Is there a "successor" metric in topology?

Question

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?

Answer 1

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.

Examples:

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 $A$.

Example:

4. The interior of $[a,b)$ in the topological space (1) is $(a,b)$.
5. 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 $A$.

Example:

6. The closure of the example (4) is $[a,b]$
7. 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.

Example:

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

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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 $\{a,b\}$.

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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 $B(a,r)=\{x\in X:\ d(a,x)<r\}$.

Examples:

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

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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.

Answer 2

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.