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Rudin 2.41, why does an unbounded infinite set have no limit points?

  1. Nov 4, 2011 #1
    Please let me know if this should be in the Homework section.

    Part of Rudin 2.41 says that for a set E in ℝk... "If E is not bounded, then E contains points xn with |xn|>n (n = 1, 2, 3,...)." I can understand the argument this far. I do not get the next sentence "The set S consisting of these points xn is infinite and clearly has no limit point in ℝk". I can understand that S is infinite but to me S clearly does have limit points in ℝk. For example, take the real line, n = 1, S to be the set of points greater than 1 and ε>0. Every neighborhood of p = 2 with radius ε would contain a point in S making p a limit point. This contradicts what Rudin states as 'clearly'. What am I missing here?


  2. jcsd
  3. Nov 4, 2011 #2
    It's not true that an unbounded infinite set has no limit points.

    But it's not an arbitrary such set. It's a sequence, so a countable set. A finite set has no limit points. So, any finite subset has no limit points and it's going off to infinity. If there were a limit point, that would contradict the fact that each x_n is bigger than n because there would be some subsequence that converges. A convergent sequence is bounded.
  4. Nov 6, 2011 #3
    Pretty much what the previous poster said, but you're missing the point that rudin is not defining S as ALL points greater than some n, but rather each x_n is ONE point greater than n: in other words, an infinite sequence.

    Rudin defines this sequence using distinct points, thus while there are an infinite amount, the points are "discrete" objects, if you will. They do not form a connected set, there is distance between each one. Hence, there is no point such that EVERY neighborhood contains a point of the sequence.
  5. Nov 12, 2011 #4
    I get it now. Back in 2.7 he defines a sequence and whenever you see x_n for (n=1, 2, 3,...) you should read this as a sequence.

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