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
Prove that every metric space is a Hausdorff space.
Let $X$ be a metric space. The open sets in $X$ are therefore the any set that is the union of a collection of open balls with respect to the metric $d$ defined on $X$.
Let $x, y \in X$ where $x \neq y$. Then $d[x, y] > 0$ and so if we let $r = \frac{d[x, y]}{2}$ then the ball centered at $x$ with radius $r$ and the ball centered at $y$ with radius $r$ are disjoint, that is:
[1]\begin{align} \quad B[x, r] \cap B[y, r] = \emptyset \end{align}
Furthermore if we set $U = B[x, r]$ and $V = B[y, r]$ we have that $U$ and $V$ are open sets of $X$ with respect to the metric $d$. Therefore any metric space is a Hausdorff space.
Example 2
Prove that if $[X, \tau]$ is a Hausdorff space then for every $x \in X$, the singleton set $\{ x \}$ is closed.
Suppose that $[X, \tau]$ is a Hausdorff space and let $x \in X$. We will show that $X \setminus \{ x \}$ is therefore open. Let $y \in X \setminus \{ x \}$. Since $[X, \tau]$ is Hausdorff there exists open neighbourhoods of $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$. But since $x \in U$ we must have that $x \not \in V$, so $y \in V \subseteq X \setminus \{ x \}$.
So for every $y \in X \setminus \{ x \}$ there exists a $V \in \tau$ such that $y \in V \subseteq X \setminus \{ x \}$ so $\mathrm{int} [X \setminus \{ x \}] = X \setminus \{ x \}$, i.e., $X \setminus \{ x \}$ is open, and so, $\{ x \}$ is closed.
Example 3
Use Example 2 to show that the set $X = \{ a, b, c, d \}$ with the topology $\tau = \{ \emptyset, \{ a \}, \{a, b \}, \{a, c \}, \{a, b, c \}, X \}$ is not a Hausdorff space.
The contrapositive of the result from Example 2 says that if there exists a singleton set $\{ x \}$ which is open then $[X, \tau]$ is not a Hausdorff space.
With the topology above we see that the singleton set $\{ a \}$ is open. For the element $b \in X$ we see that the only open neighbourhoods of $b$ are $\{ a, b \}$, $\{a, b, c \}$, and $X$. Furthermore, all of the open neighbourhoods of $a$ are $\{a \}$, $\{ a, b \}$, $\{a, c \}$, $\{a, b, c \}$, $X$.
The intersection of any of these open neighbourhoods is nonempty, and so there does not exist any open neighbourhoods $U$ of $a$ and $V$ of $b$ such that $U \cap V = \emptyset$, so $[X, \tau]$ is not a Hausdorff space.