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I have been able to show that

  1. $d(x,y) \geq 0 $

  2. If $d(x,y) = 0$ , then $x=y$

  3. $d(x,y)=d(y,x)$

  4. For the fourth property,

Let $z \in X$, then $d(x,y) = |x-y|^3 =|x-z+z-y|^3$

(by triangle inequality)

$\leq (|x-z| + |z-y|)^3$ $= |x-z|^3 + |z-y|^3 + 3|x-z||z-y|(|x-z|+|z-y|)$

After this, since the given expression can't always be $\leq|x-z|^3 + |z-y|^3$, d is not a metric. To justify I have to take one example but which??

David C. Ullrich
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Binod Kumar
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  • A metric on what? The real numbers? In any case, a single counterexample is all you need to disprove a result. – lulu Sep 01 '22 at 13:13
  • Have you tried even one example? Seems to me that any $x,y,z$ with $x<z<y$ should suffice... – David C. Ullrich Sep 01 '22 at 13:15
  • See for example https://math.stackexchange.com/q/3820282/42969 – Martin R Sep 01 '22 at 13:16
  • See Here. – SoG Sep 01 '22 at 13:17
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    @CarlosAdir In one dimension that $p$-norm works out to $||v||_p=|v_1|\ne|v_1|^p$. – David C. Ullrich Sep 01 '22 at 13:18
  • You just need one example to prove it is not a metric. Is $d(0,2)\leq d(0,1)+d(1,2)?$ – Thomas Andrews Sep 01 '22 at 13:20
  • $f:\Bbb{R}\to \Bbb{R}$ induced a metric on $\Bbb{R}$ via isometry iff $f$ is injective.See Here. – SoG Sep 01 '22 at 13:23
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    Note, your argument only shows that you can't prove the triangle inequality this way. "since the given expression can't always be ..., $d$ is not a metric." That is wrong, because you haven't shown that the given expression can't always be...." – Thomas Andrews Sep 01 '22 at 13:24
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    Your proof can be made correct by assuming that $z$ is between $x$ and $y.$ Then $|x-y|=|x-z|+|z-y|,$ and you get $d(x,y)=d(x,z)+d(z,y)+k(x,y,z)$ for some non-negative $k(x,y,z),$ where $k$ is not zero if $x,y,z$ are distinct. That lets you find an example easily. – Thomas Andrews Sep 01 '22 at 13:28

1 Answers1

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$d(1, 2) = 1, d(2, 3) = 1$ but $d(1, 3) = 8$. Therefore, $d(1, 3) > d(1, 2) + d(2, 3)$

David Lui
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