# Equivalence of Definitions of Adherent Point

## 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 in Topological Space $\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.

$\Box$

### 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:

$A \cap V \ne \O$
$A \cap N \ne \O$

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:

$A \cap U \ne \O$

$\blacksquare$