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Created page with "In mathematics, a '''limit point''' (or '''cluster point''' or '''accumulation point''') of a set <math>S</math> in a topological space <math>X</math> is a point <math>x</math..."

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

[[File:Rational sequence with 2 accumulation points.svg|thumb|400px|With respect to the usual Euclidean topology, the sequence of rational numbers <math>x_n=(-1)^n \frac{n}{n+1}</math> 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 <math>S = \{x_n\}.</math>]]

== Definition ==

=== Accumulation points of a set ===

Let <math>S</math> be a subset of a topological space <math>X.</math>
A point <math>x</math> in <math>X</math> is a '''limit point''' or '''cluster point''' or '''accumulation point of the set''' <math>S</math> if every neighbourhood of <math>x</math> contains at least one point of <math>S</math> different from <math>x</math> 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 <math>X</math> is a <math>T_1</math> space (such as a metric space), then <math>x \in X</math> is a limit point of <math>S</math> if and only if every neighbourhood of <math>x</math> contains infinitely many points of <math>S.</math> In fact, <math>T_1</math> spaces are characterized by this property.

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

The set of limit points of <math>S</math> is called the derived set of <math>S.</math>

==== Types of accumulation points ====

If every neighbourhood of <math>x</math> contains infinitely many points of <math>S,</math> then <math>x</math> is a specific type of limit point called an '''ω-accumulation point''' of <math>S.</math>

If every neighbourhood of <math>x</math> contains uncountably many points of <math>S,</math> then <math>x</math> is a specific type of limit point called a '''condensation point''' of <math>S.</math>

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

=== Accumulation points of sequences and nets ===

[[File:Diagonal argument.svg|thumb|A sequence enumerating all positive rational numbers. Each positive real number is a cluster point.]]
In a topological space <math>X,</math> a point <math>x \in X</math> is said to be a '''cluster point''' or '''accumulation point of a sequence''' <math>x_{\bull} = \left(x_n\right)_{n=1}^{\infty}</math> if, for every neighbourhood <math>V</math> of <math>x,</math> there are infinitely many <math>n \in \mathbb{N}</math> such that <math>x_n \in V.</math>
It is equivalent to say that for every neighbourhood <math>V</math> of <math>x</math> and every <math>n_0 \in \mathbb{N},</math> there is some <math>n \geq n_0</math> such that <math>x_n \in V.</math>
If <math>X</math> is a metric space or a first-countable space (or, more generally, a Fréchet–Urysohn space), then <math>x</math> is cluster point of <math>x_{\bull}</math> if and only if <math>x</math> is a limit of some subsequence of <math>x_{\bull}.</math>
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 <math>x</math> to which the sequence converges (that is, every neighborhood of <math>x</math> 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 <math>f : (P,\leq) \to X,</math> where <math>(P,\leq)</math> is a directed set and <math>X</math> is a topological space. A point <math>x \in X</math> is said to be a '''cluster point''' or '''accumulation point of a net''' <math>f</math> if, for every neighbourhood <math>V</math> of <math>x</math> and every <math>p_0 \in P,</math> there is some <math>p \geq p_0</math> such that <math>f(p) \in V,</math> equivalently, if <math>f</math> has a subnet which converges to <math>x.</math> 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 <math>x_{\bull} = \left(x_n\right)_{n=1}^{\infty}</math> in <math>X</math> is by definition just a map <math>x_{\bull} : \mathbb{N} \to X</math> so that its image <math>\operatorname{Im} x_{\bull} := \left\{ x_n : n \in \mathbb{N} \right\}</math> can be defined in the usual way.

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

* 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 <math>\omega</math>-accumulation point of the associated set <math>\operatorname{Im} x_{\bull}.</math>

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

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

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

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

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

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

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

A corollary of this result gives us a characterisation of closed sets: A set <math>S</math> is closed if and only if it contains all of its limit points.

Proofs:
:''Proof'' 1: <math>S</math> is closed if and only if <math>S</math> is equal to its closure if and only if <math>S=S\cup L(S)</math> if and only if <math>L(S)</math> is contained in <math>S.</math>

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

:No isolated point is a limit point of any set.

Proof:
:If <math>x</math> is an isolated point, then <math>\{x\}</math> is a neighbourhood of <math>x</math> that contains no points other than <math>x.</math>

:A space <math>X</math> is discrete if and only if no subset of <math>X</math> has a limit point.

Proof:
:If <math>X</math> is discrete, then every point is isolated and cannot be a limit point of any set. Conversely, if <math>X</math> is not discrete, then there is a singleton <math>\{ x \}</math> that is not open. Hence, every open neighbourhood of <math>\{ x \}</math> contains a point <math>y \neq x,</math> and so <math>x</math> is a limit point of <math>X.</math>

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

Proof:
:As long as <math>S \setminus \{ x \}</math> is nonempty, its closure will be <math>X.</math> It is only empty when <math>S</math> is empty or <math>x</math> is the unique element of <math>S.</math>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Limit_point Limit point, Wikipedia] under a CC BY-SA license