Is there a "next" or "successor" metric in topology?

Following on from this question (the answer to which turned out to be a little more complicated than I expected, and seems to have created a certain amount of disagreement!), asking what is the boundary of a sequence of integers in the space $\mathbb{N}$, I was naturally intrigued to find out if there's a metric which says that two points are "close" if they are next to each other, i.e. one's an immediate successor of the other or there's no other point in-between.

The motivation behind this is that even the metrics and topological spaces which make no reference to $\mathbb{R}$, $\mathbb{Q}$ or $\mathbb{Z}_p$ etc., still seem to have the in-built assumption that there's a void between any two integers, an assumption seemingly without basis in the circumstances, and this seems to make the topological definition of the boundaries of a sequence of integers counterintuitive.

If we had a metric which said that only immediate successor was "in the neighborhood", or alternatively the restrict the metric to integer values: $\lvert a-b\rvert\in\mathbb{N}_{>0}$ then surely first and last elements of such a sequence could be defined as the boundary of a closed set and the integers one greater and one less than those as the boundaries of an open set.

This would seem to be a natural way to describe the topology of a discrete sequence of integers, or am I missing something? Is there some clear advantage to the way it's done?

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    $\begingroup$ To restrict the metric to integer values, all you need to do is take $d(a,b)=|a-b|$ as usual, but let the set component of the metric space be the natural numbers, so that the metric space is $(\mathbb N,d)$. Then the only elements are integers, and there is no "in between", and so a neighborhood would consist of a sequence of integers. $\endgroup$ Aug 2, 2017 at 18:23
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    $\begingroup$ Can you formulate the question more precisely? Like: Given a set $S$ with [some structure], does there necessarily exist a metric $d$ having the property that for all $x,y\in S$, we have [some condition on $d(x,y)$] if and only if [some condition on $x,y$ in the originally provided structure]? $\endgroup$ Aug 2, 2017 at 18:27
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    $\begingroup$ You have a huge misunderstanding on the role of neighborhoods to define a topology and therefore to define boundary. Then you are trying to unify the vague understanding you have one that with ideas that you probably have seen when computing boundary of intervals in $\mathbb{R}$. $\endgroup$
    – Hellen
    Aug 2, 2017 at 18:27
  • $\begingroup$ @ChrisCulter define the metric space $(\mathbb{N},d)$ with the metric $d(a,b)=\lvert a-b\rvert$. In this topology what is the boundary of the closed set $\{3,4,5\}$, and what about the same set viewed as an open set? Is there a valid reason why this apparently more intuitive definition of the topological boundaries of a sequence of integers is not used? $\endgroup$ Aug 2, 2017 at 18:38
  • $\begingroup$ @Hellen what is my apparent misunderstanding? $\endgroup$ Aug 2, 2017 at 18:40

2 Answers 2


Succinct definition of boundary starting from open sets:

Let $X$ be a set and its topology is just a collection $T\subset P(X)$ of some of its subsets that are going to be called open. The collection is required to have some properties which you can check in Wikipedia's article for topology.

The set $X$ with the collection $T$ is called a topological space.


  1. $X=\mathbb{R}$ and $T$ consist of the sets that can be obtained by making arbitrary unions of intervals of the form $(a,b)$.
  2. $X=\mathbb{N}$ and $T$ consists of all possible subsets of $X$.
  3. $X=\{3,4,5\}$ and $T$ consists of all possible subsets of $X$.

Interior: Given a topological space $X$ with topology $T$ and given a subset $A\subset X$, the interior of $A$ is the union of all elements of $T$ that are subsets of $A$.


  1. The interior of $[a,b)$ in the topological space (1) is $(a,b)$.
  2. The interior of $\{3,4,5\}$ in the topological space (2) is $\{3,4,5\}$.

Closure: Given a topological space $X$ with topology $T$ and given a subset $A\subset X$, the closure of $A$ is the complement of the union of all elements of $T$ that are completely outside of $A$.


  1. The closure of the example (4) is $[a,b]$
  2. The closure of the example (5) is $\{3,4,5\}$

Boundary: Given a topological space $X$ with topology $T$ and given a subset $A\subset X$, the boundary of $A$ is the difference of its closure minus its interior.


  1. The boundary of example (4) is $\{a,b\}$
  2. The boundary of example (5) is empty

The case of boundary of intervals in $\mathbb{R}$

Given an finite interval that includes or not its extreme points ($A=(a,b)$, $[a,b)$, $(a,b]$, or $[a,b]$) it is always the case that its boundary is the set of its extreme points.

Observe that the interior is the interval excluding its extreme points $(a,b)$. This is an element of the topology (see example 1) and we cannot fit any more intervals without extreme points inside $A$.

The closure is always $[a,b]$. Observe that in the complement of $A$ we can fit the intervals $(b,+\infty)$ and $(-\infty,a)$. Any other is inside those two. The complement of $(-\infty,a)\cup(b,+\infty)$ is $[a,b]$, and that is the closure.

Finally the boundary is closure $[a,b]$ minus interior $(a,b)$. That is $\{a,b\}$.

Topology from metric

Given a set $X$ and a metric $d$ on it. One can define a topology $T$ to consist of arbitrary unions of balls $B(a,r)=\{x\in X:\ d(a,x)<r\}$.


  1. The usual topology in $\mathbb{R}$ is of this form. Observe that the intervals $(a,b)$ are just the balls $B(\frac{a+b}{2},\frac{b-a}{2})$.

Impossibility of a topology on $\mathbb{N}$ such that the boundary of finite intervals consists of its extreme points

Assume we want to put a topology on $\mathbb{N}$ such that for every finite interval (finitely many consecutive integers) the boundary is its extreme points.

If the boundary of $A=\{a,a+1,a+2,...,a+n\}$ ought to be $\{a,a+n\}$. Then its interior would have to be $\{a+1,a+2,...,a+n-1\}$. Since $a$ and $n$ were arbitrary, that means (taking $n=1$) that all sets of the form $\{a\}$ are open (elements of the topology). By a property required for topologies, which you are supposed to read, this means that all subsets are open.

But if all subsets are open, then $A$ is open. Therefore, the interior of $A$ would have to be $A$ and its complement would have to be $A$ as well. This causes that the boundary is empty.

Thereofore, there is no such topology as we wanted. In particular, there is no such topology defined by a metric.


Topologically, in the usual metric, every subset of $\mathbb N$ has an empty boundary because every subset is open. This is not really a useful concept in $\mathbb N$.

To say the $\{3,4,5\}$ has boundary $\{3,5\}$ could perhaps be interpreted order-theoretically. The set is an interval, and what you call the boundary is its set of endpoints. Every subset of $\mathbb N$ (or any totally ordered set) is a union of intervals, though possibly not in an interesting way (the intervals could all be singletons). A boundary in your sense could be the set of endpoints of maximal intervals in the subset.

  • $\begingroup$ Thanks. The metric I had in mind would give the more than 2 numbers for a discontinuous sequence of integers. $\endgroup$ Aug 3, 2017 at 3:14

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