Show that $d:\mathbb{C}\times\mathbb{C}\rightarrow[0,\infty[$ is a metric on $\mathbb{C}$.

Question

Let $d:\mathbb{C}\times\mathbb{C}\rightarrow[0,\infty[$ be $$d(x,y)=\begin{cases} |x-y|&,\text{if }x,y\neq0\text{ and }x=ty\text{ for some }t\in\mathbb{R}\\ |x|+|y|&,\text{else.} \end{cases}$$ Show that $d$ is a metric on $\mathbb{C}$.

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So I need to show that (1) $d(x,y)=0\Leftrightarrow x=y$, (2) $d(x,y)=d(y,x)$ and (3) $d(x,z)\leq d(x,y)+d(y,z)$. Do the criteria have to satisfy both cases or is it sufficient to show either one? I suppose it's both. Anyways, here is my sorry attempt.

|x-y|:

(1.1): $|x-y|=0 \Leftrightarrow x-y=0 \Leftrightarrow x=y$

(2.1): $|x-y|=|-(x-y)|=|y-x|$

(3.1): $|x-z|=|x-y+y-z|\leq|x-y|+|y-z|$

|x|+|y|:

(1.2): $|x|+|y|=0\Leftrightarrow|x|=|y|\Leftrightarrow x=y$

(2.2): $|x|+|y| = |y| + |x|$

(3.2): $|x|+|z| \leq |x|+|y|+|y|+|z| (?)$

Answer 1

Do the criteria have to satisfy both cases or is it sufficient to show either one?

The cases hold for different values of $x$ and $y$. You need to show the criteria for all required values. For a given $x$, and $y$, you have to show it for the criterion those values satisfy. For instance, you start with trying to show $d(x,y) = 0 \iff x = y$. You do this as follows:

Let $(x, y) \in \Bbb C^2$.

• If $x = y$, then $x = 1\cdot y$, so $d(x,y) = |x - y| = |0| = 0$.
• Suppose $d(x,y) = 0$. If $x = ty$, then $d(x,y) = |x - y| = 0$, so $x - y = 0$ and $x = y$. If $ x \ne ty$, then $0 = |x| + |y|$, so $|x| = -|y|$, but $|x| \ge 0$, and $-|y| \le 0$. Hence the common value can only be $0$. Therefore $x = 0$ and $y = 0$, and so $x = y$. In all cases, $d(x,y) = 0 \implies x = y$.

The other two criteria should be handled similarly. For those it is not enough to just plug "$\iff$" between some statements. You have to actually show that the statements are true for the exact same values of $x$ and $y$.