1. Let's consider the set of real numbers R and two topologies defined on it:

T1 - standard real line topology;

T2 - lower-limit topology.

Therefore, I see that:

T1 = {empty set, R, all intervals (a,b)};

T2 = {empty set, R, all intervals [a,b)}.

At the same time, it is considered that T2 is

**strictly finer**than T1. That actually implies that T1<T2 which assumes that T2 contains all elements of T1. But T2 does not have intervals (a,b)? Does it?

2. Let's consider again the set R of real numbers and the lower-limit topology T2. Apparently, with this topology R is Hausdorff space, since every two distinct points can be separated by neighborhoods. There is a property that

**Every singleton in Hausdorff space is closed.**

At the same time, according to the definition of closed set as complement of open set, I can see that only closed set in T2 are just

empty set and R (they are actually both open and closed in every topology);

[a,b) (they are actually both open and closed in T2);

(-inf,a).

But I can't think of example where singleton could be a complement of open set in T2. Unless, as it was discussed in question 1, the intervals (a,b) are also open in T2.

What you think about that?