# Showing a collection of closed subsets of the set X is a topology

Let C be a collection of subsets of the set X. Suppose that "empty set" and X are in C, and that finite unions and arbitrary intersections of elements of C are in C. Show that the collection T = {X-C : C in C} is a topology on X.

I am having issues in using the right form to answer this question. Or maybe i am completely off:

To show this set is a topology i have to show the 4 traits of a topology are existant:

1) X - "empty set" = X since empty set is close --> X is open

2) X - X = empty set, since X is also closed, empty set is also open

3) X - (finite intersections) Ai = U(unions) (X - Ai)

4) X - (arbitrary unins) Ai = (intersectins) (X - Ai)

Am i even treating the topology in the right way? This has to do with the section on closed sets in Munkres section 17. Thanks

• You seem to be on the right track. Do you have a question about a particular piece of the proof, or using terminology to write your ideas as a comprehensible proof? – aschepler Oct 14 '14 at 1:34
• The use of the right sets. Was i right to substitute the respective sets into C to satisfy the 4 needed properties. I ask because it seems almost identical to showing that a set X is a topological space under the closed train of thought. But this topology is defined as X - C not just X alone. – dc3rd Oct 14 '14 at 1:45
• This is nothing more than the application of DeMorgan's Law – Jose Antonio Oct 14 '14 at 2:18

(3) should be $$\bigcup_{\alpha\in A}(X-A_\alpha)=X-\bigcap_{\alpha\in A}A_\alpha.$$
(4) should be $$\bigcap_{k=1}^n(X-A_k)=X-\bigcup_{k=1}^nA_k.$$
it is true for indiscrete topology where the smallest possible topology, i.e $$\{X,\emptyset\}$$, since $$X$$ and $$\emptyset$$ are clopen, hence we can call that a special case but generally the collection of closed subsets of $$X$$ doesn't form a topology.