Topological Basis

Last updated Nov 1, 2022

Let $X, \tau$ be a Topological Space. Then $\mathcal{B} \subset \tau$ is a Topological Basis for $X$ if $$\tau = {\bigcup\limits_{B \in \mathcal{C}} B : \mathcal{C} \subset \mathcal{B}}$$

1. Even if $\emptyset \not\in \mathcal{B}$, $\mathcal{B}$ can still be a Topological Basis because $\emptyset = \bigcup\limits_{B \in \emptyset} B$.
2. To show $\mathcal{B}$ is a Topological Basis, it is sufficient to show the Sets are Open in $X$ and every Open $U \subset X$ can be written as a $U = \bigcup\limits_{B \in \mathcal{C}} B$ for some $\mathcal{C} \subset \mathcal{B}.$ The first criterion ensures $\mathcal{B} \subset \tau$, so $\tau(\mathcal{B}) \subset \tau$. The second criterion ensures $\tau(\mathcal{B}) \supset \tau$.