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
Best Answer
(1) and (2) are right.
(3) should be $$\bigcup_{\alpha\in A}(X-A_\alpha)=X-\bigcap_{\alpha\in A}A_\alpha.$$
(4) should be $$\bigcap_{k=1}^n(X-A_k)=X-\bigcup_{k=1}^nA_k.$$