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.
Every finite interval or bounded interval that contains an infinite number of points must have at least one point of accumulation.
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.
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.