Limit Points (or Cluster Points) in Higher Dimensions

In mathematics, a limit point (or cluster point or accumulation point) of a set $S$ in a topological space $X$ is a point $x$ that can be "approximated" by points of $S$ in the sense that every neighborhood of $x$ with respect to the topology on $X$ also contains a point of $S$ other than $x$ itself. A limit point of a set $S$ does not itself have to be an element of $S.$ There is also a closely related concept for sequences. A cluster point or accumulation point of a sequence $(x_{n})_{n\in \mathbb {N} }$ in a topological space $X$ is a point $x$ such that, for every neighborhood $V$ of $x,$ there are infinitely many natural numbers $n$ such that $x_{n}\in V.$ 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 (respectively, a limit point of a sequence, 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 neighborhood of $x$ contains a point of $S$. Unlike for limit points, this point of $S$ may be $x$ 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, $0$ is a boundary point (but not a limit point) of set $\{0\}$ in $\mathbb {R}$ with standard topology. However, $0.5$ is a limit point (though not a boundary point) of interval $[0,1]$ in $\mathbb {R}$ with standard topology (for a less trivial example of a limit point, see the first caption).

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 $x_{n}=(-1)^{n}{\frac {n}{n+1}}$ 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 $S=\{x_{n}\}.$

Definition

Accumulation points of a set

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

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

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

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

The set of limit points of $S$ is called the derived set of $S.$

Types of accumulation points

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

If every neighborhood of $x$ contains uncountably many points of $S,$ then $x$ is a specific type of limit point called a condensation point of $S.$

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

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 $X,$ a point $x\in X$ is said to be a cluster point or accumulation point of a sequence $x_{\bullet }=\left(x_{n}\right)_{n=1}^{\infty }$ if, for every neighborhood $V$ of $x,$ there are infinitely many $n\in \mathbb {N}$ such that $x_{n}\in V.$ It is equivalent to say that for every neighborhood $V$ of $x$ and every $n_{0}\in \mathbb {N} ,$ there is some $n\geq n_{0}$ such that $x_{n}\in V.$ If $X$ is a metric space or a first-countable space (or, more generally, a Fréchet–Urysohn space), then $x$ is cluster point of $x_{\bullet }$ if and only if $x$ is a limit of some subsequence of $x_{\bullet }.$ 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 $x$ to which the sequence converges (that is, every neighborhood of $x$ 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 $f:(P,\leq )\to X,$ where $(P,\leq )$ is a directed set and $X$ is a topological space. A point $x\in X$ is said to be a cluster point or accumulation point of a net $f$ if, for every neighborhood $V$ of $x$ and every $p_{0}\in P,$ there is some $p\geq p_{0}$ such that $f(p)\in V,$ equivalently, if $f$ has a subnet which converges to $x.$ 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 $x_{\bullet }=\left(x_{n}\right)_{n=1}^{\infty }$ in $X$ is by definition just a map $x_{\bullet }:\mathbb {N} \to X$ so that its image $\operatorname {Im} x_{\bullet }:=\left\{x_{n}:n\in \mathbb {N} \right\}$ can be defined in the usual way.

• If there exists an element $x\in X$ that occurs infinitely many times in the sequence, $x$ is an accumulation point of the sequence. But $x$ need not be an accumulation point of the corresponding set $\operatorname {Im} x_{\bullet }.$ For example, if the sequence is the constant sequence with value $x,$ we have $\operatorname {Im} x_{\bullet }=\{x\}$ and $x$ is an isolated point of $\operatorname {Im} x_{\bullet }$ and not an accumulation point of $\operatorname {Im} x_{\bullet }.$

• 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 $\omega$-accumulation point of the associated set $\operatorname {Im} x_{\bullet }.$

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

• Any $\omega$-accumulation point of $A$ is an accumulation point of any of the corresponding sequences (because any neighborhood of the point will contain infinitely many elements of $A$ and hence also infinitely many terms in any associated sequence).

• A point $x\in X$ that is not an $\omega$-accumulation point of $A$ cannot be an accumulation point of any of the associated sequences without infinite repeats (because $x$ has a neighborhood that contains only finitely many (possibly even none) points of $A$ 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 $\operatorname {cl} (S)$ of a set $S$ is a disjoint union of its limit points $L(S)$ and isolated points $I(S)$:

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

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

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, $x$ is a limit point of $S,$ if and only if every neighborhood of $x$ contains a point of $S$ other than $x,$ if and only if every neighborhood of $x$ contains a point of $S\setminus \{x\},$ if and only if $x$ is in the closure of $S\setminus \{x\}.$

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

Proof

("Left subset") Suppose $x$ is in the closure of $S.$ If $x$ is in $S,$ we are done. If $x$ is not in $S,$ then every neighbourhood of $x$ contains a point of $S,$ and this point cannot be $x.$ In other words, $x$ is a limit point of $S$ and $x$ is in $L(S).$

("Right subset") If $x$ is in $S,$ then every neighbourhood of $x$ clearly meets $S,$ so $x$ is in the closure of $S.$ If $x$ is in $L(S),$ then every neighbourhood of $x$ contains a point of $S$ (other than $x$), so $x$ is again in the closure of $S.$ This completes the proof.

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

Proof

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

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

No isolated point is a limit point of any set.

Proof

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

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

Proof

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

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

Proof

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

Licensing

Content obtained and/or adapted from:

• Limit point, Wikipedia under a CC BY-SA license