basis for a topological space Excuse me can you see my question
Let (X,T) be a topological space . Let H be the collection of closed sets in X . Let C be the collection of closed set satisfying  every $$V \in H$$ is an intersection of element of C . Show that $$B={A^{c} : A \in C}$$ is a base for T ??
For the seconed condition that i take B1 And B2 in B s.t $$x \in B1 \cap B2$$ there exist $$B3=B1 \cap B2 $$s.t $$ x\in B3 \subset B1 \cap B2 $$
But i can not make the first condition << can you help me please
 A: You have to show that elements of $B$ cover $X$. So let $x$ be an element of $X$. Let $U$ be an open subset of $X$ such that $x\in U\subseteq X$. You can always find such $U$, at the very least choose $U=X$. Then $U^c$ is a closed subset of $X$. By definition of $C$, you may find elements $C_i$ in $C$ such that
$$U^c=\bigcap_{i}C_i$$
as a consequence,
$$U=(U^c)^c=\left(\bigcap_{i} C_i\right)^c=\bigcup_{i} C_i^c$$
But $C_i^c$ is an element of $B$ for all $i$, hence $x$ lies in the union of a family of elements of $B$, as wanted.
A: You only need to prove by definition. The idea of proof is quite straightforward. Recall that a collection of sets $B$ is called as a base of topology $T$ if and only if for every open set $O \in T$, there is an indexed family of sets $A_i$ in $B$, such that $O = \bigcup_{i \in \Gamma} A_i$. Given this definition, for this specific question we have:
For any open set $O$, $O^c \in H$ because it is a closed set, therefore, there exists a family of sets $M_i$ in $C$, such that $O^c = \bigcap_{i \in \Gamma} M_i$, which means $O = (\bigcap_{i \in \Gamma} M_i)^c = \bigcup_{i \in \Gamma} (M_i^c)$. But every $M_i^c \in B$, by definition above, so we have proved $B$ is a base for $T$.
