# Countable Complement Topology

1. Sep 29, 2010

### 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?

2. Sep 30, 2010

### VeeEight

This is not true.

This is not true either

To show it is a topology, set up your usual arbitrary union of open sets and take the complement in X. There is a set theory identity you use here to show that is it countable. Do the same for the finite intersection of open sets. This is essentially the same set up as the finite complement topology.