> [!definition] > > Let $X$ be a non-empty [[Set|set]] of points. A family of subsets $\ct \subseteq \pow{X}$ forms a **topology** on $X$ if: > - $\emptyset, X \in \ct$ > - Arbitrary unions of elements of $ct$ is another element of $\ct$. > - Any *finite* intersection in $\ct$ is another element of $\ct$. > > The pair $(X, \ct)$ is called a **topological space**, and the elements of $\ct$ are the [[Open Set|open sets]] of $\ct$. > [!definition] > > Let $X$ be a non-empty set of points and $\topo_1, \topo_2$ be topologies on $X$. If $\topo_1 \subset \topo_2$, then $\topo_1$ is **weaker/coarser** than $\topo_2$, and $\topo_2$ is **stronger/finer** than $\topo_1$. > [!definition] > > Let $X$ be a non-empty set and $\ce \subset \pow{X}$ be a family of sets. Then the intersection of all topologies containing $\ce$, $\topo(\ce)$, is the topology **generated by** $\ce$, and $\ce$ is a *subbase* of $\topo(\ce)$. > [!theorem] > > Let $X$ be a topological space and $\ce \subset \pow{X}$ be a family of sets. Then the topology $\topo(\ce)$ generated by $\ce$ consists of $\emptyset$, $X$, and unions of finite intersections of $\ce$. > > *Proof*. Let > $ > \cf = \bracs{\bigcap_{i = 1}^{n}V_i: \bracs{V_i}_1^n \subset \ce, n \in \nat} > $ > be the collection of finite intersections of $\ce$, then for any $U, V \in \cf$ where $U = \bigcap_{i = 1}^{m}U_i$ and $V = \bigcap_{i = 1}^{n}V_i$, > $ > U \cap V = \bigcap_{i = 1}^{m}U_i \cap \bigcap_{i = 1}^{n}V_i > $ > is also a finite intersection of $\ce$ sets. Therefore $\cf$ is closed under intersections. > > Let > $ > \topo = \bracs{\bigcup_{V \in \mathcal{V}}V: \mathcal{V} \subset \cf} \cup \bracs{\emptyset, X} > $ > be the collection of all unions of $\cf$ sets. Then for any $U, V \in \ct$, where $U = \bigcup_{U' \in \mathcal{U}}U'$ and $V = \bigcup_{V' \in \mathcal{V}}V'$, > $ > \begin{align*} > U \cap V &= \paren{\bigcup_{U' \in \mathcal{U}}U'} \cap \paren{\bigcup_{V' \in \mathcal{V}}V'} \\ > &= \bigcup_{U' \in \mathcal{U}}\paren{U' \cap \bigcup_{V' \in \mathcal{V}}V'} \\ > &= \bigcup_{U' \in \mathcal{U}}\bigcup_{V' \in \mathcal{V}}U' \cap V' > \end{align*} > $ > As $\cf$ is closed under intersections, $U' \cap V' \in \cf$ and $U \cap V$ is still a union of $\cf$ sets, so $U \cap V \in \ct$. And for any $\bracs{U_i}_{i \in I}$ where each $U_i = \bigcup_{U' \in \mathcal{U_i}}U'$, > $ > \bigcup_{i \in I}U_i = \bigcup_{i \in I}\bigcup_{U' \in \mathcal{U_i}}U' = \bigcup_{U' \in \bigcup_{i \in I}\mathcal{U}_i}U' > $ > where $\bigcup_{i \in I}\mathcal{U_i} \subset \cf$ and $\bigcup_{i \in I}U_i \in \topo$. Since $\topo$ is closed under finite intersections and arbitrary unions, $\topo$ is a topology and therefore contains $\topo(\ce)$. > > Since $\topo(\ce) \supset \ce$ is closed under unions and finite intersections, $\topo(\ce) \supset \cf$ contains the collection $\cf$ of all finite intersections of $\ce$ sets, and by extension the collection $\topo$ of all unions of $\cf$ sets. Therefore $\topo(\ce) = \topo$.