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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 adherent points for which every neighbourhood 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 should also not be confused with Boundary points. For example, $0$ is a boundary point (but not a limit point) of set $\$ in $\mathbb$ with standard topology. However, $0.5$ is a limit point (though not a boundary point) of interval $, 1/math> in\mathbbwith standard topology \left(for a less trivial example of a limit point, see the first caption\right).This concept profitably generalizes the notion of alimitand is the underpinning of concepts such asclosed setandtopological closure. Indeed, a set is closed if and only if it contains all of its limit points, and the topological closure operation can be thought of as an operation that enriches a set by uniting it with its limit points.$ There is also a closely related concept for sequences. A cluster point (or accumulation point) of a sequence $\left(x_n\right)_$ 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. 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 $X$ 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 \$ 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 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 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 $x_ = \left\left(x_n\right\right)_^$ if, for every neighbourhood $V$ of $x,$ there are infinitely many $n \in \mathbb$ such that $x_n \in V.$ It is equivalent to say that for every neighbourhood $V$ of $x$ and every $n_0 \in \mathbb,$ 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 $x_$ if and only if $x$ is a limit of some subsequence of $x_.$ 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 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 $f : \left(P,\leq\right) \to X,$ where $\left(P,\leq\right)$ 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 $f\left(p\right) \in V,$ 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 $x_ = \left\left(x_n\right\right)_^$ in $X$ is by definition just a map $x_ : \mathbb \to X$ so that its image $\operatorname x_ := \left\$ 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 $\operatorname x_.$ For example, if the sequence is the constant sequence with value $x,$ we have $\operatorname x_ = \$ and $x$ is an isolated point of $\operatorname x_$ and not an accumulation point of $\operatorname x_.$ * 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 $\operatorname x_.$ 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 $x_$ that will satisfy $A = \operatorname x_.$ * 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 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 $\operatorname\left(S\right)$ of a set $S$ is a disjoint union of its limit points $L\left(S\right)$ and isolated points $I\left(S\right)$: :$\operatorname \left(S\right) = L\left(S\right) \cup I\left(S\right), L\left(S\right) \cap I\left(S\right) = \varnothing.$ A point $x \in X$ is a limit point of $S \subseteq X$ if and only if it is in the closure of $S \setminus \.$ 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 $S \setminus \,$ if and only if $x$ is in the closure of $S \setminus \.$ If we use $L\left(S\right)$ 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 $L\left(S\right).$ This fact is sometimes taken as the of closure. ("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 $L\left(S\right).$ ("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 $L\left(S\right),$ 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'' 1: $S$ is closed if and only if $S$ is equal to its closure if and only if $S=S\cup L\left(S\right)$ if and only if $L\left(S\right)$ 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 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. If $x$ is an isolated point, then $\$ is a neighbourhood 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. 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 $\$ that is not open. Hence, every open neighbourhood of $\$ 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 $S.$ As long as $S \setminus \$ is nonempty, its closure will be $X.$ It is only empty when $S$ is empty or $x$ is the unique element of $S.$

See also

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Citations

References

* {{springer|title=Limit point of a set|id=p/l058880 Category:Topology Category:General topology