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 $|a|\cdot||v||$). This implies $d(0,2)=2$, a contradiction.

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burak kılıç
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burak kılıç

Updated on December 05, 2022

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  • burak kılıç
    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
      mathcounterexamples.net about 6 years
      Same question than for your previous post... What have you tried? What are the axioms of a norm?
    • Henricus V.
      Henricus V. about 6 years
      This metric is not homogeneous.
    • burak kılıç
      burak kılıç about 6 years
      What would happen if you helped? if you can not help ,Do not answer unnecessarily.
    • Asinomás
      Asinomás about 6 years
      You 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
      burak kılıç about 6 years
      you right.I agree but first i have tried and I just started using this site
    • layman
      layman about 6 years
      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.
    • layman
      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 $c$, $||u|| = |c| \cdot||u||$. So, take for example, $c = 2$. Then if $x \neq y$, we have $2x \neq 2y$, right?
    • layman
      layman about 6 years
      (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.
    • layman
      layman about 6 years
      Sorry, in my second to last comment, I wrote $||u|| = |c| \cdot ||u||$, but I should have written $||cu|| = |c| \cdot ||u||$.
    • Watson
      Watson about 6 years