The usual topology is the smallest topology containing the upper and lower topology

rlkeenan

Trying to prove:
The usual topology is the smallest topology for R containing Tl and Tu.
NOTE: for e>0
The usual topology: TR(R)={A<R|a in A =>(a-e,a+e)<A}
The lower topology: Tl(R)={A<R|a in A =>(-∞ ,a+e)<A}
The upper topology: Tu(R)={A<R|a in A =>(a-e, ∞)<A}

3. The attempt at a solution
claim 1: If T is a topology for R s.t. Tl<T and Tu<T then TR<T
proof: let Tl<T and Tu<T
claim 1.1: If p is in TR then p is in T
proof: let p=(a,b) for any a,b in R
Then p is in TR by definition of TR
We know (-∞,b) is in Tl<T and (a,∞) is in Tu<T therefore (-∞,b),(a,∞) are in T
and (-∞,b)^(a,∞)=(a,b) is in T since T is a topology
therefore p is in T
therefore TR<T
...not sure where to go after here maybe show that if there is a T'<TR s.t. Tu<T' and Tl<T' then T'=TR

Eynstone

The smallest topology containing the two would be generated by the intersections of sets open in the two topologies , i.e., sets of the form (-∞ ,b)
and (a, ∞).This is either empty or just the interval (a,b).

rlkeenan

I get that but that implies that TR={(a,b)|a,b in R} so you get TR<Tl and TR<Tu and we already know that Tl<TR and Tu<TR so then you have that TR=Tl and TR=Tu so then you have Tl=Tu which can't be right