# Limit point

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 neighbourhood 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 $S$ .

Limit points should 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 $\{x_{n}\}$ .

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 neighbourhood $V$ of $x$ , there are infinitely many natural numbers $n$ such that $x_{n}\in V$ . This concept generalizes to nets and filters.

## Definition

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 $S$  if every neighbourhood of $x$  contains at least one point of $S$  different from $x$  itself.

Note that it doesn't 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 $X$  is a $T_{1}$  space (which all metric spaces are), then $x\in X$  is a limit point of $S$  if and only if every neighbourhood 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 limit point

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 neighborhood $U$  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 $S$ .

## For sequences and nets

In a topological space $X$ , a point $x\in X$  is said to be a cluster point (or accumulation point) of a sequence $(x_{n})_{n\in \mathbb {N} }$  if, for every neighbourhood $V$  of $x$ , there are infinitely many $n\in \mathbb {N}$  such that $x_{n}\in V$ . It is equivalent to say that for every neighbourhood $V$  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 $(x_{n})_{n\in \mathbb {N} }$  if and only if $x$  is a limit of some subsequence of $(x_{n})_{n\in \mathbb {N} }$ . 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 the net $f$  if, for every neighbourhood $V$  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 the related topic of filters.

### Properties

• Every limit of a non-constant sequence is an accumulation point of the sequence.

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

To each sequence $(x_{n})_{n\in \mathbb {N} }$  in a topological space $X$  we can associate the set $A=\{x_{n}:n\in \mathbb {N} \}$  consisting of all the elements in the sequence.

• If there a 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 $A$ . For example, if the sequence is the constant sequence with value $x$ , we have $A=\{x\}$  and $x$  is an isolated point of $A$  and not an accumulation point of $A$ .
• 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 $A$ .

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 that will have $A$  as associated set of elements.

• 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 (even none) points of $A$  and that neighborhood can only contain finitely many terms of such sequences).

## Selected facts

• We have the following characterization of limit points: $x$  is a limit point of $S$  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 characterisation of closed sets: A set $S$  is closed if and only if it contains all of its limit points.
• Proof: $S$  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 $S$ .
• Another proof: 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 neighbourhood of $x$ . Since $x$  is a limit point of $S$ , any open neighbourhood 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 neighbourhood 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 neighbourhoods 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 neighbourhood of $x$  that contains no points other than $x$ .
• The closure $cl(S)$  of a set $S$  is a disjoint union of its limit points $L(S)$  and isolated points $I(S)$ :
$cl(S)=L(S)\cup I(S),L(S)\cap I(S)=\emptyset .$
• 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 neighbourhood 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's only empty when $S$  is empty or $x$  is the unique element of $S$ .
• By definition, every limit point is an adherent point.