Find a metric $d$ on $\mathbb{N}$ Find a metric $d$ on $\mathbb{N}$ such that for any $n \in \mathbb{N}$ and any $\epsilon >0$ there exists an $m \in \mathbb{N},m\neq n$ such that $d(n,m)<\epsilon$.
From this definition, all the numbers are the limits of some sequence in the set $\mathbb{N}$. But how could this be? How could all the elements be arbitrarily close to each other but not equal at the same time? 
 A: Set up a bijection mapping the natural numbers to the positive rationals. Then pull back the metric from the positive rationals to the positive naturals
Produce an explicit bijection between rationals and naturals?
A: The rational numbers   are countable. Let $f$ be a bijection from the natural number to the rationals and define $d(m,n) = |f(m)-f(n)|$. Since the metric on the rationals has the property you wish, so does this translation to the natural numbers. This is not a very nice metric.
A: Hint: You may not necessarily be able to find a sequence such that the epsilon condition could be met. Epsilons are everywhere in Analysis, but it’s not always a sequence.
It’s helpful to think of the epsilon condition as a bound on the metric: the lower bound is $\epsilon >0$, for any $\epsilon$. Essentially make it as small as possible - $\inf \epsilon = 0$, so one essentially has to force the metric to align to this condition.
The metric has to take 2 natural numbers, and make the result 0, in some way. 
A: A different, more explicit approach:
Choose some irrational number $c$. For real $x$, let $f(x)$ be the distance from $x$ to the nearest integer; this ranges from $0$ at the integers to $\frac12$ at the midpoints between them.
Then $d(m,n)=f(cm-cn)$ is a metric satisfying these properties. By a pigeonhole argument, for any positive integer $N$, we can always find some $n<N$ with $f(cn)\le \frac1N$.
In essence, we sprinkle the integers around a circle, and pull back distances from there. Because $c$ is irrational, we never repeat a point, and the points are dense in the circle.
Many variations on this theme are possible. For one of the simplest to write down, there's $d(m,n)=|\sin(m-n)|$ (angles measured in radians).
