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

  1. Oct 2, 2011 #1
    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
     
  2. jcsd
  3. Oct 3, 2011 #2
    Re: The usual topology is the smallest topology containing the upper and lower topolo

    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).
     
  4. Oct 3, 2011 #3
    Re: The usual topology is the smallest topology containing the upper and lower topolo

    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
     
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