# 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 | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle U\) | \(\subseteq\) | \(\displaystyle \paren {A^-}^{\complement}\) | 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:

- $A \cap U \ne \O$

$\blacksquare$