2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $F(A)\subset F(C(A))=C(A).$

$\endgroup$
  • $\begingroup$ 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

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.