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Homework Help: K topology strictly finer than standard topology

  1. Sep 20, 2006 #1
    I would like a little clarification in how to prove that the k topology on R is strictly finer than the standard topology on R. They have a proof of this in Munkres' book. I know how to prove that its finer, but the part that shows it to be strictly finer im not sure. It says given the basis element B = (-1,1) - K for T'' (the k topology), there is no open interval that contains 0 and lies in B. If what it says is what i think then i can think of many counterexamples, for example: use the element 1/2 of K. (-1,1)-K = (-3/2,1/2). Use the open interval (-1/3, 1/3) which lies in B right?
  2. jcsd
  3. Sep 20, 2006 #2


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    No, all the elements 1, 1/2, 1/3, 1/4, 1/5, ... are outside of B. Any open interval around 0 has to have one of these fractions. Think about it. An open interval around 0 must be of the form (a, b) with a < 0 < b. If b > 1, then choose n = 2. Clearly, 1/2 is in (a, b) but it's not in B. If b < 1, take n = floor(1/b). Then 1/n is in (a,b) but not in B.
  4. Sep 22, 2006 #3
    ok i think i know what my problem was. I took (-1,1) - K to mean the set of all x-1/n between -1 and 1, where n is a positive integer and x is real. I guess the minus K actually means exclude any 1/n for any positive integer n from the interval (-1,1). Yes, in that case, any open interval in there would have to contain some 1/n 's, and therefore is not in B.
  5. Sep 22, 2006 #4


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    Yes, (-1,1) is a set, and K is a set, and (-1,1) - K is their set difference.
  6. Sep 6, 2009 #5
    sorry for bumping this old topic. i'm reading this section right now and i'm very confused.

    can some one give me a notation for definition of k-topology? may be an example? the book said basis is the interval (a,b), along with sets (a,b) - K where K is stated above. What the difference between interval and set? isn't open interval of real numbers is uncountable set?
    so is it (a,b) U [(a,b) - K] ? or just (a,b) - K? or maybe if 0 not in (a,b) then it's just the interval (a,b), and if 0 in (a,b) then it is (a,b) - K?

  7. Nov 23, 2009 #6
    Thanks for clearing this up.

    I was also thinking that (-1,1) - K meant the set of all x-1/n between -1 and 1 (basically, the open interval (-2,1) in R).

    Hopefully this will help becu:

    We're looking at a basis element in the K-topology,
    B = {x in R: -1 < x < 1}\{1/n: n is a natural number}.
    So, if we look at any open interval in R (in the standard topology) containing 0, we cannot find that interval in the R_K topology, since this excludes all numbers of the form 1/n: n is in N, but every open interval containing 0 in R contains a number of the form 1/n (archimedean principle). Thus the interval in R (std.) contains elements which are not in R_K, so by definition the interval cannot be a subset of B.
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