Which of the following define a metric on $\Bbb{R}^2$:
a) $d_1 \Bigl((x,y),(x',y')\Bigr)=\min\{\vert x-x' \vert, \vert y-y' \vert \}$
b) $d_2 \Bigl((x,y),(x',y')\Bigr)=\vert x \vert+\vert y \vert+\vert x' \vert+\vert y' \vert$
c)$D \Bigl((x,y),(x',y')\Bigr)=d(x,x')+d(y,y')$ where $d$ is a metric in $\Bbb{R}$.
My try:
a) $d_1$ is not a metric, since $d_1 \Bigl((1,2),(1,1)\Bigr)=0$ but $(1,2)\neq(1,1)$
c) $D$ is a metric follows from the fact that $d$ is a metric on $\Bbb{R}$.
What about b?
I think $d_2$ is indeed a metric, for, it is clear that
$$d_2 \Bigl((x,y),(x',y')\Bigr) =0 \Leftrightarrow (x,y)=(x',y')$$
$$d_2 \Bigl((x,y),(x',y')\Bigr) =d_2 \Bigl((x',y'),(x,y)\Bigr)$$
Also for the triangle inequality,
$$d_2 \Bigl((x,y),(x',y')\Bigr) =\vert x \vert+\vert y \vert+\vert x' \vert+\vert y' \vert$$ $$\leq \Bigl(\vert x \vert+\vert y \vert+\vert s \vert+\vert t \vert \Bigr) + \Bigl(\vert s \vert+\vert t \vert + \vert x' \vert+\vert y' \vert \Bigr) =d_2 \Bigl((x,y),(s,t)\Bigr) + d_2 \Bigl((s,t),(x',y')\Bigr)$$
But the answer given in a booklet is only c). What I'm doing wrong in b)? Any ideas?
b) fails as a non-zero vector has non-zero distance to itself, e.g. $d_2((1,1),(1,1)) = 4 > 0$.
c) can be shown, and is in fact one of the standard choices for a metric compatible with the product topology. You can reduce each property we need for $D$ to the corresponding one for $d$. Try it.