# Difference between cluster points and limit points?

I have a weird doubt about these terms "cluster points" and "limit points". I am just calling it weird because I don't know where I am going wrong as I have seen somewhere that cluster points and limit points are one and the same, and somewhere else that they have very slight difference in their meaning. The set of all limit points of $\mathbb{Z}$ is empty. What about the set of cluster points of $\mathbb{Z}$? Please clear my doubt of their definitions, thanks in advance.

• In metric spaces, the two terms coincide. In spaces that don't satisfy the $T_1$ separation axiom, the two concepts differ. Oct 7, 2015 at 17:10
• Sorry, but I dont know the T1 seperation axiom Oct 7, 2015 at 17:11
• I suspected that. Then just remember that there are spaces where the two concepts are different, but you will not encounter any such space for a while. Oct 7, 2015 at 17:16
• I had to google cluster point. My interpretation is that a cluster point is in reference to a sequence and that every neighborhood has an infinite number of terms of the sequence. (Which would make it a limit point of the sequence?) I had to google T1. A T1 space is one where every two distinct points will have neighborhoods not containing the other. So every metric space is T1. A space not T1 would have two distinct a $\ne$ b but b is in every neighborhood of b. Which would clearly have consequence.. I'd need to know the precise definition of cluster point to continue. Oct 7, 2015 at 17:26
• Does this answer your question? what is the difference between cluster point and limit point? Dec 25, 2021 at 13:59

I might be utterly wrong so take this with a grain of salt.

A limit point of A is a point in which every neighborhood has at least one point other than itself of A.

A cluster point of A is a point in which every neighborhood has an infinite number of points of A.

In a metric space these are the equivalent. (For $a_0$ in the neighborhood of x, find the neighborhood of x with radius $d(a_0, x)/2$ and $a_0$ is not in it so another $a_1$ must be in it. Inductively repeat. Thus all limit points are cluster points. [If you accept countable axiom of choice.])

Metric spaces have what is called the T1 axiom which is, apparently (I googled this), that if $a \ne b$ then there exists neighborhoods around each which don't contain the other. (I interpret this to mean that in a metric space $d(a, b)> 0$ so the neighborhoods of radius $d(a, b)/2$ around each point will not have either point in the other neighborhoods.) A non-metric space without the T1 axiom might have two $a \ne b$ such that $b$ is in every neighborhood of $a$! (Clearly impossible in metric spaces.) Thus if $b \in B$ then $a$ is a limit point of B because $b$ is in every neighborhood. But $b$ could be the only point in a neighborhood. Thus $a$ isn't necessarily a cluster point of B.

Am I right? If I'm not, ignore me.

• This is weird. I'm trying to edit but every time I press the edit button I get the text of an entirely different post of mine. So errata: A limit point every neighbor hood as at least one point other than itself. (obviously.) Oct 7, 2015 at 17:50
• @Bungo. Absolutely right! I was being sloppy and forgot to state that. I'm trying to edit that. Other than that, am I right? Oct 7, 2015 at 17:52
• @Bungo And Gamelin and Greene define adherent points as including isolated points and limit points. (Now why can't I edit. This is weird!) Oct 7, 2015 at 17:55
• @Bungo axiom of countable choice. Yeah, me too. Wouldn't all proofs by induction be invalid without it? Oct 7, 2015 at 17:57
• Oh, good point. Oct 7, 2015 at 17:59

If you are talking about numeric sequences, a point is a cluster point of a sequence if it a limit of a subsequence of that sequence. If every subsequence has the same limit then that point is a limit point for the sequence. The sequence 1, 1/2, 1/4, 3/4, 1/5, 4/5, 1/6, 5/6, ... has no limit but has both 0 and 1 as cluster points. But if you are talking about general sets in a metric space, limit points and cluster points are the same thing- every open neighborhood of the point contains at least one other point of the set.

In both senses, the set of all integers, with the discrete topology, has no limit points and no cluster points.