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Point set topology

  1. Jul 15, 2012 #1
    Why is it that the set A={1/n:n is counting number} is not a closed set?

    We see that no matter how small our ε is, ε-neighborhood will always contain a point not in A (one reason is that Q* is dense in ℝ), thus, all the elements in A is boundary point, and we know that by definition, if bd(A)≤A, then A is closed (this is what Steven R. Lay used in his book). (≤-subset). A good friend of mine told me that A does not contain cluster point and that made A not a closed set, he said (and i know) that closed set always contain cluster points. is this some sort of contradiction?
  2. jcsd
  3. Jul 15, 2012 #2
    Sure, all points in A are boundary points. This means that [itex]A\subseteq bd(A)[/itex]. What you want is the reverse inclusion!! So you have to show that all boundary points are exactly in A. This is not true here, there is a boundary point of A that is not in A.
  4. Jul 17, 2012 #3


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    Re the cluster points, it is true that a closed set contains all its cluster points. Maybe your friend was referring to your set A: does it contain all its cluster points?
  5. Jul 18, 2012 #4


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    As micromass said, the fact that all points in A are boundary points is irrelevant. In order to be closed, all boundary points must be in A. Since this is a sequence of points converging to 0, 0 is as boundary point but is not in A. That is what your friend was saying.
  6. Jul 18, 2012 #5
    The set 1/n (n = 1,2,...) doesn't contain any limit points (can you see why?), but it certainly has a limit point (can you see what the limit point is?) and so from the definition we see that this set is not closed.
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