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Protter-Morrey Problem 6.1.17

Let $S$ be a set and $d$ a function from $S \times S$ into $\mathbb{R}^1$ with the properties:
(i) $d(x,y) = 0$ if and only if $x = y$
(ii) $d(x,z) \le d(x,y) + d(z,y)$ for all $x$, $y$, $z$ $\in S$.
Show that $d$ is a metric and hence that $(S,d)$ is a metric space.

The only thing I have not figured out how to derive is that $d(x,y) = d(y,x)$ for all $x$, $y$ $\in S$.
Could someone give me some hint or guidance?

taiwanjizhan
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  • The only trick I know is to try setting $z$ equal to $x$ and/or $y$ in (ii) and see what you can do with that. – JonathanZ Nov 07 '20 at 21:58

1 Answers1

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Just pay attention to the order of points: use the triangle inequality twice to get $$d(x,y) \leq d(x,z)+d(y,z)\leq d(x,z) + d(y,x)+d(z,x)$$for all $x,y,z\in S$. Choose $z = x$ to get $d(x,y)\leq d(y,x)$ for all $x,y\in S$. Since $x$ and $y$ were arbitrary, switching their roles gives $d(y,x)\leq d(x,y)$ as well. Thus, we have symmetry.

Lastly, for positivity, we use the triangle inequality again: $$0 = d(x,x) \leq d(x,y)+d(x,y) = 2d(x,y)\implies d(x,y)\geq 0.$$

Ivo Terek
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    Slightly simpler: Property (ii) says that $$d(\color{red}x,\color{blue}z) \le d(\color{red}x,\color{green}y) + d(\color{blue}z,\color{green}y).$$ Whence $$d(\color{red}a,\color{blue}b) \le d(\color{red}a,\color{green}a) + d(\color{blue}b,\color{green}a)$$ and $$d(\color{red}b,\color{blue}a) \le d(\color{red}b,\color{green}b) + d(\color{blue}a,\color{green}b).$$ For positivity, $$d(\color{red}x,\color{blue}x) \le d(\color{red}x,\color{green}y) + d(\color{blue}x,\color{green}y).$$ – Steve Kass Nov 07 '20 at 22:26