# If $C$ contains $\emptyset$ and $X$ and is stable by finite unions and arbitrary intersections then $\{X\setminus Q; Q\in C\}$ is a topology on $X$

$C$ is a collection of subsets of $X$ such that $\emptyset$ and $X$ are in $X$. Finite union and arbitrary intersections of elements $Q$ of $C$ are in $C$. Show that the coollection $T = \{X-Q; Q\in C\}$ is a topology in $X$.

For that, I need to verify that $\emptyset, X$ are in $T$. But since $X$ and $\emptyset$ are in $C$, then $X-X = \emptyset \in Q$ and $X-\emptyset = X\in Q$

Now, I need to show that an arbitrary union of elements of $T$, but what are elements of $T$? I know they are elements of the form $X-C$. So I need to pick union of elements $X-C$? How can I show that this arbitrary union is still in $T$?

• It might be good to know the basic set-theoretic relationships between union, intersection, and complement. – Justin Young Sep 8 '16 at 19:40

Suppose $\{A_{i}\mid i\in I\}$ is a collection of sets in $T$. Each $A_{i}$ is of the form $A_{i}=X-Q_{i}$ for some $Q_{i}\in C$. Now, by DeMorgan's Law,

$$\underset{i\in I}{\bigcup}A_{i}=\underset{i\in I}{\bigcup}(X-Q_{i})=X-\underset{i\in I}{\bigcap}(X-(X-Q_{i}))=X-\underset{i\in I}{\bigcap}Q_{i}\in T,$$

since $\underset{i\in I}{\bigcap}Q_{i}\in C$.

For finite intersections, suppose you have $A_{1},\ldots,A_{n}\in T$. Then, applying DeMorgan's Law again, you obtain,

$$\bigcap_{1\leq i\leq n}A_{i}=\bigcap_{1\leq i\leq n}(X-Q_{i})=X-\bigcup_{1\leq i\leq n}(X-(X-Q_{i}))=X-\bigcup_{1\leq i\leq n}Q_{i}\in T,$$

since $\bigcup_{1\leq i\leq n}Q_{i}\in C$.

Okay, so here's a hint, even though the answer has basically been posted by someone else.

Consider an arbitrary union $\bigcup \limits_{i \in I} A_{i}$ of elements of $T$ (i.e., $A_{i} \in T$ for each $i$). We want to show $\bigcup \limits_{i \in I} A_{i} \in T$.

What needs to happen for $\bigcup \limits_{i \in I} A_{i}$ to be in $T$? By the definition of $T$, you need the set's complement to be in $\mathcal{C}$.

But let's look at that complement. $$X - \bigcup \limits_{i \in I} A_{i} = \bigcap \limits_{i \in I} (X - A_{i}).$$

We got this by one of DeMorgan's laws. It shouldn't be too hard to convince yourself that the complement of a union is the intersection of the complements.

Ok, so we got our union to be in the form $\bigcap \limits_{i \in I} (X - A_{i})$. How does this help us?

Well, since $A_{i} \in T$ for each $i$, what does that tell you about the complement of $A_{i}$ (denoted $X - A_{i}$)?

What does this imply about the arbitrary intersection $\bigcap \limits_{i \in I} (X - A_{i})$?

What does that tell you about the complement of this intersection, which is $\bigcup \limits_{i \in I} A_{i}$?