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Math Help - Basic topology

  1. #1
    Member mohammadfawaz's Avatar
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    Basic topology

    Hello,

    I have the following problem:
    Prove that the set of limit points E' of a subset E is a closed set.

    Help please,

    Thank you,
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  2. #2
    Member mabruka's Avatar
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    What definition of closed are you using?

    I think the complement of an open set right?
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  3. #3
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    Quote Originally Posted by mohammadfawaz View Post
    Prove that the set of limit points E' of a subset E is a closed set.
    This may not be true in a general space.
    It depends upon the nature of the space.
    What sort of topological space is this set in?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mohammadfawaz View Post
    Hello,

    I have the following problem:
    Prove that the set of limit points E' of a subset E is a closed set.

    Help please,

    Thank you,
    This is true in T_1 spaces. Maybe that is what the OP meant?
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  5. #5
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    Quote Originally Posted by mohammadfawaz View Post
    Hello,

    I have the following problem:
    Prove that the set of limit points E' of a subset E is a closed set.

    Help please,

    Thank you,
    Let (xn) be a convergent sequence in E', (xn) --> x. We want to show x is in E'. I.e, we want to show x is a limit point of E.

    We know x is a limit point of E if given any e>0 there is a point in E that is in the ball of radius e around x. That is, there exists a p in E such that:
    d(x,p) < e

    Heres the proof:

    Since (xn) --> x, then there is a k where d(x,xk) < e/2 *

    and since xk in in E', it is a limit point of E, so there is a point p in E such that: d(xk,p) < e/2 **

    I claim: d(x, p) < e

    Pf.

    d(x,p) <= d(x,xk) + d(xk, p) < e/2 + e/2 (by * and **)
    so d(x, p) < e

    Edit: Put Metric d instead of absolute value, if that's not what you mean, just replace every d(x,y) with lx - yl
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by southprkfan1 View Post
    Let (xn) be a convergent sequence in E', (xn) --> x. We want to show x is in E'. I.e, we want to show x is a limit point of E.

    We know x is a limit point of E if given any e>0 there is a point in E that is in the ball of radius e around x. That is, there exists a p in E such that:
    lx - pl < e

    Heres the proof:

    Since (xn) --> x, then there is a k where l x-xk l < e/2 *

    and since xk in in E', it is a limit point of E, so there is a point p in E such that: l xk - p l < e/2 **

    I claim: lx - pl < e

    Pf.

    lx - pl <= lx-xkl +lxk - pl < e/2 + e/2 (by * and **)
    so lx- pl < e
    ....in a metric space.
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