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I am supposed to prove that: $d(x,y)=\left\{\begin{matrix} 0 & x=y \\ \frac{1}{x+y} & x\neq y \end{matrix}\right.$

is not a metric space, $\forall x,y \in \mathbb{N}$

The first two properties for the metric space holds, also the triangle inequality, if the two elements are equal. But how to prove that it if:

$x\neq y \neq z $, then $d(x,z)\nleqslant d(x,y)+d(y,z)$?

Thank for any help.

Shelley
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    Fortunately, you don't have to prove that $d(x,z)\not\le d(x,y)+d(y,z)$. All you have to do is find one instance where the triangle inequality fails.Maybe you could plug in some numbers at random and try to get lucky? Have you tried that? – bof Mar 22 '20 at 11:08
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    You can't prove that because it's not true in general. You need to find a single counter-example: $x,y,z$ such that $d(x,z)\nleqslant d(x,y)+d(y,z)$. – TonyK Mar 22 '20 at 11:08
  • @bof oh, okay, I thought that I should prove it in general, not just by finding counter exammple. In that way, it is easy :), thank you – Shelley Mar 22 '20 at 11:11
  • @MartinR: That doesn't work, because $x=z$. But $(x,y,z)=(1,n,2)$ works for any $n\ge 5$. – TonyK Mar 22 '20 at 11:12
  • @TonyK Why is it a problem that $x=z$ ? – Maximilian Janisch Mar 22 '20 at 11:12
  • @TonyK thanks, I thought that I have to prove it in general not just by finding counter example, thanks – Shelley Mar 22 '20 at 11:13
  • @TonyK: You are right. That is what I meant to say. – Martin R Mar 22 '20 at 11:13
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    @MaximilianJanisch: because if $x=z$, then $d(x,z)=0$. – TonyK Mar 22 '20 at 11:14

3 Answers3

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Observe that $$d(1,2)=\frac{1}{3}$$, $$d(1,5)=\frac{1}{6}$$, $$d(5,2)=\frac{1}{7}$$

$$d(1,2)=\frac{1}{3}>\frac{1}{6}+\frac{1}{7}=d(1,5)+d(5,2)$$.

So the triangle inequality is not satisfied and $d$ is not a metric.

Akash Yadav
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It is not true that $x\neq y\neq z\implies d(x,z)\nleqslant d(x,y)+d(y,z)$ for all $x,y,z\in\mathbb N$, as $x=1,y=2,z=3$ shows. However there are some $x,y,z$ such that the triangle inequality doesn't hold.

For example $$d(1,10)+d(10,5)=\frac{52}{330}<\frac{55}{330}=\frac16=d(1,5).$$

In fact, for distinct $x,y,z$, we have $$d(x,y)+d(y,z)\le d(x,z)$$ if and only if $y^2\geq x^2+x y+ x z+y z+z^2$. So by choosing $y$ large and keeping $x,z$ relatively small, you get counter-examples.

Maximilian Janisch
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In order to prove that some statement doesn't hold it suffices to find a counterexample. In this case for instance let's consider $x,z$ small, i.e. $x=1$, $z=3$, and $y$ some bigger natural number, such as $y=9$:

$$ d(x,z) = \frac{1}{x+z} = \frac{1}{4}, $$

$$ d(x,y)+d(y,z) = \frac{1}{x+y}+\frac{1}{y+z} = \frac{1}{10}+\frac{1}{12} = \frac{11}{60} < \frac{15}{60} = \frac{1}{4} = d(x,z). $$

Víctor
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