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Is this the correct way to prove that this is metric function ?

let $d_1:\Bbb Z \times \Bbb Z \rightarrow \Bbb R^+_0$

$d(m,n):=|m-n|^3$

i. $d(x,y)\geq 0$ with equality iff x=y

$|m-n|\geq0$ because it is an absolute value

if $m \neq n$ $|m-n|>0$ if m=n $|m-n|=0$

ii.$ d(m,n)=|m-n|^3=|n-m|^3=d(n,m)$

iii.

$d(m,p)=|m-p|^3$

$d(m,n)=|m-n|^3$

$d(n,p)=|n-p|^3$

$|m-p|^3\leq |m|^3+|p|^3$

$|m-n|^3\leq |m|^3+|n|^3$

$|n-p|^3\leq |n|^3+|p|^3$

so

$|m-n|^3+|n-p|^3 =|m-n+n-p|^3=d(m,p)$

and so $d(m,p)\leq d(m,n)+d(n,p)$

excalibirr
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    Take $m=-1$ and $p=1$. Then $|m-p|^3 = 2^3 > 1^3 + 1^3 = |m|^3 + |p|^3$. – Daniel Xiang Aug 06 '18 at 13:04
  • @DanielXiang can we just say that we know $|m-p| \leq |m-n|+|n-p|$ because of the triangle inequality on $\Bbb R$ and then use this fact to say their cubes must also follow this ? – excalibirr Aug 06 '18 at 13:13
  • No, because $(|m-n|+|n-p|)^3$ is not $(|m-n|^3+|n-p|^3)$. – Mark Aug 06 '18 at 13:15
  • No. Suppose $m-n = -1$ and $n-p = 1$. Then $|m-p|^3 > |m-n|^3 + |n-p|^3$. – Daniel Xiang Aug 06 '18 at 13:15
  • @exodius btw, why did you delete the other question on metric spaces that you posed ten minutes ago? The question was alright in my opinion. –  Aug 06 '18 at 13:43
  • @Brahadeesh Because it was being down voted and you had already answered my question in the comment so I didn't see the point of keeping it up. – excalibirr Aug 06 '18 at 13:46
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    @exodius in general, it is bad form to delete a question once it has been answered. It sends the feeling that the OP does not care for the time and effort that the community puts in to answer the question. In this case, the answer was simple enough that a comment was satisfactory for you, but try not to delete your questions once you have got an answer just for the sake of removing downvotes. A few downvotes will not matter in the long run if you continue to ask good questions (formatted properly, and showing your work, etc.). All the best. –  Aug 06 '18 at 13:49

1 Answers1

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The last part goes completely wrong as Daniels comment suggests.
You need to show $$ |a-b|^3 \leq |a-c|^3 +|c-b|^3 $$ which is not true. Note that if $x$ close to $0$, then $x^3$ is much and much closer to $0$. An easy counter example would thus be to take $a=0$, $b=2$ and $c=1$ to get $$ 2^3 > 1^3 + 1^3 $$ which is thus exactly the other way around.

Stan Tendijck
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