Limit point in topology Examples
From Knowino Show
In topology, a limit point (or "accumulation point") of a subset S of a topological space X is a point x that cannot be separated from S. [edit] DefinitionFormally, x is a limit point of S if every neighbourhood of x contains a point of S other than x itself. A limit point of S need not belong to S, but may belong to it. [edit] Metric spaceIn a metric space (X,d), a limit point of a set S may be defined as a point x such that for all ε > 0 there exists a point y in S such that This agrees with the topological definition given above. [edit] Properties
[edit] Derived setThe derived set of S is the set of all limit points of S. A point of S which is not a limit point is an isolated point of S. A set with no isolated points is dense-in-itself. A set is perfect if it is closed and dense-in-itself; equivalently a perfect set is equal to its derived set. [edit][edit] Limit point of a sequenceA limit point of a sequence (an) in a topological space X is a point x such that every neighbourhood U of x contains all points of the sequence with numbers above some n(U). A limit point of the sequence (an) need not be a limit point of the set {an}. [edit] Adherent pointA point x is an adherent point or contact point of a set S if every neighbourhood of x contains a point of S (not necessarily distinct from x). A point x is an ω-accumulation point of a set S if every neighbourhood of x contains infinitely many points of S. [edit] Condensation pointA point x is a condensation point of a set S if every neighbourhood of x contains uncountably many points of S. [edit] References
From ProofWiki
DefinitionLet $T = \struct {S, \tau}$ be a topological space. Limit Point of SetLet $A \subseteq S$. Definition from Open NeighborhoodA point $x \in S$ is a limit point of $A$ if and only if every open neighborhood $U$ of $x$ satisfies: $A \cap \paren {U \setminus \set x} \ne \O$That is, if and only if every open set $U \in \tau$ such that $x \in U$ contains some point of $A$ distinct from $x$.
$\forall U\in \tau :x\in U \implies A \cap \paren {U \setminus \set x} \ne \O\text{.}$ Definition from ClosureA point $x \in S$ is a limit point of $A$ if and only if $x$ belongs to the closure of $A$ but is not an isolated point of $A$.Definition from Adherent PointA point $x \in S$ is a limit point of $A$ if and only if $x$ is an adherent point of $A$ but is not an isolated point of $A$. Definition from Relative ComplementA point $x \in S$ is a limit point of $A$ if and only if $\left({S \setminus A}\right) \cup \left\{{x}\right\}$ is not a neighborhood of $x$. Limit Point of PointThe concept of a limit point can be sharpened to apply to individual points, as follows:
That is, it is a limit point of the singleton $\set a$. Limit Point of SequenceLet $T = \struct {S, \tau}$ be a topological space. Let $A \subseteq S$. Let $\sequence {x_n}$ be a sequence in $A$. Let $\sequence {x_n}$ converge to a value $\alpha \in S$.
ExamplesEnd Points of Real IntervalThe real number $a$ is a limit point of both the open real interval $\openint a b$ as well as of the closed real interval $\closedint a b$. It is noted that $a \in \closedint a b$ but $a \notin \openint a b$. Union of Singleton with Open Real IntervalLet $\R$ be the set of real numbers. Let $H \subseteq \R$ be the subset of $\R$ defined as: $H = \set 0 \cup \openint 1 2$Then $0$ is not a limit point of $H$. Real Number is Limit Point of Rational Numbers in Real NumbersLet $\R$ be the set of real numbers. Let $\Q$ be the set of rational numbers. Let $x \in \R$.
Zero is Limit Point of Integer Reciprocal SpaceLet $A \subseteq \R$ be the set of all points on $\R$ defined as: $A := \set {\dfrac 1 n : n \in \Z_{>0} }$Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Also
Let $(X,\tau)$ be a topological space where $X = \{a,b,c,d\}$, $\tau=\{\emptyset,X,\{a\},\{a,b\},\{a,c\},\{a,b,c\}\}$. Then what are limit points of the set $A = \{a,c,d\}$? Is it true that $b$, $c$ and $d$ are limit points? See also: Limit of a function and Limit of a sequence In mathematics, a limit point (or cluster point or accumulation point) of a set
S
{\displaystyle S}
The limit points of a set should not be confused with adherent points for which every neighbourhood of
x
{\displaystyle x}
contains a point of
S
{\displaystyle S}
. Unlike for limit points, this point of
S
{\displaystyle S}
may be
x
{\displaystyle x}
itself. A limit point can be characterized as an adherent point that is not an isolated point.
Limit points of a set should also not be confused with boundary points. For example,
0
{\displaystyle 0}
is a boundary point (but not a limit point) of set
{
0
}
{\displaystyle \{0\}}
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. Let S {\displaystyle S} be a subset of a topological space X . {\displaystyle X.} A point x {\displaystyle x} in X {\displaystyle X} is a limit point or cluster point or accumulation point of the set S {\displaystyle S} if every neighbourhood of x {\displaystyle x} contains at least one point of S {\displaystyle S} different from x {\displaystyle 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 {\displaystyle X} is a T 1 {\displaystyle T_{1}} space (such as a metric space), then x ∈ X {\displaystyle x\in X} is a limit point of S {\displaystyle S} if and only if every neighbourhood of x {\displaystyle x} contains infinitely many points of S . {\displaystyle S.} [6] In fact, T 1 {\displaystyle T_{1}} spaces are characterized by this property. If X {\displaystyle X} is a Fréchet–Urysohn space (which all metric spaces and first-countable spaces are), then x ∈ X {\displaystyle x\in X} is a limit point of S {\displaystyle S} if and only if there is a sequence of points in S ∖ { x } {\displaystyle S\setminus \{x\}} whose limit is x . {\displaystyle x.} In fact, Fréchet–Urysohn spaces are characterized by this property. The set of limit points of S {\displaystyle S} is called the derived set of S . {\displaystyle S.} Types of accumulation pointsIf every neighbourhood of x {\displaystyle x} contains infinitely many points of S , {\displaystyle S,} then x {\displaystyle x} is a specific type of limit point called an ω-accumulation point of S . {\displaystyle S.} If every neighbourhood of x {\displaystyle x} contains uncountably many points of S , {\displaystyle S,} then x {\displaystyle x} is a specific type of limit point called a condensation point of S . {\displaystyle S.} If every neighbourhood U {\displaystyle U} of x {\displaystyle x} satisfies | U ∩ S | = | S | , {\displaystyle \left|U\cap S\right|=\left|S\right|,} then x {\displaystyle x} is a specific type of limit point called a complete accumulation point of S . {\displaystyle S.} Accumulation points of sequences and netsA sequence enumerating all positive rational numbers. Each positive real number is a cluster point. In a topological space X , {\displaystyle X,} a point x ∈ X {\displaystyle x\in X} is said to be a cluster point or accumulation point of a sequence x ∙ = ( x n ) n = 1 ∞ {\displaystyle x_{\bullet }=\left(x_{n}\right)_{n=1}^{\infty }} if, for every neighbourhood V {\displaystyle V} of x , {\displaystyle x,} there are infinitely many n ∈ N {\displaystyle n\in \mathbb {N} } such that x n ∈ V . {\displaystyle x_{n}\in V.} It is equivalent to say that for every neighbourhood V {\displaystyle V} of x {\displaystyle x} and every n 0 ∈ N , {\displaystyle n_{0}\in \mathbb {N} ,} there is some n ≥ n 0 {\displaystyle n\geq n_{0}} such that x n ∈ V . {\displaystyle x_{n}\in V.} If X {\displaystyle X} is a metric space or a first-countable space (or, more generally, a Fréchet–Urysohn space), then x {\displaystyle x} is a cluster point of x ∙ {\displaystyle x_{\bullet }} if and only if x {\displaystyle x} is a limit of some subsequence of x ∙ . {\displaystyle x_{\bullet }.} 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 {\displaystyle x} to which the sequence converges (that is, every neighborhood of x {\displaystyle x} contains all but finitely many elements of the sequence). That is why we do not use the term limit point 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 , ≤ ) → X , {\displaystyle f:(P,\leq )\to X,} where ( P , ≤ ) {\displaystyle (P,\leq )} is a directed set and X {\displaystyle X} is a topological space. A point x ∈ X {\displaystyle x\in X} is said to be a cluster point or accumulation point of a net f {\displaystyle f} if, for every neighbourhood V {\displaystyle V} of x {\displaystyle x} and every p 0 ∈ P , {\displaystyle p_{0}\in P,} there is some p ≥ p 0 {\displaystyle p\geq p_{0}} such that f ( p ) ∈ V , {\displaystyle f(p)\in V,} equivalently, if f {\displaystyle f} has a subnet which converges to x . {\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 filters. Every sequence x ∙ = ( x n ) n = 1 ∞ {\displaystyle x_{\bullet }=\left(x_{n}\right)_{n=1}^{\infty }} in X {\displaystyle X} is by definition just a map x ∙ : N → X {\displaystyle x_{\bullet }:\mathbb {N} \to X} so that its image Im x ∙ := { x n : n ∈ N } {\displaystyle \operatorname {Im} x_{\bullet }:=\left\{x_{n}:n\in \mathbb {N} \right\}} can be defined in the usual way.
Conversely, given a countable infinite set A ⊆ X {\displaystyle A\subseteq X} in X , {\displaystyle X,} we can enumerate all the elements of A {\displaystyle A} in many ways, even with repeats, and thus associate with it many sequences x ∙ {\displaystyle x_{\bullet }} that will satisfy A = Im x ∙ . {\displaystyle A=\operatorname {Im} x_{\bullet }.}
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 cl ( S ) {\displaystyle \operatorname {cl} (S)} of a set S {\displaystyle S} is a disjoint union of its limit points L ( S ) {\displaystyle L(S)} and isolated points I ( S ) {\displaystyle I(S)} : cl ( S ) = L ( S ) ∪ I ( S ) , L ( S ) ∩ I ( S ) = ∅ . {\displaystyle \operatorname {cl} (S)=L(S)\cup I(S),L(S)\cap I(S)=\varnothing .} A point x ∈ X {\displaystyle x\in X} is a limit point of S ⊆ X {\displaystyle S\subseteq X} if and only if it is in the closure of S ∖ { x } . {\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, x {\displaystyle x} is a limit point of S , {\displaystyle S,} if and only if every neighborhood of x {\displaystyle x} contains a point of S {\displaystyle S} other than x , {\displaystyle x,} if and only if every neighborhood of x {\displaystyle x} contains a point of S ∖ { x } , {\displaystyle S\setminus \{x\},} if and only if x {\displaystyle x} is in the closure of S ∖ { x } . {\displaystyle S\setminus \{x\}.} If we use L ( S ) {\displaystyle L(S)} to denote the set of limit points of S , {\displaystyle S,} then we have the following characterization of the closure of S {\displaystyle S} : The closure of S {\displaystyle S} is equal to the union of S {\displaystyle S} and L ( S ) . {\displaystyle L(S).} This fact is sometimes taken as the definition of closure. Proof ("Left subset") Suppose x {\displaystyle x} is in the closure of S . {\displaystyle S.} If x {\displaystyle x} is in S , {\displaystyle S,} we are done. If x {\displaystyle x} is not in S , {\displaystyle S,} then every neighbourhood of x {\displaystyle x} contains a point of S , {\displaystyle S,} and this point cannot be x . {\displaystyle x.} In other words, x {\displaystyle x} is a limit point of S {\displaystyle S} and x {\displaystyle x} is in L ( S ) . {\displaystyle L(S).} ("Right subset") If x {\displaystyle x} is in S , {\displaystyle S,} then every neighbourhood of x {\displaystyle x} clearly meets S , {\displaystyle S,} so x {\displaystyle x} is in the closure of S . {\displaystyle S.} If x {\displaystyle x} is in L ( S ) , {\displaystyle L(S),} then every neighbourhood of x {\displaystyle x} contains a point of S {\displaystyle S} (other than x {\displaystyle x} ), so x {\displaystyle x} is again in the closure of S . {\displaystyle S.} This completes the proof. A corollary of this result gives us a characterisation of closed sets: A set S {\displaystyle S} is closed if and only if it contains all of its limit points. Proof Proof 1: S {\displaystyle S} is closed if and only if S {\displaystyle S} is equal to its closure if and only if S = S ∪ L ( S ) {\displaystyle S=S\cup L(S)} if and only if L ( S ) {\displaystyle L(S)} is contained in S . {\displaystyle S.} Proof 2: Let S {\displaystyle S} be a closed set and x {\displaystyle x} a limit point of S . {\displaystyle S.} If x {\displaystyle x} is not in S , {\displaystyle S,} then the complement to S {\displaystyle S} comprises an open neighbourhood of x . {\displaystyle x.} Since x {\displaystyle x} is a limit point of S , {\displaystyle S,} any open neighbourhood of x {\displaystyle x} should have a non-trivial intersection with S . {\displaystyle S.} However, a set can not have a non-trivial intersection with its complement. Conversely, assume S {\displaystyle S} contains all its limit points. We shall show that the complement of S {\displaystyle S} is an open set. Let x {\displaystyle x} be a point in the complement of S . {\displaystyle S.} By assumption, x {\displaystyle x} is not a limit point, and hence there exists an open neighbourhood U {\displaystyle U} of x {\displaystyle x} that does not intersect S , {\displaystyle S,} and so U {\displaystyle U} lies entirely in the complement of S . {\displaystyle S.} Since this argument holds for arbitrary x {\displaystyle x} in the complement of S , {\displaystyle S,} the complement of S {\displaystyle S} can be expressed as a union of open neighbourhoods of the points in the complement of S . {\displaystyle S.} Hence the complement of S {\displaystyle S} is open. No isolated point is a limit point of any set. Proof If x {\displaystyle x} is an isolated point, then { x } {\displaystyle \{x\}} is a neighbourhood of x {\displaystyle x} that contains no points other than x . {\displaystyle x.} A space X {\displaystyle X} is discrete if and only if no subset of X {\displaystyle X} has a limit point. Proof If X {\displaystyle X} is discrete, then every point is isolated and cannot be a limit point of any set. Conversely, if X {\displaystyle X} is not discrete, then there is a singleton { x } {\displaystyle \{x\}} that is not open. Hence, every open neighbourhood of { x } {\displaystyle \{x\}} contains a point y ≠ x , {\displaystyle y\neq x,} and so x {\displaystyle x} is a limit point of X . {\displaystyle X.} If a space X {\displaystyle X} has the trivial topology and S {\displaystyle S} is a subset of X {\displaystyle X} with more than one element, then all elements of X {\displaystyle X} are limit points of S . {\displaystyle S.} If S {\displaystyle S} is a singleton, then every point of X ∖ S {\displaystyle X\setminus S} is a limit point of S . {\displaystyle S.} Proof As long as S ∖ { x } {\displaystyle S\setminus \{x\}} is nonempty, its closure will be X . {\displaystyle X.} It is only empty when S {\displaystyle S} is empty or x {\displaystyle x} is the unique element of S . {\displaystyle S.}
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