Bases of a Topology
Table of Contents

Bases of a Topology

We will now focus our attention at a special type of subset of a topology called a basis for $\tau$ which we define below.

Definition: Let $(X, \tau)$ be a topological space. A Basis for the topology $\tau$ is a collection $\mathcal B$ of subsets from $\tau$ such that every $U \in \tau$ is the union of some collection of sets in $\mathcal B$.

Note that by definition, $\mathcal B = \tau$ is a basis of $\tau$ - albeit a rather trivial one! The emptyset is also obtained by an empty union of sets from $\mathcal B$.

For example, consider any nonempty set $X$ with the discrete topology $\tau = \mathcal P (X)$. Consider the collection:

(1)
\begin{align} \quad \mathcal B = \{ \{ x \} : x \in X \} \end{align}

We claim that $\mathcal B$ is a basis of the discrete topology $\tau$. Let's verify this. First, since $\tau$ is the discrete topology we see that every subset of $X$ is contained in $\tau$. For each $B = \{ x \} \in \mathcal B$ we therefore have that:

(2)
\begin{align} \quad B = \{ x \} \in \mathcal B \in \tau \end{align}

For the second condition, let $U \in \tau$. Then since $\tau$ is the discrete topology, we have that $U \subseteq X$. For all $x \in U$, we have that $U$ can be expressed as the union of some collection of sets in $\mathcal B$. In particular, for each $U \in \tau$ we have that:

(3)
\begin{align} \quad U = \bigcup_{x \in U} \{ x \} \end{align}

Therefore $\mathcal B = \{ \{ x \} : x \in X \}$ is a basis of the discrete topology.

For another example, consider the set $X = \{ a, b, c, d \}$ and the following topology on $X$:

(4)
\begin{align} \quad \tau = \{ \emptyset, \{ a \}, \{d \}, \{a, d \}, \{ b, c\}, \{a, b, c \}, \{ b, c, d \}, X \} \end{align}

Consider the collection of open sets $\mathcal B = \{ \{ a \}, \{ d \}, \{b, c \} \}$. We claim that $\mathcal B$ is a basis of $\tau$. Clearly all of the sets in $\mathcal B$ are contained in $\tau$, so every set in $\mathcal B$ is open.

For the second condition, we only need to show that the remaining open sets in $\tau$ that are not in $\mathcal B$ can be obtained by taking unions of elements in $\mathcal B$. The $\emptyset \in \tau$ can be obtained by taking the empty union of elements in $\mathcal B$. Furthermore:

(5)
\begin{align} \quad & \{ a \} \cup \{ d \} = \{ a, d \} \\ \quad & \{ a \} \cup \{ b, c \} = \{ a, b, c \} \\ \quad & \{ a \} \cup \{ b, c \} \cup \{ d \} = X \end{align}

Therefore every $U \in \tau$ is the union of some collection of sets from $\mathcal B$, so $\mathcal B$ is a basis of $\tau$.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License