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?

  • $\begingroup$ Why $d_2 \Bigl((x,y),(x',y')\Bigr) =0 \Leftrightarrow (x,y)=(x',y')$ ? $\endgroup$ – yoyo Jun 30 '18 at 3:04
  • $\begingroup$ $\vert x \vert+\vert y \vert+\vert x' \vert+\vert y' \vert=0$ implies all $x,y,x',y'$ are zero , since $\vert . \vert \geq 0$ $\endgroup$ – user444830 Jun 30 '18 at 3:06
  • 2
    $\begingroup$ According to (b), the distance from $(1,1)$ to itself is $4$... $\endgroup$ – G Tony Jacobs Jun 30 '18 at 3:11
  • $\begingroup$ oops...! I only think on the only if part..But the other way it is false...! Thanks ! $\endgroup$ – user444830 Jun 30 '18 at 3:15
  • $\begingroup$ (c) indeed follows from $d$ being a metric, but this is not obvious. If you have not done so, I suggest that you work through demonstrating that $D$ satisfies all the qualifications of a metric (making sure of course to examine both directions of those "if and only if"s ;-). $\endgroup$ – Paul Sinclair Jun 30 '18 at 17:39

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.


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