K topology strictly finer than standard topology

In summary, the R_K topology 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 I am 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/
  • #1
ak416
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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 I am 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?
 
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  • #2
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.
 
  • #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.
 
  • #4
Yes, (-1,1) is a set, and K is a set, and (-1,1) - K is their set difference.
 
  • #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?

thanks.
 
  • #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.
 

FAQ: K topology strictly finer than standard topology

What is the K topology and how does it differ from the standard topology?

The K topology is a topology on a set that is strictly finer than the standard (or Euclidean) topology. This means that the K topology contains more open sets than the standard topology, making it a more refined or "finer" structure.

How is the K topology defined?

The K topology is defined by taking the collection of all open intervals (a,b) in the standard topology and adding to it the collection of all sets of the form (a,b) ∪ {x}, where x is any real number. This results in a topology that is strictly finer than the standard topology.

Can you give an example of a set that has different open sets in the K topology compared to the standard topology?

One example is the set [0,1], which has only the open intervals (a,b) in the standard topology. However, in the K topology, it also has sets of the form (a,b) ∪ {x}, where x is any real number. So for example, the set (0,1) ∪ {0.5} is open in the K topology but not in the standard topology.

How does the K topology affect the continuity of functions?

The K topology is strictly finer than the standard topology, which means that more sets are considered open. This can affect the continuity of functions, as a function that is continuous in the standard topology may not be continuous in the K topology. In particular, the K topology is often used to study functions that are discontinuous in the standard topology, such as the Dirichlet function.

Are there any applications of the K topology in real-world scenarios?

The K topology has applications in various areas of mathematics, such as real analysis, topology, and functional analysis. It has also been used in economics and game theory to study equilibrium concepts and decision making. Additionally, the K topology has been applied in physics to study the topology of space-time and the behavior of particles in non-Euclidean spaces.

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