|Feb9-09, 07:13 PM||#1|
The question is
Show that the topologies of Rl (which is the lower limit topology) and then.... RK (which is the set of all numbers of the form 1/n, for n E [tex]Z[/tex]+ meaning that they are all 1/n for all the positive integers)
So i can compare this to the standard topology which is...
Given a basis element (a,b) for the standard topology (i will call this T) and a point x of (a,b), the basis element [x,b)for Rl contains x and lies in (a,b). ON the other hand, given the basis element [x,b) for Rl there is no open interval (a,b) that contians x and lies in [x,d). Thus Rl is strictly finer than T.
same applies to Rk.
Given a basis element (a,b) for the standard topology (i will call this T) and a point x of (a,b), the basis element for Rk contains x. ON the other hand, given the basis element B=(-1,2) - K for Rk and the point 0 of B, There is no open interval that contains 0 and lies in B.
I don't know how i can compare the two together... i might be like blind... and that i might be right under my nose... but still.. i'm struggling.
|Feb9-09, 08:01 PM||#2|
Not all topologies are comparable. If you want to prove that one is more comparable than the other, you prove that all open sets in one are open sets in the other. To prove that two topologies aren't comparable, find an open set in one that isn't open in the other, then find one that is open in the other that isn't open in the first one.
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