K topology strictly finer than standard topology

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Homework Help Overview

The discussion revolves around the k topology on the real numbers, specifically addressing how it is strictly finer than the standard topology. Participants are examining the definitions and implications of the k topology as presented in Munkres' book, particularly focusing on the basis elements and their properties.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are exploring the definition of the k topology and its basis elements, questioning the meaning of set differences and the implications for open intervals. There is a discussion about whether certain intervals can be contained within the basis element B and the nature of the elements excluded from it.

Discussion Status

The discussion is ongoing, with participants clarifying their understanding of the k topology and its basis. Some have provided insights into the implications of excluding certain elements from intervals, while others are still seeking clarification on notation and definitions.

Contextual Notes

Participants are grappling with the definitions of open intervals and sets in the context of topology, particularly how the exclusion of specific elements (like 1/n) affects the properties of the k topology. There is also mention of confusion regarding the notation used in the definitions provided in the textbook.

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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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.
 
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.
 
Yes, (-1,1) is a set, and K is a set, and (-1,1) - K is their set difference.
 
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.
 
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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