A metric on the natural numbers

Question

Does there exist a complete metric on the set of natural numbers such that $\{n,\,n+1,\,n+2,\,\cdots\}$ is a closed ball for each $n$?

Answer 1

Yes, there are many. For example, consider the metric $d(m, n)=\vert {1\over 2^m}-{1\over 2^n}\vert+1$ for $m\not=n$, and $d(m, m)=0$. Given any $n$, the set $\{n+k: k\in\mathbb{N}\}$ is the closed ball of radius $1+{1\over 2^{n+1}}$ centered at $n+1$. Meanwhile, it's trivially complete: since every distinct pair of points are at distance at least 1, there are no non-eventually-constant Cauchy sequences.

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Note that if $M=(X, d)$ is a metric space, then $M^+=(X, d^+)$ is also a metric space, where $d^+(a, b)=d(a, b)+1$ for $a\not=b$ and $d(a, a)=0$. The only nontrivial axiom to check is the Triangle Inequality: if $a, b, c$ are distinct, then $$d^+(a, b)+d^+(b, c)=[d(a, b)+d(b, c)+1]+1\ge [d(a, c)+1]+1>d^+(a, c).$$ (If $a=b$ or $b=c$, things are even easier.) This construction can be a useful source of counterexamples.

Note that this is just a special case of the sum of two metrics: if $d_0, d_1$ are metrics on $X$, then so is $d(x, y)=d_0(x, y)+d_1(x, y)$. Here we're taking the sum of a metric which satisfies the hypotheses of your question except for completeness, and a discrete metric (which trivializes the completeness requirement while preserving the other relevant property).

Answer 2

As pointed out there are such metrics. However if you tightens the requirements one could see that one could forbid metrics fulfilling the requirements.

As mentioned in the example $d(j,k) = \left|2^{-j}-2^{-k}\right|+1$ has the property that every Cauchy-sequence will be eventually constant. This is because every point is distant to the rest of the set.

If we one for example adds the requirement that every point should be a limit point of the rest of the set I don't believe it is possible:

We would construct a subsequence $a_k$ that is Cauchy, but not convergent, the aim is to create a subsequence so that the distance of subsequent elements are less than half of the previous (which will make it a Cauchy-sequence), but still so that subsequent elements can't get near lower integers.

We use the fact that we for every $a_k$ can find infinitely many integers arbitrarily near $a_k$ (which means that some of them are larger than $a_k$). That is we can select $a_k$ such that $\delta_k=d(a_k, a_{k-1})$ is as small as we would desire.

Now if for example $\delta_{k+1}<\delta_k/3$ we have due to triangle inequality that $d(a_l, a_{k-1}) < \sum_k^l\delta_j < \delta_k\sum_0^{l-k}3^{-j} < 3\delta_k/2$. So if we select $\delta_k$ in such a manner and also that $2\delta_k$ is less than $d(a_{k-1}, j)$ for all $j<a_{k-1}$ we would be guaranteed that $a_k$ can't converge to any $l<a_{k-1}$ and since that's eventually true for all integers we have a non-convergent Cauchy-sequence.