Equivalence of Definitions of Limit Point/Definition (1) iff Definition (4)
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 1
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 4
A 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 $x$.
Proof 1
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.
$\blacksquare$
Proof 2
The following equivalence holds:
There exists an open neighborhood $U$ of $x$ such that $A \cap \paren {U \setminus \set x} = \O$
\(\ds \O\) | \(=\) | \(\ds A \cap \paren {U \setminus \set x}\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \O\) | \(=\) | \(\ds \paren {U \cap A} \setminus \set x\) | \(\quad\) Intersection with Set Difference is Set Difference with Intersection | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \O\) | \(=\) | \(\ds \paren {A \cap U} \setminus \set x\) | \(\quad\) Intersection is Commutative | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \O\) | \(=\) | \(\ds U \cap \paren {A \setminus \set x}\) | \(\quad\) Intersection with Set Difference is Set Difference with Intersection | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \O\) | \(=\) | \(\ds U \cap \map \complement { \map \complement {A \setminus \set x} }\) | \(\quad\) Complement of Complement | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds U\) | \(\subseteq\) | \(\ds \map \complement {A \setminus \set x}\) | \(\quad\) Intersection with Complement is Empty iff Subset | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds U\) | \(\subseteq\) | \(\ds \map \complement {A \cap \map \complement {\set x} }\) | \(\quad\) Set Difference as Intersection with Complement | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds U\) | \(\subseteq\) | \(\ds \map \complement A \cup \map \complement {\map \complement {\set x} }\) | \(\quad\) De Morgan's Laws (Set Theory)/Set Complement | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds U\) | \(\subseteq\) | \(\ds \map \complement A \cup \set x\) | \(\quad\) Complement of Complement |
By Definition of Neighborhood of Point, $\map \complement A \cup \set x$ is a neighborhood of $x$
The result follows from the Rule of Transposition.
$\blacksquare$