# The discrete metric on $\mathbb R$ is not induced by a norm

1,399

Suppose it is possible, the norm of $$1$$ would have to be $$1$$ because $$d(0,1)=1$$ but then the norm of $$2$$ would have to be $$2$$ (since $$||av||$$ is equal to $$|a|\cdot||v||$$). This implies $$d(0,2)=2$$, a contradiction.

Share:
1,399

Author by

### burak kılıç

Updated on December 05, 2022

• burak kılıç 22 days

Show that discrete metric on $\mathbb R$ i.e

$d(x,y)=0$ if $x=y$ and $d(x,y)=1$ if $x\neq y$ cannot be derived from a norm.

Same question than for your previous post... What have you tried? What are the axioms of a norm?
• Henricus V. about 6 years
This metric is not homogeneous.
• burak kılıç about 6 years
What would happen if you helped? if you can not help ,Do not answer unnecessarily.
First, you have to understand that if you have a norm $|| \cdot ||$, then how do you get a metric/distance function from it? Well, you subtract two elements! So the distance between elements $a$ and $b$ under the metric induced by the norm $|| \cdot ||$ is defined as $||a - b||$. You should verify that the function $d(a,b) := ||a - b||$ is actually a metric. Now that you understand how a norm induces/gives rise to a metric, you have to show that the metric $d(x,y) := \begin{cases} 1 & x \neq y \\ 0 & x = y \end{cases}$ is not induced by a norm.
(cont.) Suppose that it were induced by a norm $|| \cdot ||$. Then we know $d(x,y) = ||x - y||$. But we also have $d(x,y) =1$ if $x \neq y$, and $0$ if $x = y$. Well, if $x = y$, $||x - y|| = 0$ by properties of a norm, so there's no problem there. But what if $x \neq y$? Then does $||x - y|| = 1$ always? No because we know that a norm satisfies for any real number $c$, $||u|| = |c| \cdot||u||$. So, take for example, $c = 2$. Then if $x \neq y$, we have $2x \neq 2y$, right?
(cont.) But then $d(x,y) = 1$ and $d(2x,2y) = 1$. But if $d(x,y) = ||x-y||$, then we have $||x-y|| = 1$ and $||2x - 2y|| = 1$. But this means $||x - y|| = ||2x - 2y||$. But we can pull the $2$ out on the right hand side, giving $||x - y|| = 2||x - y||$. Dividing both sides by $||x-y||$ (which is nonzero since $x \neq y$ by assumption) gives $1 = 2$, which is a contradiction! Thus, the metric $d$ is not induced by a norm.
Sorry, in my second to last comment, I wrote $||u|| = |c| \cdot ||u||$, but I should have written $||cu|| = |c| \cdot ||u||$.