A Suff. Condition for a Collection of Sets to be a Basis of a Topology

A Sufficient Condition for a Collection of Sets to be a Basis of a Topology

Recall from the Bases of a Topology that if $(X, \tau)$ is a topological space that a basis $\mathcal B$ of this topological space is a collection of subsets from $\tau$ such that every $U \in \tau$ is the union of some collection of sets from $\mathcal B$.

That said, not every collection $\mathcal B$ of subsets from $\tau$ is necessarily a basis of $(X, \tau)$, so we would like to develop a sufficient condition to verify that $\mathcal B$ is indeed a basis for this topological space. The following theorem grants us the sufficient condition we desire.

Theorem 1: Let $(X, \tau)$ be a topological space and let $\mathcal B$ be a collection of subsets from $\tau$. Then $\mathcal B$ is a basis of $\tau$ if and only if for all $U \in \tau$ and for all $x \in U$ there exists a $B \in \mathcal B$ such that $x \in B \subseteq U$.
  • Proof: $\Rightarrow$ Suppose that $\mathcal B$ is a basis of $\tau$. Then every $U \in \tau$ is the union of some collection of sets from $\mathcal B$, that is, there exists a subset $\mathcal B^* \subseteq \mathcal B$ such that:
(1)
\begin{align} \quad U = \bigcup_{B \in \mathcal B^*} B \end{align}
  • Let $x \in U$. Then $x \in \bigcup_{B \in \mathcal B^*} B$. So $x \in B \subseteq U$ for some $B \in \mathcal B^* \subseteq \mathcal B$. $\blacksquare$
  • $\Leftarrow$ Suppose that for all $U \in \tau$ and for all $x \in U$ there exists a $B_x \in \mathcal B$ such that $x \in B_x \subseteq U$. Then for each $U \in \tau$ we have that:
(2)
\begin{align} \quad U = \bigcup_{x \in U} B_x \end{align}
  • Then every $U \in \tau$ is the union of a collection of sets $\{ B_x \}_{x \in U} \subseteq \mathcal B$. So $\mathcal B$ is a basis of $\tau$. $\blacksquare$

Example 1

Let $X = \{ a, b, c, d \}$ and let $\tau = \{ \emptyset, \{ a \}, \{a, b \}, \{a, b, c \}, X \}$. Determine if $\mathcal B = \{ \{a \}, \{a, b, c \} \}$ is a basis of $\tau$.

By Theorem 1 above, if $\mathcal B$ is a basis of $\tau$ then for every $U \in \tau$ and for all $x \in U$ there exists a $B \in \mathcal B$ such that $x \in B \subseteq U$.

Consider the set $U = \{a, b \} \in \tau$. For all $x \in U$ does there exist a $B \in \mathcal B$ such that $x \in B \subseteq U$?

For $a \in U$ we have that $B_a = \{ a \} \in \mathcal B$ is such that $a \in B_a \subseteq U$. However, for $b \in U$ we have that $b \not \in \{a \}$ and $b \in \{a, b, c \} \not \subseteq U$. Therefore $\mathcal B$ cannot possibly be a basis of $\tau$!

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