In mathematics, a limit point of a set S in a topological space X is a point x (which is in X, but not necessarily in S) 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 uniting it with its limit points.
Contents
 1 Definition
 2 Types of limit points
 3 Some facts
 4 References
 5 External links
Definition
Let S be a subset of a topological space X. A point x in X is a limit 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.
This is equivalent, in a T_{1} 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 FréchetUrysohn, we may say that x ∈ X is a limit point of S if and only if there is an ωsequence of points in S \ {x} whose limit is x; hence, x is called a limit point.
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 x_{n} = (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.

If every open set containing x contains infinitely many points of S then x is a specific type of limit point called an ωaccumulation 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 or accumulation point of a sequence (x_{n})_{n ∈ N} if, for every neighbourhood V of x, there are infinitely many natural numbers n such that x_{n} ∈ V. If the space is Fréchet–Urysohn, this is equivalent to the assertion that x is a limit of some subsequence of the sequence (x_{n})_{n ∈ N}. The set of all cluster points of a sequence is sometimes called a limit set.
The concept of a net generalizes the idea of a sequence. Let $n:(P,\leq )\to X$ be a net, where $(P,\leq )$ is a directed set. The point $a\in X$ is said to be a cluster point of the net $n$ if for any neighborhood $U$ of $a$ and any $p\in P$, there is some $x\geq p$ such that $n(x)\in U$, equivalently, if $n$ has a subnet which converges to $a$. Cluster points in nets encompass the idea of both condensation points and ωaccumulation points. Clustering and limit points are also defined for the related topic of filters.
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). [This fact appears to be just the definition, as stated in the closure. It might become less trivial, if other definition of closure is used.]
 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 neighborhood of x. Since x is a limit point of S, any open neighborhood of x should have a nontrivial intersection with S. However, a set can not have a nontrivial 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 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.
References
 Hazewinkel, Michiel, ed. (2001), "Limit point of a set", Encyclopedia of Mathematics, Springer, ISBN 9781556080104
External links
"limit point". PlanetMath.