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Real Analysis Problem

  1. Oct 16, 2007 #1
    For my homework, I have to find a counterexample for this: (with S being a subset of the reals.)
    If P is the set of all isolated points of S, then P is a closed set.

    I don't quite understand the concept of isolated points, which might be why I can't figure out a counterexample.
  2. jcsd
  3. Oct 16, 2007 #2


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  4. Feb 24, 2011 #3
    If it > 0 is any real number and x < y, show that there exists a rational number r such that
    x < ru < y. (Hence the set (ru: r e Q) is dense in R.)
  5. Feb 25, 2011 #4


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    An "isolated" is just a point that is "away" from other points in the set. More technically, there is some neighborhood (open set) containing the point that does not contain any other point of the set. For example, [itex][0, 1]\cup {2}[/itex], the set of all numbers between 0 and 1 (inclusive) and the number 2, has "2" as an isolated point. If a set is a sequence, say, {1, 2, 3, 4, 5}, then every point is an isolated point.

    Think about a sequence of numbers that converge to some limit. What are its isolated points?
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