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Math Help - Limit Points

  1. #1
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    Limit Points

    Definition: Suppose that S is a set of real numbers and that x is any real number. We say that the number x is a limit point of the set S if for every number \delta>0 we have (x-\delta, x+\delta)\cap S \ \{x\} is not equal to \emptyset. The set of limit points of a given set S is written as L(S).

    Problem: Given any set S of real numbers, prove that the set L(S) must be closed.
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    Let x1,x2,... be a convergent sequence of L(S) and let x be its limit. We must prove that x belongs to L(S). This means proving that x is a limit point of S. Fix epsilon>0. Take x_n such that | x_n - x | < epsilon/2. Since x_n is a limit point of S, there is a y in S such that | y - x_n | < epsilon/2. Hence y is in the epsilon neighborhood of x (apply triangle inequality). Since epsilon was arbitrary, this proves that any neighborhood of x has a point of S, hence x is a limit point of S. Therefore x is in L(S) and L(S) is closed.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Slazenger3 View Post
    Definition: Suppose that S is a set of real numbers and that x is any real number. We say that the number x is a limit point of the set S if for every number \delta>0 we have (x-\delta, x+\delta)\cap S \ \{x\} is not equal to \emptyset. The set of limit points of a given set S is written as L(S).

    Problem: Given any set S of real numbers, prove that the set L(S) must be closed.
    An alternative way of thinking about it is this:

    Let s be a limit point of L(s) then for every neighborhood N of s there is a point s' of L(s), but N is also a neighborhood s' and so it must contain infinitely points of S in particular it must contain a point of S different from s...so
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