Show there is only one topology for X such that only closed sets in X are A $\subset$ X, A=S(A).

Question

I've been trying this exercice for a while...

Let $X$ be a set. Supose that for any $A \subset X$ we can associate a subset $S(A) \subset X$ such that:

1) $S(\emptyset) = \emptyset$

2) $A \subset S(A)$

3) $S(A \cup B) = S(A) \cup S(B)$

4) $S(S(A)) = S(A)$

Show there is only one topology in $X$ such that his closed sets are $A \subset X$ with $S(A) = A$.

I've tried to define a base $B$ of $X$ such that:

1) $\cup B_i = X$

2) $\forall U, V \in B$ and $\forall x \in U \cap V$, $\exists W \in B$ such that $x \in W \subset U \cap V$.

So I've only one topology for $X$. But I don't really know how to define that $B$.

Any help? Thanks!

Answer 1

Existence:

the empty set is closed since $S(\psi)=\psi$.

$X$ is closed since $X\subset S(X)\subset X$, implies that $S(X)=X$.

$A_1,A_2$ closed implies that $S(A_1\cup A_2)=S(A_1)\cup S(A_2)=A_1\cup A_2$ implies that $A_1\cup A_2$ is closed, suppose that if $A_1,...,A_{n-1}$ is closed so is $A_1\cup..\cup A_{n-1}$. Consider closed subsets $A_1,...,A_n$, $S(A_1\cup..\cup A_n)=S((A_1\cup..\cup A_{n-1})\cup A_n)=S(A_1\cup..\cup A_{n-1})\cup S(A_n)=A_1..\cup A_n$.

Let $(A_i)_{i\in I}$ a family of closed subsets, you can write $A_i=\cap_{i\in I}A_i\cup B_i$ this implies that $S(A_i)=S(\cap_{i\in I}A_i)\cup S(B_i)=S(A_i)=A_i$, we deduce that $S(\cap_{i\in I}A_i)\subset A_i$, this implies that $S(\cap_{i\in I}A_i)\subset \cap_{i\in I}A_i$. Since by hypothesis, $\cap_{i\in I}A_i\subset S(\cap_{i\in I}A_i)$, we deduce that $\cap_{i\in I}A_i=S(\cap_{i\in I}A_i)$.

Answer 2

Hint: Consider the set $\tau=\left\{A\subset X\,\middle|\,S\left(A^\complement\right)=A^\complement\right\}$. Prove that $\tau$ is the topology that you're after.