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Homework Help: Topology/Analysis: example of a particular set

  1. Aug 16, 2010 #1

    Esd

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    1. The problem statement, all variables and given/known data
    Give an example of a nonempty set A in R such that A = Bd(A) =
    Lim(A) = Cl(A).

    Bd(A) is the boundary of A, Lim(A) is the set of limit points of A, Cl(A) is the closure of A.

    2. Relevant equations
    Bd(A) = A - Int(A), Int(A) is the interior of A
    Lim(A) = is the set of all such points in R, such that any size neighborhood around any of those points contains at least one point of A different from the point itself.
    Cl(A) = Lim(A)U(A)

    3. The attempt at a solution
    Since Bd(A) = A - Int(A), then for A = Bd(A) => Int(A) is empty
    for A = Lim(A), not only are we looking for a closed set (due to it containing its boundary points), but one which is uncountable. For example, if A = integers, then none of the points in A are in Lim(A), in fact Lim(A) becomes empty. This is what makes me think that in between any 2 points in A, there must always be another point of A. And since Cl(A) = Lim(A) U (A), it will instantly follow once we get Lim(A) = A.

    Some of my attempts have been the singleton set, which makes Lim(A) empty.
    Then I tried the irrationals, but since they're dense, any rational number would become a limit point of A, thus making Lim(A) = R. This leads me to also restrict A to not be dense. Such would also rule out the set of rational numbers.

    After this, I can't deduce any other necessary and therefore neither any sufficient properties for A.

    I would appreciate a couple hints and/or pointing out any error(s) in my reasoning.

    Thank you for your time.
     
  2. jcsd
  3. Aug 16, 2010 #2

    Hurkyl

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    True. Possibly the first important thing to notice, imagination-wise.

    I agree with closed. But why uncountable? :confused:

    Sounds good. Dense would mean A = Lim(A) = R, and this would violate Int(A) = 0.

    So, I think this is the only part of what you said that you haven't really thought through?

    (Incidentally, the only sets with that property are empty, singleton, or dense. If you can prove A must have it, then you would have mostly finished a proof there is no example)
     
  4. Aug 16, 2010 #3

    Esd

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    Hurkyl,

    Thank you for replying. My reasoning for restricting it to be uncountable is as follows. If A was countable, then we have points of A (say p and q), in between which there are no points of A. Thus,for p and q, we would be able to find neighborhoods which wouldn't contain any other points of A. Thus those points (p and q) would not be included in Lim(A), even though they belong to A. So we wouldn't satisfy A= Lim(A).
     
  5. Aug 16, 2010 #4

    Hurkyl

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    That doesn't follow. The rational numbers are a counterexample.

    The union of [0,1] and [2,3] is a counter-example.
     
  6. Aug 16, 2010 #5

    Esd

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    That's correct. I had a faulty definition for "uncountable" in my head. I drop that claim.

    I'm still stuck on how to construct such a set A though.
     
  7. Aug 16, 2010 #6

    Hurkyl

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    For the solution I came up with, I started with a single point, then, I proceeded to modify A so that that point in A, Bd(A), Lim(A), and Cl(A).

    I didn't end up with a pretty answer, but was an answer nonetheless.
     
  8. Aug 16, 2010 #7

    Hurkyl

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    Oh, PS: you mentioned that A must have empty interior, and that A is uncountable. While A doesn't have to be uncountable, those two notions juxtaposed does suggest something, and I'm pretty sure it's an example too.
     
  9. Aug 16, 2010 #8

    Esd

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    Well, the irrationals have an empty interior and are uncountable. But due to being dense, don't work. I'm thinking it's maybe some subset of the irrationals, one which excludes a countable subset of the irrationals. But I don't think there exists a countable subset of the irrationals.
     
  10. Aug 16, 2010 #9

    Hurkyl

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    There's a famous closed uncountable set with empty interior. But you don't need to remember it to be able to do the problem.



    On an unrelated note, why don't you think there are countable subsets of the irrationals?
     
  11. Aug 16, 2010 #10

    Esd

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    I don't think that anymore. An example of a countable subset of irrationals would be the following:

    root(2),root(5), root(2)root(5), root(2)root(2)root(5)..: multiplying root(2) by root(5), such that at least one of root(2) or root(5) appears an odd amount of times.
     
  12. Aug 16, 2010 #11

    Esd

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    I'm still hopeless for this problem. Nor do I know of the famous set you're hinting at.
     
  13. Aug 16, 2010 #12

    Hurkyl

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    Well, let's start from the top. A is nonempty, so let's put a point in it. I like zero.

    Uh-oh, for this choice of A, 0 is not in Lim(A)! How might we change things?
     
  14. Aug 16, 2010 #13

    Esd

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    Well, for 0 to be in Lim(A),the immediate neighborhood of 0 (at least in one direction) must also be in A.
     
  15. Aug 16, 2010 #14

    Hurkyl

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    Ah, but that has non-empty interior! So that won't do.

    Ah, the problem is that we added too many points. Certainly it was enough to get 0 into Lim(A), but it was also enough to point points into Int(A). So, we need to find a much more conservative set of points with 0 as a limit point.

    (Alternatively, we could take this idea and then try removing points to fix things. I think that's hard. But maybe it's up your alley?)
     
  16. Aug 16, 2010 #15

    Esd

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    I knew that I filled the interior once I did that. However, I can't think of anything that would solve the problem of making 0 a limit point, without breaking anything else. Earlier I was thinking of having 0 and then adding numbers of the form 1/n. n being an integer. Such would make 0 a limit point, as no matter what distance from 0, we'd find n large enough so that 1/n fits in that neighborhood. The interior is still empty. However, in between 1/(n-1), 1/n, 1(n+1), there are no other points of such form. So 1/n is not a limit point.
     
  17. Aug 17, 2010 #16

    Esd

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    I was just reading the description of the Cantor set. It seems to fit. All the points are boundary points, as the open intervals are taken out. Since it's a 1/3 each time of the remaining, the distance between any given point to the next iteration of a point created next to it only get's smaller. Thus making the points limit points. No other points, but those created by this process can be limit points, as the proximity to any fixed point taken out, does not converge to 0. This makes Cantor = Bd(Cantor) = Lim(Cantor) = Cl(Cantor). It seems that in order to create other such sets, we can change the fraction of 1/3 to something else. We could even alternate different fractions for each iteration. Is this finally a correct example for the particular set A?
     
  18. Aug 18, 2010 #17

    Hurkyl

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    So then, for each n, fix that. :smile:



    Anyways, yes, the Cantor set is another example.
     
  19. Aug 18, 2010 #18

    HallsofIvy

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    What is the simplest possible non-empty set? Will that satisfy these conditions?
     
  20. Aug 18, 2010 #19

    Esd

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    If by simplest, you mean the singleton set - then no. Because Lim(singleton) is empty.
     
  21. Aug 19, 2010 #20

    HallsofIvy

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    Ah- I had overlooked that condition- sorry.
     
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