# The Open Neighbourhoods of Points in a Topological Space Examples 2

Recall from The Open Neighbourhoods of Points in a Topological Space page that if $(X, \tau)$ is a topological space then an open neighbourhood of the point $x \in X$ is any open set $U \in \tau$ such that $x \in U$.

We will now look at some more examples of open neighbourhoods of points in topological spaces.

## Example 1

**Consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the countable complement topology. Give two examples of open neighbourhoods of $1 \in \mathbb{R}$.**

The open neighbourhoods of $1 \in \mathbb{R}$ are the open sets containing $1$.

Consider the set $A = ( \mathbb{R} \setminus \mathbb{N} ) \cup \{ 1 \}$. Then $A^c = \mathbb{N} \setminus \{ 1 \}$ which is a countable set. Therefore $A \in \tau$ and $1 \in A$, so $A$ is an open neighbourhood of $1$.

For another example, consider the set $B = \mathbb{R} \setminus \{ 2 \}$. Then $B^c = \{ 2 \}$ which is a countable set. Therefore $B \in \tau$ and $1 \in B$, so $B$ is an open neighbourhood of $1$.

In fact, for any $b \in \mathbb{R}$ where $b \neq 1$, then if we define $B_b = \mathbb{R} \setminus \{ b \}$ then $B_b^c = \{ b \}$ which is countable so $B_b \in \tau$, and so $B_b$ is an open neighbourhood of $1$ since $1 \in B_b$.

## Example 2

**Consider the set $X = \{ a, b, c, d, e \}$. Find a non-discrete topology $\tau$ on $X $] such that [[$ a, b \in X$ share exactly $3$ open neighbourhoods.**

If $a$ and $b$ share exactly $3$ open neighbourhoods then there must be $3$ open sets containing both $a$ and $b$. Consider the following topology:

(1)Then $a, b \in U_1 = \{a, b, c \} \in \tau$, $a, b \in U_2 = \{a, b, c, d \} \in \tau$ and $a, b \in U_3 = X \in \tau$ so $a$ and $b$ share exactly $3$ open neighbourhoods.

## Example 3

**Let $X$ be a set and let $\tau_1$ and $\tau_2$ both be topologies on $X$. Consider the topological space $(X, \tau_1 \cap \tau_2)$. Show that if $U$ is an open neighbourhood of $x \in X$ with respect to the topology $\tau_1 \cap \tau_2$ then $U$ is also an open neighbourhood of $x$ with respect to $\tau_1$ and with respect to $\tau_2$.**

Suppose that $U$ is an open neighbourhood of $x \in X$ with respect to the topology $\tau_1 \cap \tau_2$. Then $x \in U \in \tau_1 \cap \tau_2$. Since $U \in \tau_1 \cap \tau_2$ we have that $U \in \tau_1$ and $U \in \tau_2$. Therefore $x \in U \in \tau_1$ and $x \in U \in \tau_2$. So $U$ is an open neighbourhood of $x$ with respect to $\tau_1$ and with respect to $\tau_2$.