Cl(A) if and only if each element of base on $X$ meets A.

Question

Let $(X, \tau)$ topological space, $A \subseteq X$ Prove Cl(A) $ if and only if each element of base on $X$ meets $A$.

Answer 1

I think $CI(A)$ is supposed to mean $Cl(A)$ or the closure of $A$, $\overline{A}$.

$\implies$ suppose $\overline{A} = X$. If there is a basic open set $U$ such that $U \cap A = \emptyset$, then for any $x \in U$, $x \not \in \overline{A} = X$, a contradiction.

$\impliedby$ Suppose every basic open set meets $A$. Then given any $x \in X$, then pick any open set $U$ containing $x$. Then there is a basic open $V \subseteq U$ containing $x$. Then by assumption $V \cap A \not = \emptyset$, which means that $U \cap A \not = \emptyset$, meaning $x \in \overline{A}$ and thus $X = \overline{A}$.