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


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 $A \cap B = \emptyset$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.