Topology Generated by a Basis

Last updated Nov 1, 2022

Let $X$ be a Set. Let $\mathcal{B} \subset \mathcal{P}(X)$ be such that:

1. $\forall x \in X$ there exists $B \in \mathcal{B}$ so that $x \in B$.
2. Let $B_{1}, B_{2} \in \mathcal{B}$. $\forall x \in B_{1} \cap B_{2}$ there exists $B_{3}$ s.t.
   1. $x \in B_{3}$
   2. $B_{3} \subset B_{1} \cap B_{2}$

Then $\mathcal{B}$ is a Topological Basis for $\tau(B)$. We refer to $\tau(B)$ as the topology generated by basis $\mathcal{B}$.

Let $U \in \tau(\mathcal{B})$. Denote $\mathcal{B}’ := {\bigcap\limits_{i=1}^{n} B_{i} \subset X : B_{i} \in \mathcal{B}, i \in [n], n \in \mathbb{N} \cup {0}}$. By The Topology Generated by a Set has a Basis of Finite Intersections, $\mathcal{B}’$ is a Topological Basis for $\tau(\mathcal{B})$, so we only need to show $\mathcal{B}’ \subset {\bigcup\limits_{B \in \mathcal{S}}B \subset X : \mathcal{S} \subset \mathcal{B}} \subset \tau(\mathcal{B})$ . The latter inclusion follows because $\mathcal{B} \subset \tau(\mathcal{B})$ by definition of Topology Generated by a Set and $\tau(\mathcal{B})$ is closed under arbitrary Set Union by definition of Topological Space.

So let $A \in \mathcal{B}’$. If $A = \bigcap\limits_{i=1}^{0} = X$, then because of (1), $${\bigcup\limits_{B \in \mathcal{S}}B \subset X : \mathcal{S} \subset \mathcal{B}} \ni \bigcup\limits_{B \in \mathcal{B}} B = X = \bigcap\limits_{i=1}^{0} \in \mathcal{B}’$$ Otherwise $A$ is a nonzero finite Set Intersection of elements of $\mathcal{B}$. By assumption (2) we can inductively build $B_{A}^{(x)}$ s.t. $x \in B_{A}^{(x)}$ and $B_{A}^{(x)} \subset A$. Thus $A = \bigcup\limits_{x \in A} B_{A}^{(x)}$ and we have our result. $\blacksquare$

1. This is really just Topology Generated by a Set