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Simple topological argument

  1. May 22, 2013 #1
    Determine the interior, boujdary, and closure for the set:

    { 1/n : n is in the positive integers Z}.

    Attempt: two things bothering me.

    1) if i am in the set of positive integers, how does 1/n even exist?

    2) now let's say it does exist, then the inteior would be empty because every ball drawn around an interior point would contain points that are not from the integers i.e. rationals and reals.

    Moving to the boundary, the boundary would contain all of the points from the positive integers Z but again how would this be the case given that the function is 1/n?

    The closure is defined as the union of the set and the boundary, the set is empty, but the boundary is Z positive, so it would be Z positive?
     
  2. jcsd
  3. May 22, 2013 #2

    Dick

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    Your set doesn't consist of integers, n is an integer. The set is {1/n}={1,1/2,1/3,1/4,...}. It only contains one integer. Now try and think about those questions again.
     
  4. May 22, 2013 #3
    I had a feeling thats what it was meaning. If thats tge case,

    Interior is (0,1) , boundary is the points 0 and 1, and tge closure is the union of those 2.

    Then why would my instructor say that the set itself is the set of integers?
     
  5. May 22, 2013 #4

    Dick

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    You probably understood the instructor, the given problem doesn't say that at all. And your answers are way off. Why would you think the interior is (0,1)?? The interior has to be a subset of {1/n}. Is (0,1)?
     
  6. May 22, 2013 #5
    Oh, well now i have a feeling i get what the instructor was getting at. The interior would be empty because between every integer n, there will be a rational number or an irrational number in which case any ball around a
    1 /n would have a rational or irrational in it.

    The boundry would be all of the integers Z of the form 1/n.

    Closure: is the union of the two above, so all the integers Z of the form 1/n
     
  7. May 22, 2013 #6

    Dick

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    Yes, the interior is empty. If you take any point 1/n then a ball around it contains other points that don't have the form 1/(some integer). The rest of what you are saying is nonsense. Did you draw a sketch of the set of points {1/n}? Does it have any limit points? What's your definition of 'boundary'?
     
  8. May 22, 2013 #7
    Hey Dick, I have a question about this as well.

    Would it be true that the set is its own boundary? That is, that each element of the set {1/n : n is a positive integer} would be a boundary point? If that is true is the set not already a closed set, and therefore the closure of the set is the set itself?

    I'm not sure what I'm talking about as I have very little background (ie. no background) in topology.
     
  9. May 23, 2013 #8

    Dick

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    The set is not a closed set. There's point in the closure of the set that's not in the set. Just one. Where is it?
     
  10. May 23, 2013 #9
    Would it be 0?

    EDIT: My reasoning for this is that as you increase n you get increasingly close to 0 but never actually reach 0. I feel like this is "intuitively" correct, but am unsure how I would justify this "rigorously".
     
    Last edited: May 23, 2013
  11. May 23, 2013 #10

    Dick

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    It would. Explain why? To make it rigorous use the definition of closure.
     
  12. May 23, 2013 #11
    I believe the definition of closure is as follows.

    The closure of a set S is the union of S with its own boundary, or:

    [itex]\overline{S} = S \cup \delta S[/itex]


    So we know that S contains all elements of [itex]\delta[/itex] S with the exception of the point 0, which is contained within the boundary because the value 1/n converges to this point as n approaches infinity? So the difference between 1/∞ and 0 is simply that 1/∞ is contained within the set but is not actually a boundary point.

    I feel as if I'm just repeating myself in a convoluted manner now...
     
  13. May 23, 2013 #12

    Dick

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    That is getting convoluted. My definition says that a point is a boundary point if every neighborhood of that point contains points in S and outside of S.
     
  14. May 23, 2013 #13
    At the moment we haven't discussed limit points, but my definitiin of a boundary point is any ball drawn around a boundary point contains elements from the set and elements from the complement of the set, i.e. the ball around our boundary point that intersects the set is non empty and the same ball intersects with the complement is nonempty.

    But with that being the case, every ball around the point of the form 1/n from Z would obviously include points not in our set. Particularly all points not of 1/n not in Z, but as well wouldn't the point 1/n be in the set?
    Hence why I say that the boundary points are the set of 1/n where n = Z.
     
  15. May 23, 2013 #14

    Dick

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    Yes, all of the points 1/n are boundary points. But there's also a boundary point that is not in your set. BTW don't say 'integers of the form 1/n'. They aren't integers. n is an integer, 1/n isn't.
     
    Last edited: May 23, 2013
  16. May 23, 2013 #15
    I just realized you had semt a reply to @Tsuno, so if my conclusion is correct for the boundary and moving to the closure, are we able to have the closure as being only one point, in this case 0? I always thought that the closure had to be a the collection of both the points in the set and the boundary. Hence why i said all of the boundary is the closure.
     
  17. May 23, 2013 #16

    Dick

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    You are not being very clear here. Is 0 a boundary point or not?
     
  18. May 23, 2013 #17
    Im tempted to say 0 is a boundary point because a boundary point doesn't have to be in the set. As such any ball drawn around the point 0 would have a point in the set amd a point in the complement of the set.
     
  19. May 23, 2013 #18

    Dick

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    Yes, 0 is a boundary point. Now can you describe the boundary and the closure?
     
  20. May 23, 2013 #19
    Then the closure would be the union of the set and the boundary.

    The boundary is the set of 1/n with n from Z and 0

    The set is well like above the set of 1/n where n is fro Z

    That means the closure is essentially 1/n from Z and 0.....i.e. just the boundary points.
     
  21. May 23, 2013 #20

    Dick

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    Yes. The closure and the boundary are both the same. {1/n: n a positive integer} U {0}. Another BTW, Z is the symbol usually used to denote ALL integers, not just the positive ones. N is usual for the positive ones.
     
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