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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.

Asinomás
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burak kılıç
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  • Same question than for your previous post... What have you tried? What are the axioms of a norm? – mathcounterexamples.net Dec 19 '16 at 17:02
  • This metric is not homogeneous. – Henricus V. Dec 19 '16 at 17:03
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    What would happen if you helped? if you can not help ,Do not answer unnecessarily. – burak kılıç Dec 19 '16 at 17:05
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    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. – Asinomás Dec 19 '16 at 17:06
  • you right.I agree but first i have tried and I just started using this site – burak kılıç Dec 19 '16 at 17:09
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    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 Dec 19 '16 at 17:09
  • (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 Dec 19 '16 at 17:10
  • (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 Dec 19 '16 at 17:10
  • Sorry, in my second to last comment, I wrote $||u|| = |c| \cdot ||u||$, but I should have written $||cu|| = |c| \cdot ||u||$. – layman Dec 19 '16 at 17:16
  • Duplicate of https://math.stackexchange.com/questions/200023 – Watson Dec 19 '16 at 17:22

2 Answers2

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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.

Asinomás
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Suppose $d$ is induced by a norm, which we will denote $||\cdot||$

Then $||x|| = d(x,0)$.

Note that: $||2\times 1|| = d(2,0) = 1\neq 2 = 2\times ||1||$

fosho
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