Is Countable Complement Topology a Valid Topological Structure on R?

[Fluffman4]

Let T be the collection of subsets of R consisting of the empty set and every set whose complement is countable.

a) Show that T is a topology on R.

b) Show that the point 0 is a limit point of the set A= R - {0} in the countable complement topology.

c) Show that in A = R -{0} there is no sequence converging to 0 in the countable complement topology.



As far as proving that it's a topology, I think I was able to come up with that the empty set and T are in the topology. As for showing that the union of sets in T are open and that the intersection of sets in T are open, that's another story. I'm kind of confused as to how to go about it, if countable sets are always open. It's also kind of confusing to me since all the sets in T are the sets whose complement is countable, then does that imply that all the sets in T are uncountable?

 

[lippyka]

a) To show that T is a topology on R, we need to show that it is closed under finite intersections and unions, contains the empty set and R, and any set in T is open. Since the empty set and R are both subsets of R whose complements are countable, they are in T. Now, let A,B be in T. Then, their complements are countable, so A∩B and A∪B are also in T since their complements are the union of two countable sets, which is countable. So, T is closed under finite intersections and unions. Now, let S be an arbitrary subset of R with a countable complement. Then, S is open in T since its complement is countable. Thus, T is a topology on R. b) To show that 0 is a limit point of A=R-{0}, we need to show that every open set containing 0 intersects A in points other than 0. Let U be an open set containing 0. Then, U is in T since its complement is countable. Since U is in T, U intersects A in points other than 0, because U contains 0 and A does not. Thus, 0 is a limit point of A in the countable complement topology.c) To show that there is no sequence converging to 0 in the countable complement topology, we need to show that for any sequence {xn} in A, xn≠0 for all n∈N. Since A=R-{0}, xn≠0 for all n∈N. Thus, there is no sequence converging to 0 in the countable complement topology.