TITLE: Prove that there is a unique topology given interior operator
QUESTION [11 upvotes]: Suppose $f: \mathcal{P}(X) \to \mathcal{P}(X)$ is a function that
satisfies, for every set $A,B \subseteq X$
$(I_1): f(X) = X$
$(I_2): f(A) \subseteq A$
$(I_3): f(A \cap B) =f(A) \cap f(B)$
$(I_4): f(f(A)) = f(A)$
Prove that there exists a unique topology $\mathcal{T}$ on $X$ such
  that for all $A \subseteq X$, $int(A) = f(A)$, where $intA := \{a \in A \mid A \in \mathcal{V}(a)\}$ denotes the
  interior of the set $A$, with respect to the topology $\mathcal{T}$

My attempt:
I defined $\mathcal{T}:= \{A \subseteq X \mid f(A) = A\}$
and I managed to prove with $I_1, I_2, I_3$ that this set is a topology, and that it is unique, if it exists. So I only need to show that it exists, for which I have to prove that $int(A) = f(A)$ whenever $A \subseteq X$
I know that $I_4$ will be crucial to show this, as I did not use it yet. How would I complete the proof?
Edit: I can show that $int(A) \subseteq f(A)$. I just need the other inclusion and I'm done!

REPLY [7 votes]: Let $\mathcal T$ be any topology on $X$ with the property that for all $A\subseteq X$, $\operatorname{int}(A)=f(A)$. Then for $U\in\mathcal T$, we have $U=\operatorname{int}(U)=f(U)$, and for any $B\notin \mathcal T$, we have $B\ne\operatorname{int}(B)=f(B)$. We conclude that $\mathcal T=\{\,A\subseteq X\mid f(A)=A\,\}=:\mathcal T_0$, i.e., there is at most one topology with said property.
Next we have to show that $\mathcal T_0$ is indeed a topology.

We have $X\in\mathcal T_0$ because of $I_1$ and $\emptyset\in \mathcal T_0$ because of $I_2$
$\mathcal T_0$ is closed under finite intersection because of $I_3$
Let $\{U_i\}_{i\in I}$ be a family of sets $U_i\in\mathcal T_0$ and $U=\bigcup U_i$. We need to show $U\in\mathcal T_0$. From $I_2$, $f(U)\subseteq U$. On the other hand, for each $i\in I$,
$$U_i=f(U_i)=f(U_i\cap U)=f(U_i)\cap f(U)\subseteq f(U),$$
by $I_3$, hence $U=\bigcup U_i\subseteq f(U)$ and ultimately $f(U)=U$, $U\in \mathcal T_0$.

Finally, it remains to be shown that $\mathcal T_0$ does have the claimed property. This is where $I_4$ comes into play:
It guarantees that for each $A$, the set $f(A)$ is in $\mathcal T_0$. To finish, the other properties guarantee that there cannot be a larger open set contained in $A$: If $U\in\mathcal T_0$ for $U\subseteq A$, then $U=f(U)\subseteq f(A)$.