Prove that there is a unique topology given interior operator

Question

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!

Answer 1

For the other direction, let $B \subseteq X$ and let $y \in f(B)$. By definition of interior, it suffices to find a $U \subseteq B$ such that $y \in U$ and $f(U)=U$. Take $U=f(B)$.

Answer 2

Handy fact: $f$ is monotonic: $A \subseteq B$ implies $f(A) \subseteq f(B)$. Proof: $A \subseteq B$ implies $A \cap B = A$ so $f(A \cap B) = f(A) \cap f(B) = f(A)$ by $(I_3)$ and so $f(A) \subseteq f(B)$. QED.

Suppose $A$ is a subset of $X$. Then $f(A)$ is open in $\mathcal{T}$, because $f(f(A)) = f(A)$ by $I_4$, so $f(A)$ is a fixpoint of $f$ so a member of $\mathcal{T}$.

We also have $f(A) \subseteq A$ by $I_2$ and so $f(A) \subseteq \operatorname{int}(A)$ as the interior is the maximal open subset of $A$ by definition, and $f(A)$ is an open subset of $A$.

Also $\operatorname{int}(A) = \cup \{O \in \mathcal{T}: O \subseteq A\}$ (this is just the definition of interior in any topology), so take any open set $O$ that is a subset of $A$. But then $f(O) \subseteq f(A)$ by monotonicity but as $f(O) = O$ as $O \in \mathcal{T}$ we have $O \subseteq f(A)$ and as this holds for all $O$ in the union that defines the interior we have $\operatorname{int}(A) \subseteq f(A)$ and equality ensues.

Answer 3

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 $U=f(U)\subseteq f(A)$.