Show that $\tau := \{X \setminus \phi (A) \mid A \subset X \}$ is a topology on $X$. Let $X$ be a set and $\mathcal P (X)$ the powerset of $X$. Suppose $\phi: \mathcal P(X) \to \mathcal P(X)$ is a function such that


*

*$\phi(\emptyset) = \emptyset$,

*For all $A \subset X$: $A \subset \phi(A)$,

*For all $A \subset X$: $\phi(\phi(A)) = \phi(A)$ and

*For all $A, B \in \mathcal P(X)$: $\phi(A \cup B) = \phi(A) \cup \phi(B)$.


Show that $\tau := \{ X \setminus \phi(A) \mid A \subset X \}$ is a topology on $X$.
Progress I've made:


*

*$\emptyset = X \setminus \phi(X) \in \tau$.

*$X = X \setminus \phi(\emptyset) \in \tau$.

*Finite intersections: $(X \setminus \phi(A)) \cap (X \setminus \phi(B)) = X \setminus (\phi(A) \cup \phi(B)) = X \setminus \phi(A \cup B) \in \tau$. 

*Infinite unions: This is what I'm struggling with. How do I complete the proof?

 A: Easy fact from the axioms: $C \subseteq D$ implies $\phi(C) \subseteq \phi(D)$: 
$C \subseteq D$ iff $C \cup D=D$ which implies $\phi(C) \cup \phi(D) = \phi(C \cup D) = \phi(D)$ which holds iff $\phi(C) \subseteq \phi(D)$. QED
Suppose the subsets $B_i, i \in I$ of $X$ obey $\phi(B_i) = B_i$ for all $i$.
Then define $B=\bigcap_{i \in I} B_i$ and note that by the second axiom $B \subseteq \phi(B)$. On the other hand, for all $i$ $B \subseteq B_i$ so that by the first fact we have $\forall i: \phi(B) \subseteq \phi(B_i) = B_i$ so $\phi(B) \subseteq \bigcap_{i \in I} B_i = B$ and so $\phi(B) = B$. 
Now if $X\setminus \phi(A_i), i \in I$ are in $\tau$ then:
$$\bigcup_i (X\setminus \phi(A_i)) = X\setminus \bigcap_{i \in I} \phi(A_i)$$ and as $\phi(\phi(A_i)) = \phi(A_i)$ we apply the last paragraph's fact (with $B_i = \phi(A_i)$ on intersections to see that $\bigcap_{i \in I} \phi(A_i) = \phi\left(\bigcap_{i \in I} \phi(A_i)\right)$ and so $$\bigcup_i (X\setminus \phi(A_i)) = X \setminus \phi\left(\bigcap_{i \in I} \phi(A_i)\right) \in \tau$$
as required. The proof idea is clear: sets $\phi(A)$ are just the closed sets of the topology and thus ought to be closed under all intersections. 
