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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 \mathbb 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. There is also a closely related concept for sequences. A cluster point (or accumulation point) of a sequence (x_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. 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(x_n\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 : (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 filters.


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


Every sequence x_ = \left(x_n\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(S) of a set S is a disjoint union of its limit points L(S) and isolated points I(S): :\operatorname (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 \. 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(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 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(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'' 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 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