Equivalence of Definitions of Limit Point/Definition (1) iff Definition (4)/Proof 2

From ProofWiki
Jump to navigation Jump to search

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

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.

$\blacksquare$