Adherent point

In mathematics, an adherent point (also closure point or point of closure or contact point) of a subset A {\displaystyle A} of a topological space X , {\displaystyle X,} is a point x {\displaystyle x} in X {\displaystyle X} such that every neighbourhood of x {\displaystyle x} (or equivalently, every open neighborhood of x {\displaystyle x}) contains at least one point of A . {\displaystyle A.} A point x ∈ X {\displaystyle x\in X} is an adherent point for A {\displaystyle A} if and only if x {\displaystyle x} is in the closure of A , {\displaystyle A,} thus

x ∈ Cl X ⁡ A {\displaystyle x\in \operatorname {Cl} _{X}A} if and only if for all open subsets U ⊆ X , {\displaystyle U\subseteq X,} if x ∈ U then U ∩ A ≠ ∅ . {\displaystyle x\in U{\text{ then }}U\cap A\neq \varnothing .}

This definition differs from that of a limit point of a set, in that for a limit point it is required that every neighborhood of x {\displaystyle x} contains at least one point of A {\displaystyle A} different from x . {\displaystyle x.} Thus every limit point is an adherent point, but the converse is not true. An adherent point of A {\displaystyle A} is either a limit point of A {\displaystyle A} or an element of A {\displaystyle A} (or both). An adherent point which is not a limit point is an isolated point.

Intuitively, having an open set A {\displaystyle A} defined as the area within (but not including) some boundary, the adherent points of A {\displaystyle A} are those of A {\displaystyle A} including the boundary.

Examples and sufficient conditions

If S {\displaystyle S} is a non-empty subset of R {\displaystyle \mathbb {R} } which is bounded above, then the supremum sup S {\displaystyle \sup S} is adherent to S . {\displaystyle S.} In the interval ( a , b ] , {\displaystyle (a,b],} a {\displaystyle a} is an adherent point that is not in the interval, with usual topology of R . {\displaystyle \mathbb {R} .}

A subset S {\displaystyle S} of a metric space M {\displaystyle M} contains all of its adherent points if and only if S {\displaystyle S} is (sequentially) closed in M . {\displaystyle M.}

Adherent points and subspaces

Suppose x ∈ X {\displaystyle x\in X} and S ⊆ X ⊆ Y , {\displaystyle S\subseteq X\subseteq Y,} where X {\displaystyle X} is a topological subspace of Y {\displaystyle Y} (that is, X {\displaystyle X} is endowed with the subspace topology induced on it by Y {\displaystyle Y}). Then x {\displaystyle x} is an adherent point of S {\displaystyle S} in X {\displaystyle X} if and only if x {\displaystyle x} is an adherent point of S {\displaystyle S} in Y . {\displaystyle Y.}

Proof
By assumption, S ⊆ X ⊆ Y {\displaystyle S\subseteq X\subseteq Y} and x ∈ X . {\displaystyle x\in X.} Assuming that x ∈ Cl X ⁡ S , {\displaystyle x\in \operatorname {Cl} _{X}S,} let V {\displaystyle V} be a neighborhood of x {\displaystyle x} in Y {\displaystyle Y} so that x ∈ Cl Y ⁡ S {\displaystyle x\in \operatorname {Cl} _{Y}S} will follow once it is shown that V ∩ S ≠ ∅ . {\displaystyle V\cap S\neq \varnothing .} The set U := V ∩ X {\displaystyle U:=V\cap X} is a neighborhood of x {\displaystyle x} in X {\displaystyle X} (by definition of the subspace topology) so that x ∈ Cl X ⁡ S {\displaystyle x\in \operatorname {Cl} _{X}S} implies that ∅ ≠ U ∩ S . {\displaystyle \varnothing \neq U\cap S.} Thus ∅ ≠ U ∩ S = ( V ∩ X ) ∩ S ⊆ V ∩ S , {\displaystyle \varnothing \neq U\cap S=(V\cap X)\cap S\subseteq V\cap S,} as desired. For the converse, assume that x ∈ Cl Y ⁡ S {\displaystyle x\in \operatorname {Cl} _{Y}S} and let U {\displaystyle U} be a neighborhood of x {\displaystyle x} in X {\displaystyle X} so that x ∈ Cl X ⁡ S {\displaystyle x\in \operatorname {Cl} _{X}S} will follow once it is shown that U ∩ S ≠ ∅ . {\displaystyle U\cap S\neq \varnothing .} By definition of the subspace topology, there exists a neighborhood V {\displaystyle V} of x {\displaystyle x} in Y {\displaystyle Y} such that U = V ∩ X . {\displaystyle U=V\cap X.} Now x ∈ Cl Y ⁡ S {\displaystyle x\in \operatorname {Cl} _{Y}S} implies that ∅ ≠ V ∩ S . {\displaystyle \varnothing \neq V\cap S.} From S ⊆ X {\displaystyle S\subseteq X} it follows that S = X ∩ S {\displaystyle S=X\cap S} and so ∅ ≠ V ∩ S = V ∩ ( X ∩ S ) = ( V ∩ X ) ∩ S = U ∩ S , {\displaystyle \varnothing \neq V\cap S=V\cap (X\cap S)=(V\cap X)\cap S=U\cap S,} as desired. ◼ {\displaystyle \blacksquare }

Proof

By assumption, S ⊆ X ⊆ Y {\displaystyle S\subseteq X\subseteq Y} and x ∈ X . {\displaystyle x\in X.} Assuming that x ∈ Cl X ⁡ S , {\displaystyle x\in \operatorname {Cl} _{X}S,} let V {\displaystyle V} be a neighborhood of x {\displaystyle x} in Y {\displaystyle Y} so that x ∈ Cl Y ⁡ S {\displaystyle x\in \operatorname {Cl} _{Y}S} will follow once it is shown that V ∩ S ≠ ∅ . {\displaystyle V\cap S\neq \varnothing .} The set U := V ∩ X {\displaystyle U:=V\cap X} is a neighborhood of x {\displaystyle x} in X {\displaystyle X} (by definition of the subspace topology) so that x ∈ Cl X ⁡ S {\displaystyle x\in \operatorname {Cl} _{X}S} implies that ∅ ≠ U ∩ S . {\displaystyle \varnothing \neq U\cap S.} Thus ∅ ≠ U ∩ S = ( V ∩ X ) ∩ S ⊆ V ∩ S , {\displaystyle \varnothing \neq U\cap S=(V\cap X)\cap S\subseteq V\cap S,} as desired. For the converse, assume that x ∈ Cl Y ⁡ S {\displaystyle x\in \operatorname {Cl} _{Y}S} and let U {\displaystyle U} be a neighborhood of x {\displaystyle x} in X {\displaystyle X} so that x ∈ Cl X ⁡ S {\displaystyle x\in \operatorname {Cl} _{X}S} will follow once it is shown that U ∩ S ≠ ∅ . {\displaystyle U\cap S\neq \varnothing .} By definition of the subspace topology, there exists a neighborhood V {\displaystyle V} of x {\displaystyle x} in Y {\displaystyle Y} such that U = V ∩ X . {\displaystyle U=V\cap X.} Now x ∈ Cl Y ⁡ S {\displaystyle x\in \operatorname {Cl} _{Y}S} implies that ∅ ≠ V ∩ S . {\displaystyle \varnothing \neq V\cap S.} From S ⊆ X {\displaystyle S\subseteq X} it follows that S = X ∩ S {\displaystyle S=X\cap S} and so ∅ ≠ V ∩ S = V ∩ ( X ∩ S ) = ( V ∩ X ) ∩ S = U ∩ S , {\displaystyle \varnothing \neq V\cap S=V\cap (X\cap S)=(V\cap X)\cap S=U\cap S,} as desired. ◼ {\displaystyle \blacksquare }

Consequently, x {\displaystyle x} is an adherent point of S {\displaystyle S} in X {\displaystyle X} if and only if this is true of x {\displaystyle x} in every (or alternatively, in some) topological superspace of X . {\displaystyle X.}

Adherent points and sequences

If S {\displaystyle S} is a subset of a topological space then the limit of a convergent sequence in S {\displaystyle S} does not necessarily belong to S , {\displaystyle S,} however it is always an adherent point of S . {\displaystyle S.} Let ( x n ) n ∈ N {\displaystyle \left(x_{n}\right)_{n\in \mathbb {N} }} be such a sequence and let x {\displaystyle x} be its limit. Then by definition of limit, for all neighbourhoods U {\displaystyle U} of x {\displaystyle x} there exists n ∈ N {\displaystyle n\in \mathbb {N} } such that x n ∈ U {\displaystyle x_{n}\in U} for all n ≥ N . {\displaystyle n\geq N.} In particular, x N ∈ U {\displaystyle x_{N}\in U} and also x N ∈ S , {\displaystyle x_{N}\in S,} so x {\displaystyle x} is an adherent point of S . {\displaystyle S.} In contrast to the previous example, the limit of a convergent sequence in S {\displaystyle S} is not necessarily a limit point of S {\displaystyle S}; for example consider S = { 0 } {\displaystyle S=\{0\}} as a subset of R . {\displaystyle \mathbb {R} .} Then the only sequence in S {\displaystyle S} is the constant sequence 0 , 0 , … {\displaystyle 0,0,\ldots } whose limit is 0 , {\displaystyle 0,} but 0 {\displaystyle 0} is not a limit point of S ; {\displaystyle S;} it is only an adherent point of S . {\displaystyle S.}

Notes

Citations