Equivalence of Definitions of Limit Point of Set
Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
The following definitions of the concept of limit point are equivalent:
Definition from Open Neighborhood
A 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 $x$.
Definition from Closure
A 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 Point
A 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 Complement
A point $x \in S$ is a limit point of $A$ if and only if $\paren {S \setminus A} \cup \set x$ is not a neighborhood of $x$.
Proof
Definition from Open Neighborhood $\iff$ Definition from Closure
The closure of $A$ is defined as the union of the set of all isolated points of $A$ and the set of all limit points of $A$.
The rest then follows directly from Equivalence of Definitions of Isolated Point.
$\Box$
Definition from Closure $\iff$ Definition from Adherent Point
Follows directly from Equivalence of Definitions of Adherent Point.
$\Box$
Definition from Open Neighborhood $\iff$ Definition from Relative Complement
The following equivalence holds:
\(\ds \) | \(\) | \(\ds \) | There exists an open neighborhood $U$ of $x$ such that $A \cap \paren {U \setminus \set x} = \O$ | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \) | There exists an open neighborhood $U$ of $x$ such that $U \subseteq \paren{S \setminus A} \cup \set x$ | \(\quad\) Modus Ponendo Tollens | ||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \) | $\paren {S \setminus A} \cup \set x$ is a neighborhood of $x$ | \(\quad\) Definition of Neighborhood of Point |
The result follows from the Rule of Transposition.
$\blacksquare$