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I need to find an example of a metric space , in which

${lim}_{n→∞} (1/n)$ it's different from $0$.

I took the set of real numbers with the discrete metric space $(R,d)$ in which this limit does not exist , but I'm not sure if my problem is solved or i need to find a metric space where this limit exist but its different from $0$?

math
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  • Take $\mathbb R$ with the usual metric, and rename $0$ to "banana". So now $\lim 1/n$ is "banana" and clearly banana $\ne 0$. – Ittay Weiss Dec 02 '15 at 09:19
  • You could argue that you solved the question already. Indeed, if you consider $(\mathbb{R},d_{discrete})$, this limit does not exist, so it's certainly not zero.

    I'm wondering whether there is a metric on $\mathbb{R}$ such that this limit exists and is not zero.

     – Mathematician 42 Dec 02 '15 at 09:25

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take the real numbers but define $d(x,y) = |x-y|$ if $x$ and $y$ are not 0 or 1.

define $d(1,x) = |x|$ for all $x \neq 0,1$ and define $ d(0,y) = |y-1|$ for all $y\neq 0,1.$

In other words, swap 0 and 1. Define $d(0,1)=1.$

Mark Joshi
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  • This is a useful type of example for many similar questions. As a note of interest, you can also think of this as the metric induced by a function that sends 0 to 1, 1 to 0, and x to itself otherwise. Indeed one can always define a metric induced by any injection $f$ by setting $d(x,y)=|f(x)=f(y)|$. – Luke Hamblin Dec 02 '15 at 10:08
  • I've written the same myself, but you raced me. :) +1 – CiaPan Dec 02 '15 at 10:09
  • Doesn't a metric space necessitate that $d(x,y)=0 \iff x=y$? In this case, $d(0,1) = 0$ but $1\neq 0$. – mathochist Dec 02 '15 at 10:25
  • I've made it more precise – Mark Joshi Dec 02 '15 at 10:35