The Open Neighbourhoods of Points in a Topological Space

# 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. An Open Neighbourhood of the point $x \in X$ is any (open) set $U \in \tau$ such that $x \in U$.

For example, 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 $a, b, c \in X$:

For another example, consider the set of integers $\mathbb{Z}$ with the cofinite topology $\tau$:

(1)
\begin{align} \quad \tau = \{ \emptyset \} \cup \{ U \subseteq \mathbb{Z} : U^c \: \mathrm{is \: finite} \} \end{align}

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)
\begin{align} \left ( \{1 \} \cup (\mathbb{Z} \setminus A) \right )^c = A \setminus \{ 1 \} \end{align}

Since $A$ is finite, $A \setminus \{ 1 \}$ is finite, so $\{ 1 \} \cup (\mathbb{Z} \setminus A) \in \tau$. Furthermore, we have that:

(3)
\begin{align} 1 \in \{ 1 \} \cup (\mathbb{Z} \setminus A) \end{align}

Therefore each set of the form $\{ 1 \} \cup (\mathbb{Z} \setminus A)$ where $A \subseteq \mathbb{Z}$ is finite is an open neighbourhood of $1$.