Which of the following lists of subsets of the set form topologies.
- - Yes
- - No not in the list.
- Yes
- No , closed under union and intersection and is in the list, but is not.
Prove that a function is continuous if and only if the inverse image of every closed set is closed.
Proof:
We use the following equations from the Boolean algebra of sets.
If is any subset. then
Suppose the inverse image of every closed set is closed. Let be an open set and
Then is closed and so is closed hence is open.
Suppose that is continuous and is closed then is open since is open
but then is closed.
Let be the floor function. for
Is continuous?
Answer:
No because the supspace topology on is descrete so points are open sets but
not an open set.
Let be a set and a collection of subsets closed under finite intersection, then the set of arbitrary unions of sets of form a topology.on [ Added : and are in ]
Proof:
Again, the solution amounts to recalling some properties of Boolean Algebras. In this case
More generally, if and are indexed set of sets then
and similarly for any arbitrary collection of indexed sets.
And if and are indexed set of sets then where is either in the index set or in the index set and similarly for an arbitrary collections of index sets.
We are given that and are in . We are also given that arbitrary unions of sets of is in . This is because a union of unions of sets in is a union of sets. That a finit intersection of sets of is in is because an intersection of unions of sets in is a union of intersections of sets.