Accumulation point
In mathematics, a limit point (or accumulation point) of a set S in a topological space X is a point x in 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. Note that x does not 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 adding its limit points.
Definition
Let S be a subset of a topological space X. A point x in X is a limit point of S if every open set containing x contains at least one point of S different from x itself.
This is equivalent, in a T1 space, to requiring that every neighbourhood of x contains infinitely many points of S. (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.)
Alternatively, if the space X is sequential, we may say that x ∈ X is a limit point of S if and only if there is an ω-sequence of points in S \ {x} whose limit is x; hence, x is called a limit point.
Every finite interval or bounded interval that contains an infinite number of points must have at least one point of accumulation.
Types of limit points
If every open set containing x contains infinitely many points of S then x is a specific type of limit point called a 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.
A point x ∈ X is a cluster point of a sequence (xn)n ∈ N if, for every neighbourhood V of x, there are infinitely many natural numbers n such that xn ∈ V. If the space is sequential, this is equivalent to the assertion that x is a limit of some subsequence of the sequence (xn)n ∈ N.
The concept of a net generalizes the idea of a sequence. 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.
The set of all cluster points of a sequence is sometimes called a limit set.
Some facts
- We have the following characterisation 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 neighbourhood of the point meets the set. Now, x is a limit point of S, if and only if every neighbourhood of x contains a point of S other than x, if and only if every neighbourhood 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 characterisation of the closure of S: The closure of S is equal to the union of S and L(S).
- 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 we can find an open set around x contained entirely in the complement of S. But then this set contains no point in S, so x is not a limit point, which contradicts our original assumption. 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 neighborhood 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 neighborhoods 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 y ≠ 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 \ 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.