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.

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 (xn)n  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 xn  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 T1 space (which all metric spaces are), then x X is a limit point of S if and only if every neighbourhood of x contains infinitely many points of S. Indeed, T1 spaces are characterized by this property.

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

Types of limit points

A sequence enumerating all positive rational numbers. Each positive real number is a cluster point.
With respect to the usual Euclidean topology, the sequence of rational numbers xn = (-1)n·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 {xn}.

If every open set containing 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 open set containing x contains uncountably many points of S then x is a specific type of limit point called a condensation point of S.

If every open set U containing x satisfies |U S| = |S| then x is a specific type of limit point called a complete accumulation point of S.

For Sequences and Nets

In a topological space , a point is said to be a cluster point (or accumulation point) of a sequence if, for every neighbourhood of , there are infinitely many such that . It is equivalent to say that for every neighbourhood of and every , there is some such that . If is a metric space or a first-countable space (or, more generally, a Fréchet–Urysohn space), then is cluster point of if and only if is a limit of some subsequence of . The set of all cluster points of a sequence is sometimes called the limit set.

The concept of a net generalizes the idea of a sequence. A net is a function , where is a directed set and is a topological space. A point is said to be a cluster point (or accumulation point) of the net if, for every neighbourhood of and every , there is some such that , equivalently, if has a subnet which converges to . 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.

Some 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 \ {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 \ {x}, if and only if x is in the closure of S \ {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 ∪ 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.
  • 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 yx, 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 \ S is still a limit point of S.
    • Proof: As long as S \ {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.

References

  • Hazewinkel, Michiel, ed. (2001) [1994], "Limit point of a set", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4
  • "limit point". PlanetMath.
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