Hausdorff Topological Spaces Examples 1

# Hausdorff Topological Spaces Examples 1

Recall from the Hausdorff Topological Spaces page that a topological space $(X, \tau)$ is said to be a Hausdorff space if for every distinct $x, y \in X$ there exists open neighbourhoods $U, V \in \tau$ such that $x \in U$, $y \in V$ and $U \cap V = \emptyset$.

We also looked at two notable examples of Hausdorff spaces - the first being the set of real numbers $\mathbb{R}$ with the usual topology of open intervals on $\mathbb{R}$, and the second being the discrete topology on any nonempty set $X$.

We will now look at some problems regarding Hausdorff topological spaces.

## Example 1

Consider the set $X = \{ a, b, c ,d \}$ with the topology $\tau = \{ \emptyset, \{ a \}, \{ b \} \{ a, b \}, \{b, c \}, \{a, b, c \}, X \}$. Is $(X, \tau)$ a Hausdorff space?

Notice that for $d \in X$ that the only open neighbourhood of $d$ is the whole space. However, for $a \in X$ we have that the open neighbourhoods of $a$ are $\{ a \}$, $\{a, b \}$, $\{a, b, c \}$, and $X$, all all of these open neighbourhoods intersect the whole space. Therefore $(X, \tau)$ is not a Hausdorff space.

## Example 2

Prove that for any nonempty set $X$ that if $\tau$ is the indiscrete topology then $(X, \tau)$ is not a Hausdorff space.

Let $X$ be a nonempty set and let $\tau = \{ \emptyset, X \}$. Then for every $x, y \in X$ we have that the only open neighbourhood of $x$ is $X$ and the only open neighbourhood of $y$ is $X$. Since $X$ is nonempty, we have that $X \cap X \neq \emptyset$, and so $(X, \tau)$ cannot be a Hausdorff space.

## Example 3

Consider the topological space $(\mathbb{R}, \tau)$ where $\tau = \{ (-n, n) : n \in \mathbb{Z}, n \geq 1 \}$. Is $(\mathbb{R}, \tau)$ a Hausdorff space?

Consider the numbers $0, \frac{1}{2} \in \mathbb{R}$. We see that for all $n \in \mathbb{Z}$, $n \geq 1$ that:

(1)
\begin{align} \quad 0, \frac{1}{2} \in (-1, 1) \subset (-2, 2) \subset ... \subset (-n, n) \subset ... \end{align}

I.e., every set in $\tau$ is an open neighbourhood of $0$ and $\frac{1}{2}$. Therefore, if $U$ is any open neighbourhood of $0$ and $V$ is an open neighbourhood of $\frac{1}{2}$ then $U \cap V \neq \emptyset$. Therefore, $(\mathbb{R}, \tau)$ is not a Hausdorff space.