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Let $(E,d)$ be a metric space and assume that the metric d satisfies $d(x,z) \leq \max(d(x,y),d(y,z))$ for all $x,y,z\in E$.

Prove that if $d(x,y) \neq d(y,z)$ then $d(x,z)=\max(d(x,y),d(y,z))$.

I have spent literally hours on this and I feel like a moron because I can't figure it out. Please help! Thanks!

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You should enclose all mathematical expressions in \$'s, not just the symbols you can't write in plain text. It makes them much easier to read. Also, I don't think it is good practice to write quantifiers in symbolic form after the quantified statement. –  tomasz Sep 24 '12 at 15:27
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@Allison Cameron: Have a look at en.wikipedia.org/wiki/Ultrametric_space. Hope it helps. –  uforoboa Sep 24 '12 at 15:30
    
@uforoboa thank you! –  user39794 Sep 24 '12 at 15:33
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3 Answers

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Let $d(x,y)\leq d(y,z)\leq d(x,z)$ (WLOG)

Then, $d(x,z)\leq max(d(x,y),d(y,z))=d(y,z)\implies d(x,z)\leq d(y,z)$

Now, $d(y,z)\leq max(d(y,x),d(x,z))=d(x,z)\implies d(y,z)\leq d(x,z)$

Both of these $\implies d(x,z)=d(y,z)\implies d(x,z)= max(d(x,y),d(y,z))$

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Let $a=d(y,z)$, $b=d(x,z)$, $c=d(x,y)$.

Without loss of generality we may assume that $a\gt c$.

We have $a\le \max(b,c)$. Since $a\gt c$, we must have $a\le b$.

We have $b\le \max(a,c)$, so $b\le a$.

It follows that $b=a$.

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We already know "$\leq$", so let's assume "$<$" and hope to reach a contradiction.

Without loss of generality, let's assume $d(x, y) < d(y, z)$, so that our new assumption is that $d(x, z) < d(y, z)$.

But we have $d(y, z) \leq max(d(y, x), d(x, z)) = max(d(x, y), d(x,z))$, a contradiction to (one of) the above two inequalities.

Further reading: The inequality that your metric satisfies is called the "ultrametric inequality" or the "non-archimedean triangle inequality". A metric space with a metric that satisfies the ultrametric inequality is called an ultrametric space. You can read more about them here: http://en.wikipedia.org/wiki/Ultrametric_space

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