Trying to prove:(adsbygoogle = window.adsbygoogle || []).push({});

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

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

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