Thread: Help with coarser/finer topologies

Consider the usual topology on R (set of real nos).


U = {V ⊆ R: if x ∈ V then ∃ an open interval (a,b) s.t. x ∈ (a,b) ⊆ V}
and consider the left-hand topology on R..

L = {V ⊆ R: if x ∈ V then ∃ a,b ∈V, a < b, s.t. x ∈ [a,b) ⊆ V}

does it follow that the left hand topology is finer than the usual topology? Is any U-open set also L-open? I'm confused.

Is (0,1), which is U-open, also L-open?

Can someone offer enlightenment? Thanks.

The equation shows that every open interval is a union of L-open sets, and hence is open in the L topology. Since the open intervals generate the usual topology, it follows that U⊆L (in other words, the left-hand topology is finer than the usual topology).

In particular, if you put a=0 and b=1, the same equation tells you that (0,1) is L-open.