Recent content by 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

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...