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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 $cl(E)^c = \emptyset, cl(\emptyset)^c = E$

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A set $A$ is open if and only if its complement $E \setminus A$ is closed.

$E \setminus A$ is closed if and only if $E \setminus A = cl(E \setminus A) = B \cup (E \setminus A)$.

$E \setminus A = B \cup (E \setminus A)$ if and only if $B \subset E \setminus A$.

$B \subset E \setminus A$ if and only if $A \subset E \setminus B$ if and only if $A \cap B = \emptyset$.

So, a set $A$ is open in this topology if and only if $A \cap B = \emptyset$.

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