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Let us consider the function on the set $M$

$$d:M\times M \rightarrow \mathbb R.$$

I want to show that it defines a metric space on $M$ if the following two conditions are hold:

  1. $d(x,y) =0 \iff x=y$

  2. $d(x,y)\leq d(x,z)+d(y,z)$

So we need to prove that $d(x,y)\geq 0$ and $d(x,y)=d(y,x)$ but I don't know how to prove these two things.

Sahiba Arora
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Marc
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2 Answers2

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Let us start by proving that $d(x,y)=d(y,x)$. You know that $d(x,y)\leqslant d(x,x)+d(y,x)$, but, since $d(x,x)=0$, this means that $d(x,y)\leqslant d(y,x)$. By the same argument, $d(y,x)\leqslant d(x,y)$.

Now, suppose that, for some $x$ and some $y$ in $M$, $d(x,y)<0$. Then $$0=d(x,x)\leqslant d(x,y)+d(x,y)<0.$$

José Carlos Santos
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  • Thank you :) I actually thought that in the task there are some facts missing. – Marc Dec 04 '17 at 09:03
  • Is supposing that d(x,y)< 0 in the second part of the solution necessary? Could we not just say for arbitrary $x \neq y$ that $$0=d(x,x) \leq d(x,y)+d(x,y) \Rightarrow 0 \leq 2d(x,y) \Rightarrow 0 \leq d(x,y) $$ ? – Sofia Sep 05 '20 at 11:36
  • @Sofia Yes. That will work too. – José Carlos Santos Sep 05 '20 at 13:19
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Choose $z=x$ in 2) to learn that $d(x,y) \leq d(y,x)$ using 1). Then replacing the roles of $x$ and $y$ yields $d(y,x) \leq d(x,y)$. Altogether, $d(x,y)=d(y,x)$.

After that, choose $y=x$ in 2) to learn that $0 \leq d(x,z)+d(x,z)=2d(x,z)$. Hence $d(x,z)\geq 0$.

max_zorn
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