# 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?

• It doesn't say any two, only that if you start with one (call it $a$) you can find some $b$ with $d(a,b)< \epsilon.$ – coffeemath Dec 7 '18 at 1:04
• It's not the same $m$ for every $\epsilon$ – fleablood Dec 7 '18 at 1:05
• "all the numbers are the limits of some sequence in the set N." That's fine. Why would that be a problem? "How could all the elements be arbitrarily close to each other" They aren't. All numbers have some numbers arbitrarily close but not all of them. – fleablood Dec 7 '18 at 1:09
• Am I missing something or can't we just pick $m=n$, as $d(m,n)=0$ – Math_QED Dec 7 '18 at 7:51
• @Math_QED Yes but that seems to be the trivial answer. The question is asking for something more intelligent I suppose. I edited it to be clearer – Bunbury Dec 7 '18 at 18:31

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?

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.

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 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 $$n with $$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).