Show basis for a topology A Subbasis $\mathcal R$ of a set $X$ is defined to a collection of subsets of $X$ whose union equals $X$. Show that the collection $\mathcal B$ of all finite intersections of element of $\mathcal R$ is a topological basis of $X$.
Attempt :

Let $X$ be a set, let $\mathcal B$ a collection of subsets of $X$, namely $\mathcal B = ${$B_\alpha$}$_{\alpha \in I}$. $\mathcal B$ is called a topological basis of $X$ if satisfy :

*

*For any $x \in X$, there is an element $B_\alpha \in \mathcal B$ such that $x \in B_\alpha$.


*For any $x \in X$, if there exists $B_1, B_2 \in \mathcal B$ such that $x\in (B_1\cap B_2)$ then there is $B_3 \in \mathcal B$ with $x \in B_3 \subset (B_1 \cap B_2)$

How to show that collection $\mathcal B$ of all finite intersections of element of $\mathcal R$ satisfy the two conditions for a basis for a topology ?

$R$ satisfies $1$ because its union is $X$. The closure over finite intersections satisfies $2$ (choose $B_3$ equal to $B_1$ intersection $B_2$).
