# Bases of a Topology Examples 2

Recall from the Bases of a Topology page that if $(X, \tau)$ is a topological space then a base for the topology $\tau$ is a collection $\mathcal B \subseteq \tau$ such that every $U \in \tau$ can be written as a union of elements from $\mathcal B$, i.e., for all $U \in \tau$ we have that there exists a $\mathcal B^* \subseteq \mathcal B$ such that:

(1)We will now look at some more examples of bases for topologies.

## Example 1

**Consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology of open intervals on $\mathbb{R}$. Show that $\mathcal B = \left \{ (p, q) : p, q \in \mathbb{Q}, p < q \right \}$ forms a base of $\tau$.**

It suffices to show that every open interval $(a, b) \in \tau$ can be expressed as a union of base elements since every open set in $\tau$ is simply a union of open intervals. Let $(a, b) \in \tau$.

If $a, b \in \mathbb{Q}$ then $(a, b) \in \mathcal B$ and we're done.

If $a \in \mathbb{Q}$ and $b \not \in \mathbb{Q}$ then:

(2)Similarly, if $a \not \in \mathbb{Q}$ and $b \in \mathbb{Q}$ then:

(3)The last case occurs when $p, q \not \in \mathbb{Q}$. Then:

(4)So for every open interval $(a, b)$ there exists a $\mathcal B^* \subseteq \mathcal B$ such that $(a, b) = \bigcup_{B \in \mathcal B^*} B$, so $\mathcal B$ is a base of $\tau$.

## Example 2

**Consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology of open intervals in $\mathbb{R}$. Show that the collection $\mathcal B = \{ (w, z) : w, z \in \mathbb{Z}, w < z\}$ does not form a base of $\tau$.**

If $\mathcal B$ was a base of $\tau$ then for every set $U \in \tau$ we must be able to find a subcollection $\mathcal B^* \subseteq \mathcal B$ such that:

(5)Consider the open interval $U = \left ( 0, \frac{1}{2} \right )$. Any set in $\mathcal B$ that contains $U$ must be such that $w \leq 0$ and $\frac{1}{2} \leq z$. The smallest integer greater than $\frac{1}{2}$ is $z = 1$, and so the smallest set in $\mathcal B$ containing $U$ is $(0, 1)$.

However, there are no sets in $\mathcal B$ that are contained in $\left (0, \frac{1}{2} \right )$ and so $\mathcal B$ cannot be a base of $\tau$.