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 $\backslash $ in $\backslash mathbb$ with standard topology. However, $0.5$ is a limit point (though not a boundary point) of interval $,\; 1/math>\; in$ \backslash mathbb$with\; standard\; topology\; (for\; a\; less\; trivial\; example\; of\; a\; limit\; point,\; see\; the\; first\; caption).This\; concept\; profitably\; generalizes\; the\; notion\; of\; alimitand\; is\; the\; underpinning\; of\; concepts\; such\; asclosed\; setandtopological\; closure.\; Indeed,\; a\; set\; is\; closed\; if\; and\; only\; if\; it\; contains\; all\; of\; its\; limit\; points,\; and\; the\; topological\; closure\; operation\; can\; be\; thought\; of\; as\; an\; operation\; that\; enriches\; a\; set\; by\; uniting\; it\; with\; its\; limit\; points.$
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 $x\_n\; \backslash in\; V.$ 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 $x\; \backslash in\; X$ 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 $x\; \backslash in\; X$ is a limit point of $S$ if and only if there is a sequence of points in $S\; \backslash setminus\; \backslash $ 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 $\backslash left|\; U\; \backslash cap\; S\backslash right|\; =\; \backslash left|\; S\; \backslash right|,$ then $x$ is a specific type of limit point called a of $S.$

** For sequences and nets **

In a topological space $X,$ a point $x\; \backslash in\; X$ is said to be a cluster point (or accumulation point) of a sequence $x\_\; =\; \backslash left(x\_n\backslash right)\_^$ if, for every neighbourhood $V$ of $x,$ there are infinitely many $n\; \backslash in\; \backslash mathbb$ such that $x\_n\; \backslash in\; V.$
It is equivalent to say that for every neighbourhood $V$ of $x$ and every $n\_0\; \backslash in\; \backslash mathbb,$ there is some $n\; \backslash geq\; n\_0$ such that $x\_n\; \backslash in\; V.$
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 $f\; :\; (P,\backslash leq)\; \backslash to\; X,$ where $(P,\backslash leq)$ is a directed set and $X$ is a topological space. A point $x\; \backslash in\; X$ is said to be a cluster point (or accumulation point) of the net $f$ if, for every neighbourhood $V$ of $x$ and every $p\_0\; \backslash in\; P,$ there is some $p\; \backslash geq\; p\_0$ such that $f(p)\; \backslash in\; V,$ 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 $x\_\; =\; \backslash left(x\_n\backslash right)\_^$ in $X$ is by definition just a map $x\_\; :\; \backslash mathbb\; \backslash to\; X$ so that its image $\backslash operatorname\; x\_\; :=\; \backslash left\backslash $ can be defined in the usual way.
* If there exists an element $x\; \backslash in\; X$ 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 $\backslash operatorname\; x\_.$ For example, if the sequence is the constant sequence with value $x,$ we have $\backslash operatorname\; x\_\; =\; \backslash $ and $x$ is an isolated point of $\backslash operatorname\; x\_$ and not an accumulation point of $\backslash operatorname\; x\_.$
* 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 $\backslash omega$-accumulation point of the associated set $\backslash operatorname\; x\_.$
Conversely, given a countable infinite set $A\; \backslash subseteq\; X$ 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 $A\; =\; \backslash operatorname\; x\_.$
* Any $\backslash omega$-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 $x\; \backslash in\; X$ that is an $\backslash omega$-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 $\backslash operatorname(S)$ of a set $S$ is a disjoint union of its limit points $L(S)$ and isolated points $I(S)$:
:$\backslash operatorname\; (S)\; =\; L(S)\; \backslash cup\; I(S),\; L(S)\; \backslash cap\; I(S)\; =\; \backslash varnothing.$
A point $x\; \backslash in\; X$ is a limit point of $S\; \backslash subseteq\; X$ if and only if it is in the closure of $S\; \backslash setminus\; \backslash .$
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 $S\; \backslash setminus\; \backslash ,$ if and only if $x$ is in the closure of $S\; \backslash setminus\; \backslash .$
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 $S=S\backslash cup\; L(S)$ 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 $\backslash $ 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 $\backslash $ that is not open. Hence, every open neighbourhood of $\backslash $ contains a point $y\; \backslash neq\; x,$ 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 $X\; \backslash setminus\; S$ is a limit point of $S.$
As long as $S\; \backslash setminus\; \backslash $ 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