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Boundary of {1/n : n is a natural number}

  1. Dec 1, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm trying to figure out the the boundary of the set of all 1/n, where n is a natural number.
    Consider this as a subset of R with its usual metric, nothing fancy.

    2. Relevant equations
    There are at least two "equivalent" definitions of the boundary of a set:
    1. the boundary of a set A is the intersection of the closure of A and the closure of the complement of A.
    2. the boundary of a set A is the set of all elements x of R (in this case) such that every neighborhood of x contains at least one point in A and one point not in A.


    3. The attempt at a solution\
    The problem is that I get a different solution depending on which definition I use. Please help me find my error!

    Let A = {1/n : n is a natural number}.

    If I use the definition 1 above:
    - closure of A = {0} union A (since 0 is the only accumulation point of A)
    - complement of A = (-infty,0] union [1,infty) union ( (0,1] - {1/n: n is a natural number} )
    - closure of the complement of A = (infty, 0] union [1, infty)
    union ( (0,1] - {1/n: n is a natural number} )
    (there are no accumulation points for the complement of A in the interval (0,1] )
    - hence the boundary of A is {0,1}

    However! using definition 2 it is clear that:
    - every neighborhood of 0 contains a point of A (Archimedean property) and a point not in A
    - if x is an element of A then every neighborhood of x contains a point contains a point of A (namely x) and a point not in A.
    - hence the boundary of A is {0} union A.
     
    Last edited: Dec 1, 2009
  2. jcsd
  3. Dec 2, 2009 #2
    This sure doesn't sound right. Why isn't for example 1/2 an accumulation point for A complement?
     
  4. Dec 2, 2009 #3
    Well, it most certainly is an accumulation point for A complement. Indeed, every member of A is an accumulation point for A complement. I don't know what I was thinking. Thanks!
     
  5. Dec 2, 2009 #4

    HallsofIvy

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    So A= {1, 1/2, 1/3, 1/4, ...}. It should be obvious that, around each point in A is possible to construct a neighborhood with small enough radius (less than the distance to the next number in the sequence) that does not contain any other members of A. Thus, every point in A is a "boundary point". But there is one point [/b]not[/b] in A that is a boundary point of A. Can you think what that is?
     
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