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

1. The problem statement, all variables and given/known data
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}

2. Relevant equations

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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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