Limit point

In mathematics, a limit point (or cluster point or accumulation point) of a set ${\displaystyle S}$ in a topological space ${\displaystyle X}$ is a point ${\displaystyle x}$ that can be "approximated" by points of ${\displaystyle S}$ in the sense that every neighbourhood of ${\displaystyle x}$ with respect to the topology on ${\displaystyle X}$ also contains a point of ${\displaystyle S}$ other than ${\displaystyle x}$ itself. A limit point of a set ${\displaystyle S}$ does not itself have to be an element of ${\displaystyle S}$.

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.

With respect to the usual Euclidean topology, the sequence of rational numbers ${\displaystyle x_{n}=(-1)^{n}{\frac {n}{n+1}}}$ 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 ${\displaystyle \{x_{n}\}}$.

There is also a closely related concept for sequences. A cluster point (or accumulation point) of a sequence ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ in a topological space ${\displaystyle X}$ is a point ${\displaystyle x}$ such that, for every neighbourhood ${\displaystyle V}$ of ${\displaystyle x}$, there are infinitely many natural numbers ${\displaystyle n}$ such that ${\displaystyle x_{n}\in V}$. This concept generalizes to nets and filters.

Definition

Let ${\displaystyle S}$ be a subset of a topological space ${\displaystyle X}$. A point ${\displaystyle x}$ in ${\displaystyle X}$ is a limit point (or cluster point or accumulation point) of ${\displaystyle S}$ if every neighbourhood of ${\displaystyle x}$ contains at least one point of ${\displaystyle S}$ different from ${\displaystyle x}$ 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 ${\displaystyle X}$ is a ${\displaystyle T_{1}}$ space (which all metric spaces are), then ${\displaystyle x\in X}$ is a limit point of ${\displaystyle S}$ if and only if every neighbourhood of ${\displaystyle x}$ contains infinitely many points of ${\displaystyle S}$. In fact, ${\displaystyle T_{1}}$ spaces are characterized by this property.

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

The set of limit points of ${\displaystyle S}$ is called the derived set of ${\displaystyle S}$.

Types of limit points

If every open set containing ${\displaystyle x}$ contains infinitely many points of ${\displaystyle S}$, then ${\displaystyle x}$ is a specific type of limit point called an ${\displaystyle \omega }$-accumulation point of ${\displaystyle S}$.

If every open set containing ${\displaystyle x}$ contains uncountably many points of ${\displaystyle S}$, then ${\displaystyle x}$ is a specific type of limit point called a condensation point of ${\displaystyle S}$.

If every open set ${\displaystyle U}$ containing ${\displaystyle x}$ satisfies ${\displaystyle \left|U\cap S\right|=\left|S\right|}$, then ${\displaystyle x}$ is a specific type of limit point called a complete accumulation point of ${\displaystyle S}$.

For sequences and nets

A sequence enumerating all positive rational numbers. Each positive real number is a cluster point.

In a topological space ${\displaystyle X}$, a point ${\displaystyle x\in X}$ is said to be a cluster point (or accumulation point) of a sequence ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ if, for every neighbourhood ${\displaystyle V}$ of ${\displaystyle x}$, there are infinitely many ${\displaystyle n\in \mathbb {N} }$ such that ${\displaystyle x_{n}\in V}$. It is equivalent to say that for every neighbourhood ${\displaystyle V}$ of ${\displaystyle x}$ and every ${\displaystyle n_{0}\in \mathbb {N} }$, there is some ${\displaystyle n\geq n_{0}}$ such that ${\displaystyle x_{n}\in V}$. If ${\displaystyle X}$ is a metric space or a first-countable space (or, more generally, a Fréchet–Urysohn space), then ${\displaystyle x}$ is cluster point of ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ if and only if ${\displaystyle x}$ is a limit of some subsequence of ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$. 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 ${\displaystyle f:(P,\leq )\to X}$, where ${\displaystyle (P,\leq )}$ is a directed set and ${\displaystyle X}$ is a topological space. A point ${\displaystyle x\in X}$ is said to be a cluster point (or accumulation point) of the net ${\displaystyle f}$ if, for every neighbourhood ${\displaystyle V}$ of ${\displaystyle x}$ and every ${\displaystyle p_{0}\in P}$, there is some ${\displaystyle p\geq p_{0}}$ such that ${\displaystyle f(p)\in V}$, equivalently, if ${\displaystyle f}$ has a subnet which converges to ${\displaystyle x}$. 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.

Selected facts

• We have the following characterization of limit points: ${\displaystyle x}$ is a limit point of ${\displaystyle S}$ if and only if it is in the closure of ${\displaystyle S\setminus \{x\}}$.
  • 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, ${\displaystyle x}$ is a limit point of ${\displaystyle S}$, if and only if every neighborhood of ${\displaystyle x}$ contains a point of ${\displaystyle S}$ other than ${\displaystyle x}$, if and only if every neighborhood of ${\displaystyle x}$ contains a point of ${\displaystyle S\setminus \{x\}}$, if and only if ${\displaystyle x}$ is in the closure of ${\displaystyle S\setminus \{x\}}$.
• If we use ${\displaystyle L(S)}$ to denote the set of limit points of ${\displaystyle S}$, then we have the following characterization of the closure of ${\displaystyle S}$: The closure of ${\displaystyle S}$ is equal to the union of ${\displaystyle S}$ and ${\displaystyle L(S)}$. This fact is sometimes taken as the definition of closure.
  • Proof: ("Left subset") Suppose ${\displaystyle x}$ is in the closure of ${\displaystyle S}$. If ${\displaystyle x}$ is in ${\displaystyle S}$, we are done. If ${\displaystyle x}$ is not in ${\displaystyle S}$, then every neighbourhood of ${\displaystyle x}$ contains a point of ${\displaystyle S}$, and this point cannot be ${\displaystyle x}$. In other words, ${\displaystyle x}$ is a limit point of ${\displaystyle S}$ and ${\displaystyle x}$ is in ${\displaystyle L(S)}$. ("Right subset") If ${\displaystyle x}$ is in ${\displaystyle S}$, then every neighbourhood of ${\displaystyle x}$ clearly meets ${\displaystyle S}$, so ${\displaystyle x}$ is in the closure of ${\displaystyle S}$. If ${\displaystyle x}$ is in ${\displaystyle L(S)}$, then every neighbourhood of ${\displaystyle x}$ contains a point of ${\displaystyle S}$ (other than ${\displaystyle x}$), so ${\displaystyle x}$ is again in the closure of ${\displaystyle S}$. This completes the proof.
• A corollary of this result gives us a characterisation of closed sets: A set ${\displaystyle S}$ is closed if and only if it contains all of its limit points.
  • Proof: ${\displaystyle S}$ is closed if and only if ${\displaystyle S}$ is equal to its closure if and only if ${\displaystyle S=S\cup L(S)}$ if and only if ${\displaystyle L(S)}$ is contained in ${\displaystyle S}$.
  • Another proof: Let ${\displaystyle S}$ be a closed set and ${\displaystyle x}$ a limit point of ${\displaystyle S}$. If ${\displaystyle x}$ is not in ${\displaystyle S}$, then the complement to ${\displaystyle S}$ comprises an open neighbourhood of ${\displaystyle x}$. Since ${\displaystyle x}$ is a limit point of ${\displaystyle S}$, any open neighbourhood of ${\displaystyle x}$ should have a non-trivial intersection with ${\displaystyle S}$. However, a set can not have a non-trivial intersection with its complement. Conversely, assume ${\displaystyle S}$ contains all its limit points. We shall show that the complement of ${\displaystyle S}$ is an open set. Let ${\displaystyle x}$ be a point in the complement of ${\displaystyle S}$. By assumption, ${\displaystyle x}$ is not a limit point, and hence there exists an open neighbourhood U of ${\displaystyle x}$ that does not intersect ${\displaystyle S}$, and so ${\displaystyle U}$ lies entirely in the complement of ${\displaystyle S}$. Since this argument holds for arbitrary ${\displaystyle x}$ in the complement of ${\displaystyle S}$, the complement of ${\displaystyle S}$ can be expressed as a union of open neighbourhoods of the points in the complement of ${\displaystyle S}$. Hence the complement of ${\displaystyle S}$ is open.
• No isolated point is a limit point of any set.
  • Proof: If ${\displaystyle x}$ is an isolated point, then ${\displaystyle \{x\}}$ is a neighbourhood of ${\displaystyle x}$ that contains no points other than ${\displaystyle x}$.
• The closure ${\displaystyle cl(S)}$ of a set ${\displaystyle S}$ is a disjoint union of its limit points ${\displaystyle L(S)}$ and isolated points ${\displaystyle I(S)}$:

${\displaystyle cl(S)=L(S)\cup I(S),L(S)\cap I(S)=\emptyset .}$

• A space ${\displaystyle X}$ is discrete if and only if no subset of ${\displaystyle X}$ has a limit point.
  • Proof: If ${\displaystyle X}$ is discrete, then every point is isolated and cannot be a limit point of any set. Conversely, if ${\displaystyle X}$ is not discrete, then there is a singleton ${\displaystyle \{x\}}$ that is not open. Hence, every open neighbourhood of ${\displaystyle \{x\}}$ contains a point ${\displaystyle y\neq x}$, and so ${\displaystyle x}$ is a limit point of ${\displaystyle X}$.
• If a space ${\displaystyle X}$ has the trivial topology and ${\displaystyle S}$ is a subset of ${\displaystyle X}$ with more than one element, then all elements of ${\displaystyle X}$ are limit points of ${\displaystyle S}$. If ${\displaystyle S}$ is a singleton, then every point of ${\displaystyle X\setminus S}$ is a limit point of ${\displaystyle S}$.
  • Proof: As long as ${\displaystyle S\setminus \{x\}}$} is nonempty, its closure will be ${\displaystyle X}$. It's only empty when ${\displaystyle S}$ is empty or ${\displaystyle x}$ is the unique element of ${\displaystyle S}$.
• By definition, every limit point is an adherent point.

See also

• Derived set (mathematics)
• Adherent point
• Condensation point
• Isolated point

References

• "Limit point of a set", Encyclopedia of Mathematics, EMS Press, 2001 [1994]

External links

• "limit point". PlanetMath.