In mathematics, a limit point (or cluster point or accumulation point) of a set ${\displaystyle S}$ in a topological space ${\displaystyle X}$ is a point ${\displaystyle x}$ that can be "approximated" by points of ${\displaystyle S}$ in the sense that every neighbourhood of ${\displaystyle x}$ with respect to the topology on ${\displaystyle X}$ also contains a point of ${\displaystyle S}$ other than ${\displaystyle x}$ itself. A limit point of a set ${\displaystyle S}$ does not itself have to be an element of ${\displaystyle S.}$ There is also a closely related concept for sequences. A cluster point or accumulation point of a sequence ${\displaystyle (x_{n})_{n\in \mathbb {N} ))$ in a topological space ${\displaystyle X}$ is a point ${\displaystyle x}$ such that, for every neighbourhood ${\displaystyle V}$ of ${\displaystyle x,}$ there are infinitely many natural numbers ${\displaystyle n}$ such that ${\displaystyle x_{n}\in V.}$ This definition of a cluster or accumulation point of a sequence generalizes to nets and filters. In contrast to sets, for a sequence, net, or filter, the term "limit point" is not synonymous with a "cluster/accumulation point"; by definition, the similarly named notion of a limit point of a filter[1] (respectively, a limit point of a sequence,[2] a limit point of a net) refers to a point that the filter converges to (respectively, the sequence converges to, the net converges to).

The limit points of a set should not be confused with adherent points for which every neighbourhood of ${\displaystyle x}$ contains a point of ${\displaystyle S}$. Unlike for limit points, this point of ${\displaystyle S}$ may be ${\displaystyle x}$ itself. A limit point can be characterized as an adherent point that is not an isolated point.

Limit points of a set should also not be confused with boundary points. For example, ${\displaystyle 0}$ is a boundary point (but not a limit point) of set ${\displaystyle \{0\))$ in ${\displaystyle \mathbb {R} }$ with standard topology. However, ${\displaystyle 0.5}$ is a limit point (though not a boundary point) of interval ${\displaystyle [0,1]}$ in ${\displaystyle \mathbb {R} }$ with standard topology (for a less trivial example of a limit point, see the first caption).[3][4][5]

This concept profitably generalizes the notion of a limit and is the underpinning of concepts such as closed set and topological closure. Indeed, a set is closed if and only if it contains all of its limit points, and the topological closure operation can be thought of as an operation that enriches a set by uniting it with its limit points.

With respect to the usual Euclidean topology, the sequence of rational numbers ${\displaystyle x_{n}=(-1)^{n}{\frac {n}{n+1))}$ has no limit (i.e. does not converge), but has two accumulation points (which are considered limit points here), viz. -1 and +1. Thus, thinking of sets, these points are limit points of the set ${\displaystyle S=\{x_{n}\}.}$

## Definition

### Accumulation points of a set

Let ${\displaystyle S}$ be a subset of a topological space ${\displaystyle X.}$ A point ${\displaystyle x}$ in ${\displaystyle X}$ is a limit point or cluster point or accumulation point of the set ${\displaystyle S}$ if every neighbourhood of ${\displaystyle x}$ contains at least one point of ${\displaystyle S}$ different from ${\displaystyle x}$ itself.

It does not make a difference if we restrict the condition to open neighbourhoods only. It is often convenient to use the "open neighbourhood" form of the definition to show that a point is a limit point and to use the "general neighbourhood" form of the definition to derive facts from a known limit point.

If ${\displaystyle X}$ is a ${\displaystyle T_{1))$ space (such as a metric space), then ${\displaystyle x\in X}$ is a limit point of ${\displaystyle S}$ if and only if every neighbourhood of ${\displaystyle x}$ contains infinitely many points of ${\displaystyle S.}$[6] In fact, ${\displaystyle T_{1))$ spaces are characterized by this property.

If ${\displaystyle X}$ is a Fréchet–Urysohn space (which all metric spaces and first-countable spaces are), then ${\displaystyle x\in X}$ is a limit point of ${\displaystyle S}$ if and only if there is a sequence of points in ${\displaystyle S\setminus \{x\))$ whose limit is ${\displaystyle x.}$ In fact, Fréchet–Urysohn spaces are characterized by this property.

The set of limit points of ${\displaystyle S}$ is called the derived set of ${\displaystyle S.}$

#### Types of accumulation points

If every neighbourhood of ${\displaystyle x}$ contains infinitely many points of ${\displaystyle S,}$ then ${\displaystyle x}$ is a specific type of limit point called an ω-accumulation point of ${\displaystyle S.}$

If every neighbourhood of ${\displaystyle x}$ contains uncountably many points of ${\displaystyle S,}$ then ${\displaystyle x}$ is a specific type of limit point called a condensation point of ${\displaystyle S.}$

If every neighbourhood ${\displaystyle U}$ of ${\displaystyle x}$ satisfies ${\displaystyle \left|U\cap S\right|=\left|S\right|,}$ then ${\displaystyle x}$ is a specific type of limit point called a complete accumulation point of ${\displaystyle S.}$

### Accumulation points of sequences and nets

A sequence enumerating all positive rational numbers. Each positive real number is a cluster point.

In a topological space ${\displaystyle X,}$ a point ${\displaystyle x\in X}$ is said to be a cluster point or accumulation point of a sequence ${\displaystyle x_{\bullet }=\left(x_{n}\right)_{n=1}^{\infty ))$ if, for every neighbourhood ${\displaystyle V}$ of ${\displaystyle x,}$ there are infinitely many ${\displaystyle n\in \mathbb {N} }$ such that ${\displaystyle x_{n}\in V.}$ It is equivalent to say that for every neighbourhood ${\displaystyle V}$ of ${\displaystyle x}$ and every ${\displaystyle n_{0}\in \mathbb {N} ,}$ there is some ${\displaystyle n\geq n_{0))$ such that ${\displaystyle x_{n}\in V.}$ If ${\displaystyle X}$ is a metric space or a first-countable space (or, more generally, a Fréchet–Urysohn space), then ${\displaystyle x}$ is cluster point of ${\displaystyle x_{\bullet ))$ if and only if ${\displaystyle x}$ is a limit of some subsequence of ${\displaystyle x_{\bullet }.}$ The set of all cluster points of a sequence is sometimes called the limit set.

Note that there is already the notion of limit of a sequence to mean a point ${\displaystyle x}$ to which the sequence converges (that is, every neighborhood of ${\displaystyle x}$ contains all but finitely many elements of the sequence). That is why we do not use the term limit point of a sequence as a synonym for accumulation point of the sequence.

The concept of a net generalizes the idea of a sequence. A net is a function ${\displaystyle f:(P,\leq )\to X,}$ where ${\displaystyle (P,\leq )}$ is a directed set and ${\displaystyle X}$ is a topological space. A point ${\displaystyle x\in X}$ is said to be a cluster point or accumulation point of a net ${\displaystyle f}$ if, for every neighbourhood ${\displaystyle V}$ of ${\displaystyle x}$ and every ${\displaystyle p_{0}\in P,}$ there is some ${\displaystyle p\geq p_{0))$ such that ${\displaystyle f(p)\in V,}$ equivalently, if ${\displaystyle f}$ has a subnet which converges to ${\displaystyle x.}$ Cluster points in nets encompass the idea of both condensation points and ω-accumulation points. Clustering and limit points are also defined for filters.

## Relation between accumulation point of a sequence and accumulation point of a set

Every sequence ${\displaystyle x_{\bullet }=\left(x_{n}\right)_{n=1}^{\infty ))$ in ${\displaystyle X}$ is by definition just a map ${\displaystyle x_{\bullet }:\mathbb {N} \to X}$ so that its image ${\displaystyle \operatorname {Im} x_{\bullet }:=\left\{x_{n}:n\in \mathbb {N} \right\))$ can be defined in the usual way.

• If there exists an element ${\displaystyle x\in X}$ that occurs infinitely many times in the sequence, ${\displaystyle x}$ is an accumulation point of the sequence. But ${\displaystyle x}$ need not be an accumulation point of the corresponding set ${\displaystyle \operatorname {Im} x_{\bullet }.}$ For example, if the sequence is the constant sequence with value ${\displaystyle x,}$ we have ${\displaystyle \operatorname {Im} x_{\bullet }=\{x\))$ and ${\displaystyle x}$ is an isolated point of ${\displaystyle \operatorname {Im} x_{\bullet ))$ and not an accumulation point of ${\displaystyle \operatorname {Im} x_{\bullet }.}$
• If no element occurs infinitely many times in the sequence, for example if all the elements are distinct, any accumulation point of the sequence is an ${\displaystyle \omega }$-accumulation point of the associated set ${\displaystyle \operatorname {Im} x_{\bullet }.}$

Conversely, given a countable infinite set ${\displaystyle A\subseteq X}$ in ${\displaystyle X,}$ we can enumerate all the elements of ${\displaystyle A}$ in many ways, even with repeats, and thus associate with it many sequences ${\displaystyle x_{\bullet ))$ that will satisfy ${\displaystyle A=\operatorname {Im} x_{\bullet }.}$

• Any ${\displaystyle \omega }$-accumulation point of ${\displaystyle A}$ is an accumulation point of any of the corresponding sequences (because any neighborhood of the point will contain infinitely many elements of ${\displaystyle A}$ and hence also infinitely many terms in any associated sequence).
• A point ${\displaystyle x\in X}$ that is not an ${\displaystyle \omega }$-accumulation point of ${\displaystyle A}$ cannot be an accumulation point of any of the associated sequences without infinite repeats (because ${\displaystyle x}$ has a neighborhood that contains only finitely many (possibly even none) points of ${\displaystyle A}$ and that neighborhood can only contain finitely many terms of such sequences).

## Properties

Every limit of a non-constant sequence is an accumulation point of the sequence. And by definition, every limit point is an adherent point.

The closure ${\displaystyle \operatorname {cl} (S)}$ of a set ${\displaystyle S}$ is a disjoint union of its limit points ${\displaystyle L(S)}$ and isolated points ${\displaystyle I(S)}$:

${\displaystyle \operatorname {cl} (S)=L(S)\cup I(S),L(S)\cap I(S)=\varnothing .}$

A point ${\displaystyle x\in X}$ is a limit point of ${\displaystyle S\subseteq X}$ if and only if it is in the closure of ${\displaystyle S\setminus \{x\}.}$

Proof

We use the fact that a point is in the closure of a set if and only if every neighborhood of the point meets the set. Now, ${\displaystyle x}$ is a limit point of ${\displaystyle S,}$ if and only if every neighborhood of ${\displaystyle x}$ contains a point of ${\displaystyle S}$ other than ${\displaystyle x,}$ if and only if every neighborhood of ${\displaystyle x}$ contains a point of ${\displaystyle S\setminus \{x\},}$ if and only if ${\displaystyle x}$ is in the closure of ${\displaystyle S\setminus \{x\}.}$

If we use ${\displaystyle L(S)}$ to denote the set of limit points of ${\displaystyle S,}$ then we have the following characterization of the closure of ${\displaystyle S}$: The closure of ${\displaystyle S}$ is equal to the union of ${\displaystyle S}$ and ${\displaystyle L(S).}$ This fact is sometimes taken as the definition of closure.

Proof

("Left subset") Suppose ${\displaystyle x}$ is in the closure of ${\displaystyle S.}$ If ${\displaystyle x}$ is in ${\displaystyle S,}$ we are done. If ${\displaystyle x}$ is not in ${\displaystyle S,}$ then every neighbourhood of ${\displaystyle x}$ contains a point of ${\displaystyle S,}$ and this point cannot be ${\displaystyle x.}$ In other words, ${\displaystyle x}$ is a limit point of ${\displaystyle S}$ and ${\displaystyle x}$ is in ${\displaystyle L(S).}$ ("Right subset") If ${\displaystyle x}$ is in ${\displaystyle S,}$ then every neighbourhood of ${\displaystyle x}$ clearly meets ${\displaystyle S,}$ so ${\displaystyle x}$ is in the closure of ${\displaystyle S.}$ If ${\displaystyle x}$ is in ${\displaystyle L(S),}$ then every neighbourhood of ${\displaystyle x}$ contains a point of ${\displaystyle S}$ (other than ${\displaystyle x}$), so ${\displaystyle x}$ is again in the closure of ${\displaystyle S.}$ This completes the proof.

A corollary of this result gives us a characterisation of closed sets: A set ${\displaystyle S}$ is closed if and only if it contains all of its limit points.

Proof

Proof 1: ${\displaystyle S}$ is closed if and only if ${\displaystyle S}$ is equal to its closure if and only if ${\displaystyle S=S\cup L(S)}$ if and only if ${\displaystyle L(S)}$ is contained in ${\displaystyle S.}$

Proof 2: Let ${\displaystyle S}$ be a closed set and ${\displaystyle x}$ a limit point of ${\displaystyle S.}$ If ${\displaystyle x}$ is not in ${\displaystyle S,}$ then the complement to ${\displaystyle S}$ comprises an open neighbourhood of ${\displaystyle x.}$ Since ${\displaystyle x}$ is a limit point of ${\displaystyle S,}$ any open neighbourhood of ${\displaystyle x}$ should have a non-trivial intersection with ${\displaystyle S.}$ However, a set can not have a non-trivial intersection with its complement. Conversely, assume ${\displaystyle S}$ contains all its limit points. We shall show that the complement of ${\displaystyle S}$ is an open set. Let ${\displaystyle x}$ be a point in the complement of ${\displaystyle S.}$ By assumption, ${\displaystyle x}$ is not a limit point, and hence there exists an open neighbourhood ${\displaystyle U}$ of ${\displaystyle x}$ that does not intersect ${\displaystyle S,}$ and so ${\displaystyle U}$ lies entirely in the complement of ${\displaystyle S.}$ Since this argument holds for arbitrary ${\displaystyle x}$ in the complement of ${\displaystyle S,}$ the complement of ${\displaystyle S}$ can be expressed as a union of open neighbourhoods of the points in the complement of ${\displaystyle S.}$ Hence the complement of ${\displaystyle S}$ is open.

No isolated point is a limit point of any set.

Proof

If ${\displaystyle x}$ is an isolated point, then ${\displaystyle \{x\))$ is a neighbourhood of ${\displaystyle x}$ that contains no points other than ${\displaystyle x.}$

A space ${\displaystyle X}$ is discrete if and only if no subset of ${\displaystyle X}$ has a limit point.

Proof

If ${\displaystyle X}$ is discrete, then every point is isolated and cannot be a limit point of any set. Conversely, if ${\displaystyle X}$ is not discrete, then there is a singleton ${\displaystyle \{x\))$ that is not open. Hence, every open neighbourhood of ${\displaystyle \{x\))$ contains a point ${\displaystyle y\neq x,}$ and so ${\displaystyle x}$ is a limit point of ${\displaystyle X.}$

If a space ${\displaystyle X}$ has the trivial topology and ${\displaystyle S}$ is a subset of ${\displaystyle X}$ with more than one element, then all elements of ${\displaystyle X}$ are limit points of ${\displaystyle S.}$ If ${\displaystyle S}$ is a singleton, then every point of ${\displaystyle X\setminus S}$ is a limit point of ${\displaystyle S.}$

Proof

As long as ${\displaystyle S\setminus \{x\))$ is nonempty, its closure will be ${\displaystyle X.}$ It is only empty when ${\displaystyle S}$ is empty or ${\displaystyle x}$ is the unique element of ${\displaystyle S.}$