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What is the natural topology (or topologies) on the integers. Can we define a metric on the integers?

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    $\begingroup$ Sure you can. If you use the euclidean metric all sets are open (and closed), all points are isolated and no points are cluster points. Other metrics and topologies are possible but I don't know if any of them are significantly different. $\endgroup$
    – fleablood
    Sep 4, 2016 at 20:46
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    $\begingroup$ @fleablood: interesting and very different examples of topologies on $\Bbb{Z}$ are given by the $p$-adic topologies. $\endgroup$
    – Rob Arthan
    Sep 4, 2016 at 20:52
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    $\begingroup$ Not sure if this fits, but do check out the famous topological proof of the infinitude of primes. It's only about open and closed sets. No metric. $\endgroup$ Sep 4, 2016 at 20:53
  • $\begingroup$ @JyrkiLahtonen Although it's neat, Furstenberg's proof isn't really topological; see idmercer.com/monthly355-356-mercer.pdf. $\endgroup$ Sep 5, 2016 at 15:48
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    $\begingroup$ @Thomas To those who closed this question: can you point out what is not clear about this question? The posters below seemed to have a pretty good understanding of this and I was pretty specific in the body of the question. This seems excessive, and there is no feedback besides the vague request above, which, ironically, requires some context itself to be useful. $\endgroup$
    – user237392
    Sep 6, 2016 at 10:17

6 Answers 6

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"The integers" refers to not just a set, but a whole suite of other notions, such as the usual addition and multiplication operations.

Among these notions, there is, in fact, a topology. The usual topology on the integers is the discrete topology — the one where every subset is an open set.

One thing that needs to be internalized is that, when taken at face value, the question "Is $S$ an open set?" is utter nonsense. The meaningful questions are of the form "Is $S$ an open set in the topology $T$?"; it's just that we usually don't mention $T$ when it can be understood from context.

Anyways, one way to explain why the usual topology on the integers is that discrete topology is because that is the subspace topology relative the usual topology on the real numbers. That is (presuming $\mathbb{Z} \subseteq \mathbb{R}$),

$U \subseteq \mathbb{Z}$ is an open set${}^1$ if and only if there is an open set${}^2$ $U' \subseteq \mathbb{R}$ such that $U = U' \cap \mathbb{Z}$.

1: in the usual topology of the integers

2: in the usual topology of the real numbers

Since the usual topology on $\mathbb{R}$ can be desribed in terms of a metric, e.g. $d(x,y) = |x-y|$, the usual topology on $\mathbb{Z}$ is given by the same metric. But note that for any point $P$, the open ball of radius $1/2$ around $P$ is just the set $\{ P \}$, and consequently $\{P \}$ is an open set.

As the other answers and comments indicate, there are other topologies that are useful to put on the natural numbers.

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  • $\begingroup$ Thanks for showing me th subspace approach! Makes a lot more sense now. Does the discrete topology require any particular metric or does this space support several? $\endgroup$
    – user237392
    Sep 5, 2016 at 0:24
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    $\begingroup$ @Bey: You can play all of the usual tricks to take one metric and produce other ones that define the same topology. Additionally, one popular metric for defining discrete topologies is the metric satisfying $d(x,y) = 1$ whenever $x \neq y$. $\endgroup$
    – user14972
    Sep 5, 2016 at 0:29
  • $\begingroup$ Ok I understand now. I was thrown off because it seems that the concept of cluster point was useless in this topology but then I read that a discrete topology is defined by the absence of cluster points (among other things). $\endgroup$
    – user237392
    Sep 5, 2016 at 0:32
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    $\begingroup$ @celtschk: The point of that particular example was not to "get" the usual topology, but to show that that choice of topology fits in with other usual choices. $\endgroup$
    – user14972
    Sep 5, 2016 at 14:11
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    $\begingroup$ @celtschk starting from $\mathbb{R}$ was helpful because I have some intuition about that topology. I'm just getting into more abstract spaces where I have to question my intuition--which makes me start to doubt how much I can rely on familiar geometric notions. I see that for this case I can and its not that bad. $\endgroup$
    – user237392
    Sep 5, 2016 at 16:15
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It is clear that if you look at the topology on $\Bbb Z$ induced by the natural (Euclidean) metric on the real line you get the discrete topology: every subset is open. This is not so exciting.

But that's not forced: you can define other topologies that turn out to be more interesting and useful.

An important metric that one can define in $\Bbb Z$ (actually on $\Bbb Q$)is the $p$-adic metric which requires fixing a prime number $p$ beforehand and whose basic idea is that numbers that are highly divisible by $p$ are "small", i.e. close to $0$. There's a vast literature on the $p$-adic numbers, which are what you get completing $\Bbb Q$ under the $p$-adic metric, pretty much in the same way one can construct $\Bbb R$ as the completion of $\Bbb Q$ under the Euclidean metric.


As another example you can consider the topology in $\Bbb Z$ whose closed set are those generated by the arithmetic progressions containing $0$.

Then you can use this topology as follows: consider the union of all non-trivial closed sets. It is easy to check that this union is $$ \bigcup_{p\ \text{prime}}\{...,-2p-p,0,p,2p,...\}= \Bbb Z\setminus\{1,-1\}. $$ Since it is the complementary set of a finite subset it cannot be closed (all basic open sets are infinite for this topology). Thus the union cannot be a finite union. Hence we proved that $$ \text{there are infinitely many primes.} $$

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    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Santiago
    Sep 4, 2016 at 22:05
  • $\begingroup$ @Santiago: I had no idea the proof made it to wiki. Still a nice example though. $\endgroup$
    – AdLibitum
    Sep 4, 2016 at 23:17
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    $\begingroup$ "closed set are the arithmetic sequences containing 0": but $2\mathbb Z\cup 3\mathbb Z$ is not such a set. The version on wikipedia has a more complicated definition of the topology (and more work to show open sets are infinite) but it seems necessary. $\endgroup$
    – stewbasic
    Sep 5, 2016 at 0:06
  • $\begingroup$ you meant arithmetic progressions $\endgroup$
    – reuns
    Sep 5, 2016 at 0:07
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    $\begingroup$ Your link for the p-adic numbers goes to an Italian wiki page for a building in Baku. Copy and paste error? $\endgroup$
    – Hong Ooi
    Sep 5, 2016 at 3:07
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I think you are relying on the geometric interpretation of an open set in the case of $\mathbb{R}^n$ too much. You should think about the fact that open sets in $\mathbb{R}^n$ are derived completely from the metric on the space. Remember that a topological space $(X,\tau)$ is just a set $X$ paired with sub-collections of $X$ that obey a few axioms. If you wish to put a topology on $\mathbb{Z}$, consider the most trivial one, namely the discrete topology i.e $\{i\} \in \tau$ i.e open for all $i \in \mathbb{Z}$.

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    $\begingroup$ You are correct. I am reading "Advanced Calculus" by buck and he talks about topology concepts like open balls and accumulation points, both of which seemed unnecessary or inapplicable to a lattice of integers. $\endgroup$
    – user237392
    Sep 5, 2016 at 0:23
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    $\begingroup$ This is a very common confusion, especially for me it was. When first learning about the real line, someone just says $(a,b)$ is open and $[a,b]$ is closed and we never thought to ask why. The same can be said for someone learning about arithmetic. It isn't until (well for me) you take abstract algebra that you learn out equivalence classes, rings, fields, zero-divisors etc. And then you say; "ahh ha, it was just the consequence of a definition." $\endgroup$ Sep 5, 2016 at 0:29
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Yes, certainly there are topologies on the integers $\mathbb{Z}$ or nonnegative integers $\mathbb{N}$. For example:

The order topology

$\mathbb{Z}$ has a $<$ order, so we can define the order topology for it. In some sense the "standard" topology on $\mathbb{Z}$. For example, the same use of order will get you the topology on $\mathbb{Z}$ or $\mathbb{R}$.

This is also what you get if you consider $\mathbb{Z}$ as a metric space in the canonical way: the distance from $a$ to $b$ is $|a-b|$.

Unfortunately, this is just the discrete topology -- every subset of $\mathbb{Z}$ is open. In this sense, $\mathbb{Z}$ is not very interesting.

The co-finite topology

Another example is to the cofinite topology, which can be defined on any set but of which the natural numbers are a common example. We define it so the the closed sets are all finite sets, as well as $\mathbb{Z}$ (so the open sets are $\varnothing$ and "cofinite" sets, or complements of finite sets.) In this case: we get a compact and $T_1$, but not Hausdorff space.

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  • $\begingroup$ Does the discrete topology on the integers require the discrete metric or will the usual Euclidean metric suffice? Actually, in higher dimensions could we also use the taxicab metric? $\endgroup$
    – user237392
    Sep 5, 2016 at 0:21
  • $\begingroup$ @Bey Either the discrete metric or the usual Euclidean metric both give the discrete topology on $\mathbb{Z}$. It is the same in higher dimensions. And yes, taxicab will give the same thing as well. $\endgroup$
    – 6005
    Sep 5, 2016 at 0:24
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As long as a function $d:\Bbb N\times\Bbb N\to [0,+\infty)$ satisfies the definition of a metric, the structure $(\Bbb N,d)$ becomes a metric space with all notions that follow - induced topology, open sets, accumulation points, etc.

As an example, you can take the usual metric $d(x,y)=|x-y|$. Another example $d(x,y) = |\arctan x - \arctan y|$.

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    $\begingroup$ Both of these topologies are just the discrete topology on $\Bbb N$. (There are other possible metrizable topologies from this method, for example taking a bijection to $\Bbb Q$ to get a topology homeomorphic to $\Bbb Q$.) $\endgroup$ Sep 4, 2016 at 20:56
  • $\begingroup$ @MarioCarneiro of course, and other posters provided different examples. I was merely illustrating that "topology applies to integers". $\endgroup$ Sep 5, 2016 at 14:25
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We could add (since I don't see it or any equivalent when skimming the comments) the symmetric set topology on $\mathbb{Z}$: $U \subseteq \mathbb{Z}$ is open if $$n \in U \iff -n \in U$$ holds. This topology disconnects the integers and makes them non-compact, but does admit a countable basis (correct me if I'm messing something up here).

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