Subbases of a Topology Examples 1

# Subbases of a Topology Examples 1

Recall from the Subbases of a Topology page that if $(X, \tau)$ is a topological space then a subset $\mathcal S \subseteq \tau$ is said to be a subbase for the topology $\tau$ if the collection of all finite intersects of sets in $\mathcal S$ forms a base of $\tau$, that is, the following set is a base of $\tau$:

(1)
\begin{align} \quad \mathcal B_S = \{ U_1 \cap U_2 \cap ... \cap U_k : U_1, U_2, ..., U_k \in \mathcal S \} \end{align}

We will now look at some more examples of subbases of topologies.

## Example 1

Consider the set $X = \{ a, b, c, d, e, f \}$ with the topology $\tau = \{ \emptyset, \{ a \}, \{ c, d \}, \{a, c, d \}, \{ b, c, d, e, f \}, X \}$. Show that the subset $S = \{ \{ a \}, \{ a, c, d \}, \{ b, c, d, e, f \} \} \subset \tau$ is a subbase of $\tau$.

The collection of all finite intersections of elements from $\mathcal S$ is:

(2)
\begin{align} \quad \mathcal B_S = \{ \emptyset, \{ a \}, \{c, d \}, \{a, c, d \}, \{b, c, d, e, f \} \} \end{align}

Every set in $\tau$ apart from $X$ is a trivial union of elements in $\mathcal B_S$ and $X = \{ a \} \cup \{ b, c, d, e, f \}$, so $\mathcal B_S$ is a base of $\tau$ so $\mathcal S$ is a subbase of $\tau$.

## Example 2

Consider the set $X = \{ a, b, c, d, e \}$ with the topology $\tau = \{ \emptyset, \{ a \}, \{ b \}, \{a, b \}, \{ b, d \}, \{a, b, d \}, \{a, b, c, d \}, X \}$. Show that $\mathcal S = \{ \{ a \}, \{ b \} \{a, b \}, \{ a, b, d \}, \{a, b, c, d \}, X \} \subset \tau$ is not a subbase of $\tau$.

Consider the following set:

(3)
\begin{align} \quad \mathcal S = \{ \{ a \}, \{ b \} \{a, b \}, \{ a, b, d \}, \{a, b, c, d \}, X \} \end{align}

The set of all finite intersects of sets from $S$ is:

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

All sets except $\{ b, d \}$ can be expressed as trivial intersections. However, $\{ b, d \}$ cannot be expressed as a union of elements from $\mathcal B_S$, so $\mathcal B_S$ is not a base of $\tau$ and hence $S$ is not a subbase of $\tau$.