The Topology Generated by a Set has a Basis of Finite Intersections

Last updated Nov 1, 2022

Let $X$ be a Set and let $\mathcal{S} \subset \mathcal{P}(X)$. Let $$\mathcal{S}’ := {\bigcap\limits_{i=1}^{n} S_{i} \subset X : S_{i} \in \mathcal{S}, i \in [n], n \in \mathbb{Z}{\geq 0}}$$ and $$\mathcal{S}’’ := {\bigcup\limits{U \in \mathcal{U}}U \subset X : \mathcal{U} \subset \mathcal{S}’}$$ Then $$\tau(\mathcal{S}) = \mathcal{S}’’$$ That is, $\mathcal{S}’$ is a Topological Basis for $\tau(\mathcal{S})$.

We let $\bigcap\limits_{i=1}^{0} = X$ and $\bigcup\limits_{} \emptyset = \emptyset$. Therefore $X \in \mathcal{S}’$ and $\emptyset \in \mathcal{S}’’$.

$(\supset)$: By definition of Topology Generated by a Set, $\tau(\mathcal{S}) \supset \mathcal{S}$ and $\tau(\mathcal{S})$ is a Topological Space. Let $n \in \mathbb{Z}{\geq 0}$ , and let $S{i} \subset \mathcal{S}$ for $i \in [n]$. Then $\bigcap\limits_{i=1}^{n} S_{i} \in \tau(S)$ by definition of Topological Space. Thus $\mathcal{S}’ \subset \tau(S)$. Now let $\mathcal{U} \subset \mathcal{S}’ \subset \tau(\mathcal{S})$. Then, $\bigcup\limits_{}\mathcal{U} \in \tau(\mathcal{S})$ by definition of Topological Space. Thus $\mathcal{S}’’ \subset \tau(\mathcal{S})$. $\checkmark$

$(\subset)$: We need only show that $\mathcal{S}’’$ is a Topological Space in its own right. This will establish that $\mathcal{S}’’ \supset \tau(\mathcal{S})$ since $\mathcal{S}’’ \supset \mathcal{S}$.

1. $X = \bigcap\limits_{i=1}^{0} \in \mathcal{S}’ \subset \mathcal{S}’’$ and $\emptyset = \bigcup\limits_{} \emptyset \in \mathcal{S}’’$.
2. Suppose $\mathcal{V} \subset \mathcal{S}’’$. Then for each $V \in \mathcal{V}$, $\exists \mathcal{U}{V} \subset \mathcal{S}’$ so that $V = \bigcup\limits{} \mathcal{U}{V}$. Thus $$\bigcup\limits{V \in \mathcal{V}}V = \bigcup\limits_{V \in \mathcal{V}}\bigcup\limits_{U \in \mathcal{U}{V}}U = \bigcup\limits{U \in \bigcup\limits_{V \in \mathcal{V}}\mathcal{U}_{V}}U \in \mathcal{S}’’.$$
3. Suppose $U, V \in \mathcal{S}’’$. Then there exists $\mathcal{A}, \mathcal{B} \subset \mathcal{S}’’$ s.t. $U = \bigcup\limits_{A \in \mathcal{A}}A$ and $V = \bigcup\limits_{B \in \mathcal{B}} B$. Then $$(\bigcup\limits_{A \in \mathcal{A}}A ) \cap ( \bigcup\limits_{B \in \mathcal{B}} B)) = \bigcup\limits_{A \in \mathcal{A}, B \in \mathcal{B}} A \cap B \in \mathcal{S}’’$$ since for each $A \in \mathcal{A}, B \in \mathcal{B}$, $A \cap B \in \mathcal{S}’$.

Thus $\mathcal{S}’’$ is a Topological Space and $\mathcal{S}’’ \supset \tau(\mathcal{S})$. $\checkmark$ $\blacksquare$

1. This gives us a constructive approach to defining $\tau(\mathcal{S})$.

• Set Intersection
• Set Union
• Integer
• Empty Set
• Empty Set is a Subset of every Set