# Hausdorff Topological Spaces Examples 2

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 more problems regarding Hausdorff topological spaces.

## Example 1

**Consider the set $\mathbb{R}$ with the cofinite topology $\tau = \{ U \subseteq X : U = \emptyset \: \mathrm{or} \: U^c \: \mathrm{is \: finite} \}$. Determine if $(\mathbb{R}, \tau)$ is a Hausdorff topological space.**

First consider any open sets in $\mathbb{R}$ with respect to the topology $\tau$. We will show that none of the open sets are disjoint and hence no open neighbourhoods of any points $x, y \in \mathbb{R}$ can be disjoint.

Let $U, V \in \tau$. Since $U, V \in \tau$ we have that $U^c$ and $V^c$ are finite, so $U^c = \{ x_1, x_2, ..., x_n \}$ and $V^c = \{ y_1, y_2, ..., y_m \}$ where $x_1, x_2, ..., x_n, y_1, y_2, ..., y_m \in \mathbb{R}$.

Therefore:

(1)And hence:

(2)But $\{ x_1, x_2, ..., x_n, y_1, y_2, ..., y_m \}$ is a finite set and so $U \cap V \neq \emptyset$, so $(\mathbb{R}, \tau)$ cannot be a Hausdorff topological space.

## Example 2

**Prove that if $(X, \tau)$ is a Hausdorff space then any sequence $(x_n)_{n=1}^{\infty}$ in $X$ can converge to at most one point $x \in X$.**

Suppose that $x, y \in X$ where $x \neq y$ and let $(x_n)_{n=1}^{\infty}$ converge to $x$. Since $(X, \tau)$ is a Hausdorff topological space we have that there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$.

Since $(x_n)_{n=1}^{\infty}$ converges to $x$ we have that there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $x_n \in U$. But since $U \cap V = \emptyset$ we have that $x_n$ does not converge to $y$.

Therefore any sequence $(x_n)_{n=1}^{\infty}$ in $X$ can converge to at most one point in $x \in X$.