – Thomas Smith Jan 16 '18 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.