Let $X$ be a set and define $F:\mathcal{P}(X)\to\mathcal{P}(X)$ that satisfies the following: $$(1) F(\emptyset)=\emptyset\\ (2)\forall A\subset X:\quad A\subset F(A)\\ (3)\forall A,B\subset X:\quad F(A\cup B)=F(A)\cup F(B)\\ (4)\forall A\subset X:\quad F^2(A) = F(A) $$

Let $\tau:=\{U\subset X: F(X\setminus U)=X\setminus U\}$ be a topology on set $X$ and let $C(A)$ represent the closure of set $A$ with respect to $\tau$.

Show that for every $A\subset X\quad$ $F(A)=C(A)$

Let $A\subset B\subset X$. By (2) we have $A\subset F(B)$, therefore $F(B) = F(A)\cup F(B)$, therefore $$F(A)\subset F(B)\tag{a}$$

Because for every $A\subset X$, $X\setminus C(A)\in\tau$ we have by (a) $$C(A) = F(C(A))\supset F(A) $$

However, I'm stuck on showing $C(A)\subset F(A)$.
Assume for contradiction there exists $x\in C(A)$ such that $x\in X\setminus F(A)$. Because $x$ is an adherent point, for every $U_x\in\tau$ (a $\tau$-open set that contains $x$) we have $U_x\cap A\neq\emptyset$. By (2) $x\in X\setminus A$, so that must mean $x\in C(A)$ is exactly on the frontier (?). How do we proceed from here?

I have also attempted to use the four axioms to obtain a direct proof. For instance, by (2) we would get: $$C(A)\subset C(F(A)) $$ , but this does not imply what we need.


I assume that you've already verified that $\tau$ is a topology. By definition, a set $A\subset X$ is closed in this topology if and only if $F(A)=A.$

Since $F(F(A))=F(A),\ $ $F(A)$ is a closed set; i.e., $C(F(A))=F(A),$ so $C(A)\subset C(F(A))=F(A).$

For the other direction, since $C(A)$ is a closed set, we have $F(C(A))=C(A),$ and $A\subset C(A),$ so $F(A)\subset F(C(A))=C(A).$

  • $\begingroup$ it's so obvious !! I wasn't thinking about the 4th axiom like that at all :< smacks forehead $\endgroup$ – Alvin Lepik Oct 10 '16 at 7:48

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