Hello,

No, it has to do with the definition of convergence in a topologyLet B be the collection of subsets of R in the form [a,b) for all a < b, the empty set, and the whole set included.

Let T be the topology on R determined by taking all arbitrary unions of sets in B. Show the sequence {1+ (1/n), n= 1,2, ...} converges to one in the topology T.

I thought it was obvious that this sequence converges to 1 because 1/n converges to 0, but then I got confused by the next part.

Show that the sequence {1 - (1/n), n=1,2,...} does NOT converge to 1 in the topology T. I'm assuming this may have something to do with [a,b) where b is not included, but I"m not sure.

Here is the general definition :

Let be a sequence of points in X.

converges to if for any open neighbourhood of x :

So for the first one, you're approaching 1 by superior values. It is easy to prove that for any open set containing 1 (which is a neighbourhood), and particulary for the open sets in the form [1,a), a>1, the definition is verified.

For the second one, you're approaching 1 by inferior values. It is sufficient to take a counterexample :

[1,a) is an open set, but in no way there is an N such that for all n>N, is in this open set, because it is .

hence it doesn't converge to 1.

For the second part, to prove a set is discrete, show that any of its element and subset is an open set in the relative topology. I'm sorry I cannot help much here as I don't understand what the set exactly is