Separable Topological Spaces Examples 1

Recall from the Separable Topological Spaces page that a topological space $(X, \tau)$ is said to be separable if it contains a countable and dense subset.

We saw that if $\mathbb{R}$ is equipped with the usual topology of open intervals then the set of rational numbers $\mathbb{Q}$ is dense (and of course countable) and so $\mathbb{R}$ is a separable topological space.

We will now look at some more examples of determining whether or not a topological space is separable.

Example 1

Consider the set of real numbers $\mathbb{R}$ with the lower limit topology (Sorgenfrey line) $\tau = \{ [a, b) : a, b \in \mathbb{R}, a < b \}$. Determine whether $(\mathbb{R}, \tau)$ is a separable topological space.

We claim that the subset of rational numbers $\mathbb{Q}$ is also dense in $\mathbb{R}$ with respect to this topology.

To prove this, let $U \in \tau \setminus \{ \emptyset \}$. Then $U$ is either:

(1)

\begin{align} \quad U = \left\{\begin{matrix} [a, b) & \mathrm{for \: some \:} a, b \in \mathbb{R}, a < b\\ \mathrm{The \: union \: of \: intervals \: of \: the \: form \:} [a, b) & \mathrm{for \: each \:} a, b \in \mathbb{R}, a < b \end{matrix}\right. \end{align}

In either case $[a, b) \subseteq U$ for some $a, b \in \mathbb{R}$, $a < b$. Then for some $a, b \in \mathbb{R}$, $a < b$, suppose that $\mathbb{Q} \cap U = \emptyset$. Then this implies that $\mathbb{Q} \cap [a, b) = \emptyset$.

But then this says that for all $x$ such that $a \leq x < b$ there exists no $q \in \mathbb{Q}$ such that $a \leq < q < b$. Given any $x \in [a, b)$ there always exists a rational number $q$ such that $x < q < b$. So the assumption that for some $U \in \tau \setminus \{ \emptyset \}$ we have that $\mathbb{Q} \cap U = \emptyset$ is false.

Hence for all $U \in \tau \setminus \{ \emptyset \}$ we have that $\mathbb{Q} \cap U \neq \emptyset$, so $\mathbb{Q}$ is a dense (and countable) subset of $\mathbb{R}$, so $\mathbb{R}$ is countable with respect to the lower limit topology.

Example 2

Construct a topology $\tau$ on the set $X = \{a, b, c, d \}$ such that $(X, \tau)$ is a separable topological space.

Consider the topology $\tau = \{ \emptyset, \{a \}, \{ b \}, \{a, b \}, \{a, b, c \}, X \}$. Clearly every subset of $X$ is countable since $X$ consists of only finitely many elements. So we only need to find a dense subset of $X$.

Consider the set $A = \{a, b \}$. Then we have that:

(2)

\begin{align} \quad A \cap \{ a \} & = \{ a \} \neq \emptyset \\ \quad A \cap \{ b \} &= \{ b \} \neq \emptyset \\ \quad A \cap \{a, b \} &= \{ a, b \} \neq \emptyset \\ \quad A \cap \{a, b, c \} &= \{a, b \} \neq \emptyset \\ \quad A \cap X &= \{a, b \} \neq \emptyset \end{align}

Therefore $A$ is a countable and dense subset of $X$, so $(X, \tau)$ is a separable topological space.

Example 3

Prove that if $X$ is a finite set and $\tau$ is the discrete topology on $X$ then $(X, \tau)$ is a separable topological space.

Let $X$ be a finite set with $n$ elements. Then $X = \{ a_1, a_2, ..., a_n \}$. If $\tau$ is the discrete topology on $X$ then $\tau = \mathcal P(X)$. Clearly every subset of $X$ is countable since $X$ is a finite set, so we only need to find a dense subset of $X$.

Take $A = X$. Then for all $U \in \tau \setminus \{ \emptyset \}$ we have that $U \subseteq A$. So $A \cap U \neq \emptyset$. Hence $A$ is a countable and dense subset of $X$ so $(X, \tau)$ is a separable topological space.

Example 4

Prove that if $X$ is a countable set and $\tau$ is the discrete topology on $X$ then $(X, \tau)$ is a separable topological space.

If $X$ is countable then $X$ is either finite or countably infinite. Example 3 shows that if $X$ is finite and $\tau$ is the discrete topology on $X$ then $(X, \tau)$ is a separable topological space. We will now show that if $X$ is countably infinite then $(X, \tau)$ is also a separable topological space.

If $X$ is countably infinite then:

(3)

\begin{align} \quad X = \{ a_1, a_2, ... \} \end{align}

If we let $A = X$ then every $U \in \tau \setminus \{ \emptyset \}$ is such that $U \subseteq A$ so $A \cap U \neq \emptyset$. Therefore $A$ is a countable (and dense) subset of $X$, so $(X, \tau)$ is separable.