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Closure and interior problem

  1. Jul 13, 2010 #1

    radou

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    I'm not sure about my answers, any help is highly appreciated.

    Let (N, U) be a topological space, where N is the set of natural numbers (without 0), and U = {0} U {Oi, i is from N}, where Oi = {i, i+1, i+2, ...} and {0} is the empty set. One has to find the interior (Int) and closure (Cl) of these sets:

    (a) A = {n from N : (n - 7)/(n - 11) > 0}
    (b) A = {13, 5, 2010}

    So, for (a):

    Obviously, A = {n from N : n > 11 or n < 7} = N \ {7, 8, ... , 11}. The interior of A, Int(A), is by definition the union of all open subsets contained in A, and the open subsets in the topology (N, U) are elements of the family U or their unions and finite intersections. So Int(A) = {0} (the empty set, since no open subset in U can contain A. Now, the closure of A, Cl(A), is by definition the intersection of all closed subsets which contain A, i.e. the smallest one of them. So, Cl(A) = N, and N is closed, since its complement is the empty set, which is open. (I feel I'm missing something huge here.)

    (b) Again, Int(A) = {0}. What would Cl(A) be? The first guess is Cl(A) = {5, 6, ...}, but is this set closed? Its complement is {1, 2, 3, 4}, but this is no open set contained in U?

    I hope I didn't cause much confusion, but I need to solve this problem in order to clear out my way of thinking.
     
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  3. Jul 13, 2010 #2

    HallsofIvy

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    Are you sure that "{0}" is the empty set? That's a very strange way to represent it. Normally, "{ }" is the empty set, and "{0}" is the set whose only member is "0".
     
  4. Jul 13, 2010 #3

    radou

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    Oops, actually, I didn't find the symbol for the empty set, so I used "0" to represent it - so, we're talking about the set containing the empty set, {0} (this is exactly what's written in the problem I'm working on).
     
  5. Jul 13, 2010 #4
    The set O12 = {12, 13, ...} is contained in A and is also an element of U, and thus is open in (N, U). It is therefore contained in Int(A) as well.
     
  6. Jul 14, 2010 #5

    radou

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    OK, so Int(A) is the union of the sets Oi, where i >= 12, right? What about the closure of A?
     
  7. Jul 14, 2010 #6
    Are you using the concept of limit points, or the concept of the frontier of A: Fr(A) ?
    In either case, what do the neighborhoods of points look like in (N, U)?
     
  8. Jul 14, 2010 #7

    radou

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    Both of these concepts are defined later on in the exercise notes I'm going through, so I guess I don't need to use them.

    If i is an element of N, then ... , Oi-1, Oi, Oi+1, ... are all its neighborhoods, since they're open sets in U containing i. Now, an element i is in the closure of A iff every neighborhood of x intersects A, so Cl(A) = {1, 2, ... , 11} (since for 12 and greater elements of N, one can find neighborhoods which don't intersect A).
     
  9. Jul 14, 2010 #8
    Oi+1 does not contain i. Only the sets On where n is less than or equal to i contain i.

    Every neighborhood of 12 intersects A, since A contains all natural numbers greater than 11.
     
  10. Jul 14, 2010 #9

    radou

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    Wow, I just realized what silly stuff I wrote down. Let's start over.

    If x is an element of N, then every neighborhood of x intersects A, so CL(A) = N, right?
     
  11. Jul 14, 2010 #10
    Yep. :)
     
  12. Jul 15, 2010 #11

    radou

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    OK, and for the set {13, 5, 2010}. Its closure equals N, and its interior is the empty set, right? (Since no open set from U is contained in this set) Thanks in advance.
     
  13. Jul 15, 2010 #12
    This set's closure cannot be N, because the set O2011, an open set containing 2011, does not intersect it. The interior is correct.
     
  14. Jul 15, 2010 #13

    radou

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    Oh yes, right. Thanks a lot. :)

    So, the closure of A is {1, 2, ... 2010} or O1 \ O2011, right?
     
  15. Jul 15, 2010 #14
    Yep, that's it. :)
     
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