Let $E$ be a set and $A⊆E$. Define for all $B⊆E$ with $B\not=∅: cl(B) = B ∪ A$, while $cl(∅) = ∅$. Prove that $cl$ is a Kuratowsky closure operator. What is the associated topology?
i) closure operator :$cl(∅) = ∅ \\ \forall B \subseteq E, B \subseteq A \cup B = cl(B) \\ \forall B \subseteq E, (B \cup A) \cup A = B\cup A \Rightarrow cl(cl(B)) = cl(B) \\ B_1, B_2 \subseteq E, (B_1 \cup A) \cup (B_2 \cup A) = (B_1 \cup B_2) \cup A \implies cl(B_1) \cup cl(B_2) = cl(B_1\cup B_2) $
So $cl$ is a closure operator.
ii) associated topology: I am not sure what the associated topology is, say $\tau$ is this wanted topology on $E$
$cl(\emptyset) = \emptyset$ and $cl(E) = E \cup A = E$ so $\emptyset, E \in \tau$
but now I am not sure how to define it further as closure operator gives closed sets. Seems that taking the complements of these closures may be the idea.
EDIT: $(cl(B))^c = (A \cup B)^c = (A^c \cap B^c) $ something along these lines possibly because $cl(E)^c = \emptyset, cl(\emptyset)^c = E$