The discrete metric on $\mathbb R $ is not induced by a norm
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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 $a\cdotv$). This implies $d(0,2)=2$, a contradiction.
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burak kılıç
Updated on December 05, 2022Comments

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.

mathcounterexamples.net about 6 yearsSame question than for your previous post... What have you tried? What are the axioms of a norm?

Henricus V. about 6 yearsThis metric is not homogeneous.

burak kılıç about 6 yearsWhat would happen if you helped? if you can not help ,Do not answer unnecessarily.

Asinomás about 6 yearsYou should post your effort on a problem, when you ask questions in the future. Also, please use mathjax and proper titles when prasing questions.

burak kılıç about 6 yearsyou right.I agree but first i have tried and I just started using this site

layman about 6 yearsFirst, 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.

layman about 6 years(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 $c$, $u = c \cdotu$. So, take for example, $c = 2$. Then if $x \neq y$, we have $2x \neq 2y$, right?

layman about 6 years(cont.) But then $d(x,y) = 1$ and $d(2x,2y) = 1$. But if $d(x,y) = xy$, then we have $xy = 1$ and $2x  2y = 1$. But this means $x  y = 2x  2y$. But we can pull the $2$ out on the right hand side, giving $x  y = 2x  y$. Dividing both sides by $xy$ (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.

layman about 6 yearsSorry, in my second to last comment, I wrote $u = c \cdot u$, but I should have written $cu = c \cdot u$.

Watson about 6 yearsDuplicate of math.stackexchange.com/questions/200023
