Table of Contents

The Open Neighbourhoods of Points in a Topological Space
Recall from The Open and Closed Sets of a Topological Space page that if $(X, \tau)$ is a topological space then a set $A \subseteq X$ is said to be open if $A \in \tau$ and $A$ is said to be closed if $A^c \in \tau$. Furthermore, if $A$ is both open and closed, then we say that $A$ is clopen.
We will now look at another very important definition with regards to topological spaces known as open neighbourhoods of points $x \in X$ of a topological space $(X, \tau)$ which we define below.
Definition: Let $(X, \tau)$ be a topological space. A Neighbourhood of a point $x \in X$ is a set $N \subseteq X$ for which there exists an open set $O \subseteq N$ such that $x \in O$. An Open Neighbourhood of the point $x \in X$ is any (open) set $U \in \tau$ such that $x \in U$. 
The term "neighbourhood" is used frequently in topology to simply mean "open neighbourhood" when distinction is not important.
Before we look at examples we will look at the following proposition which gives us criterion for when a set is open.
Proposition 1 (Open Neighbourhood Criterion for Open Sets): Let $(X, \tau)$ be a topological space and let $U \subseteq X$. Then $U$ is open if and only if for every $x \in X$ there exists an open neighbourhood $U_x$ of $x$ such that $U_x \subseteq U$. 
 Proof: $\Rightarrow$ Suppose that $U$ is open. Then for each $x \in X$ let $U_x = U$. Then $U_x$ is an open neighbourhood of $x$ such that $U_x \subseteq U$.
 $\Leftarrow$ Suppose that for every $x \in X$ there exists an open neighbourhood $U_x$ of $x$ such that $U_x \subseteq U$. Then $\displaystyle{U = \bigcup_{x \in X} U_x}$. Since arbitrary unions of open sets are open, we see that $U$ is open. $\blacksquare$
Proposition 2 (Open Neighbourhood Criterion for Closed Sets): Let $(X, \tau)$ be a topological space and let $C \subseteq X$. Then $C$ is closed if and only if for every $x \in X \setminus C$ there exists an open neighbourhood $U_x$ of $x$ such that $U_x \subseteq X \setminus C$. Equivalently, $C$ is closed if and only if for every $x \in X \setminus C$ there exists an open neighbourhood $U_x$ of $x$ such that $U_x \cap C = \emptyset$. 
 Proof: Since $C$ is closed, $X \setminus C$ is open, and the result follows by proposition 1. $\blacksquare$
Example 1
Consider the finite set $X = \{ a, b, c \}$ and the nested topology $\tau = \{ \emptyset, \{ a \}, \{ a, b \}, \{ a, b, c \} \}$. Consider the element $a \in X$. The open neighbourhoods of $a$ are $\{ a \}$, $\{a, b \}$, and $\{a, b, c \}$. Now consider the element $b \in X$. The open neighbourhoods of $b$ are $\{a, b \}$ and $\{a, b, c \}$. The open neighbourhoods of $c$ are $\{a, b, c \}$. The image below illustrates the topology on $X$ and the open neighbourhoods of $a, b, c \in X$:
Example 2
For another example, consider the set of integers $\mathbb{Z}$ with the cofinite topology $\tau$:
(1)Consider the point $1 \in \mathbb{Z}$. Let $A \subseteq \mathbb{Z}$ be a finite set. Then any set of the form $\{ 1 \} \cup (\mathbb{Z} \setminus A)$ is an open neighbourhood of $1$. To prove this, we note that since $A$ is finite, that then $\mathbb{Z} \setminus A$ is an infinite set with the elements from $A$ removed. Therefore $\{ 1 \} \cup (\mathbb{Z} \setminus A)$ is also an infinite set with the elements from $A$ removed and with the element $1$ added back in. The complement of this set is:
(2)Since $A$ is finite, $A \setminus \{ 1 \}$ is finite, so $\{ 1 \} \cup (\mathbb{Z} \setminus A) \in \tau$. Furthermore, we have that:
(3)Therefore each set of the form $\{ 1 \} \cup (\mathbb{Z} \setminus A)$ where $A \subseteq \mathbb{Z}$ is finite is an open neighbourhood of $1$.
Example 3
Consider the set of real numbers $\mathbb{R}$ with the usual Euclidean topology. Given any $x \in \mathbb{R}$ and any real numbers $a, b \in \mathbb{R}$ with $a < x < b$, the interval $(a, b)$ is an open neighbourhood containing $x$.