Showing that product of two metrics is not always a metric

Question

Let $d_1$ and $d_2$ be two metrics on space $X$. Is $d_1\cdot d_2$ metric on space $X$?

I know that is satisfies first three properties of metric. How do I show that triangle inequality holds or does not hold (an example) for this?

Answer 1

Consider the real line $\mathbb R$ with metrics $d_1(x,y)=|x-y|$ and $d_2(x,y)=\min\{1,d_1(x,y)\}$. The triangle inequality should hold for $d_1\cdot d_2$ for all points $x,y,z\in\mathbb R$ for it to be a metric.

Consider the points $x=1,y=2,z=2.5$ we have that -

$d_1(1,2.5)\cdot d_2(1,2.5)=|1-2.5|\cdot\min\{1,|1-2.5|\}=1.5\cdot 1=1.5$

$d_1(1,2)\cdot d_2(1,2)+d_1(2,2.5)\cdot d_2(2,2.5)=1\cdot1+0.5\cdot0.5=1.25<1.5$

Thus triangle inequality doesn't hold.

Answer 2

The product of two metrics may or may not be a metric. I'll give two examples for these.

When the product is a metric

Consider $d_1, d_2$ be the discrete metric on $X$, then $d_1 \cdot d_2$ is again a discrete metric and so the product is a metric.

When the product is not a metric

Consider $X= \mathbb{R}$ with $d_1 = d_2$ usual metric, $i.e.$ $d_1(x,y)= |x-y|$, and $d=d_1\cdot d_2$.

Then the triangle inequality is not satisfied as

• $d(0,1)= 1^2$,
• $d(0,1/2)= 1/2^2$,
• $d(1/2,1)=1/2^2$.

and so $d(0,1) \nleq d(0,1/2)+ d(1/2,1)$, hence $d$ is not a metric.

Answer 3

I went over a couple of exercises like this and it's usually enough to just draw a triangle in order to look for counter-examples. Instead of defining metrics with a formula, define them by just listing their value on the appropriate edge: