# 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 $$X$$.

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

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.