Equivalence of Definitions of Adherent Point

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 adherent point of $A$ are equivalent:

Definition from Open Neighborhood

A point $x \in S$ is an adherent point of $A$ if and only if every open neighborhood $U$ of $x$ satisfies:

$A \cap U \ne \O$

Definition from Closure

A point $x \in S$ is an adherent point of $A$ if and only if $x$ is an element of the closure of $A$.

Definition from Neighborhood

A point $x \in S$ is an adherent point of $A$ if and only if every neighborhood $N$ of $x$ satisfies:

$A \cap N \ne \O$


Proof

Definition from Open Neighborhood is equivalent to Definition from Closure

Let $A^-$ denote the closure of $A$.


It is required to be shown that $x \in A^-$ if and only if, for every open neighborhood $U$ of $x$, the intersection $A \cap U$ is non-empty.


For a subset $H \subseteq S$, let $H^{\complement}$ denote the relative complement of $H$ in $S$.


We have that:

\(\displaystyle A \cap U\) \(=\) \(\displaystyle \O\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle A\) \(\subseteq\) \(\displaystyle U^{\complement}\) Note that $U^{\complement}$ is closed
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle A^-\) \(\subseteq\) \(\displaystyle U^{\complement}\) Set Closure is Smallest Closed Set
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle U\) \(\subseteq\) \(\displaystyle \paren {A^-}^{\complement}\) Relative Complement inverts Subsets and Relative Complement of Relative Complement
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle A^- \cap U\) \(=\) \(\displaystyle \O\)

Thus:

$x \in U \iff x \notin A^-$

The result follows from the Rule of Transposition.

$\Box$


Definition from Open Neighborhood is equivalent to Definition from Neighborhood

Necessary Condition

Let every open neighborhood $U$ of $x$ satisfy:

$A \cap U \ne \O$

Let $N$ be any neighborhood of $x$.

By definition of a neighborhood:

$\exists V \in \tau : x \in V \subseteq N$

From Set is Open iff Neighborhood of all its Points, $V$ is an open neighborhood of $x$.

Thus:

$A \cap V \ne \O$

By the contrapositive statement of Subsets of Disjoint Sets are Disjoint them:

$A \cap N \ne \O$

Since $N$ was arbitrary then every neighborhood $N$ of $x$ satisfies:

$A \cap N \ne \O$

$\Box$


Sufficient Condition

Let every neighborhood $N$ of $x$ satisfy:

$A \cap N \ne \O$

By definition, every open neighborhood $U$ of $x$ is a neighborhood of $x$.

So every open neighborhood $U$ of $x$ satisfies:

$A \cap U \ne \O$

$\blacksquare$