# The Open Neighbourhoods of Points in a Topological Space Examples 1

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 examples of open neighbourhoods of points in topological spaces.

## Example 1

**Consider the topological space $(\mathbb{R}, \tau)$ where $\tau = \{ (-n, n) : n \in \mathbb{N}, n \geq 1 \}$. Describe all of the open neighbourhoods of the point $\pi \in \mathbb{R}$.**

The collection of all open sets for this topology is:

(1)Furthermore, we see that the open sets are nested such that:

(2)We see that $\pi \not \in (-1, 1)$, $\pi \not \in (-2, 2)$, and $\pi \not \in (-3, 3)$… however, $-4 < \pi < 4$ so $\pi \in (-4, 4)$. But then $\pi (-5, 5)$, $\pi \in (-6, 6)$, …, $\pi \in (-n, n)$ for all $n \in \mathbb{N}$ such that $n \geq 4$ by the nesting $(*)$ above. Therefore the collection of all open neighbourhoods of $\pi$ is:

(3)## Example 2

**Let $X$ be a nonempty finite $n$-element set and consider the topological space $(X, \tau)$ where $\tau$ is the discrete topology on $X$. Prove that every $x \in X$ has $2^{n-1}$ open neighbourhoods.**

If $\tau$ is the discrete topology on $X$ then $\tau = \mathcal P(X)$. The size of $\tau$ is therefore $2^n$. Let $x \in X$. For each set $A \in \tau = \mathcal P (X)$ we have that either $a \in A$ or $a \not \in A$. Therefore $\mathcal P (X)$ can be divided into two groups of equal size $\frac{2^n}{2} = 2^{n-1}$ - the first group of subsets of $X$ containing $x$ and the second group of subsets of $X$ not containing $x$.

Therefore there are $2^{n-1}$ subsets of $X$ that contain $A$, but every subset of $X$ is an open set, so there are $2^{n-1}$ open neighbourhoods for every $x \in X$.

## Example 3

**Consider the topological space $(\mathbb{Z}, \tau)$ where $\tau$ is the countable complement topology. What are the open neighbourhoods of $1 \in \mathbb{Z}$? What is the smallest open neighbourhood of $1$?**

The set of open neighbourhoods of $1 \in \mathbb{Z}$ is the collection of sets in $\tau$ containing $1$.

Notice that $\mathbb{Z}$ is a countable set. So, for any $A \subseteq \mathbb{Z}$ we have that $A^c = \mathbb{Z} \setminus A$ is a countable set. Hence, every $A \subseteq \mathbb{Z}$ is an open set.

Therefore, every subset of $\mathbb{Z}$ containing $1$ is an open neighbourhood of $1$. For example, the set of odd integers $0 = \{ ..., -3, -1, 1, 3, ... \}$ is an open neighbourhood of $1$. The set of squares $S = \{1, 4, 9, 16, ... \}$ is also an open neighbourhood of $1$. Of course, the smallest open neighbourhood of $1$ is $\{ 1 \}$ itself.