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  1. #1
    Senior Member Danneedshelp's Avatar
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    Topology help

    Q: Let x\in{O}, where O is an open set. If (x_{n}) is a sequence converging to x, prove that all but a finite number of the terms of (x_{n}) must be contained in O.

    I am not sure where to begin with for this proof. Looking at it, i'd assume we need to use the definition of a cluster (limit point) somewhere in the proof. However, I am not sure. I am having a trouble breaking down the statement, so that I know what my goal really is.
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  2. #2
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    Quote Originally Posted by Danneedshelp View Post
    Q: Let x\in{O}, where O is an open set. If (x_{n}) is a sequence converging to x, prove that all but a finite number of the terms of (x_{n}) must be contained in O.

    I am not sure where to begin with for this proof. Looking at it, i'd assume we need to use the definition of a cluster (limit point) somewhere in the proof. However, I am not sure. I am having a trouble breaking down the statement, so that I know what my goal really is.


    O is open. So there exists an open ball B(x,epsilon) in O containing x, for some epsilon > 0. Pick such an epsilon. Now note that for some natural N and all n>N, |x_n - x|<epsilon and hence x_n is in B(x,epsilon) and also in O for all n>N.

    Hence. . .

    (obviously if we assume that x is not a limit point of O, that is it is in the interior of O, the assumption that O is open is unnecessary)
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by harbottle View Post
    O is open. So there exists an open ball B(x,epsilon) in O containing x, for some epsilon > 0. Pick such an epsilon. Now note that for some natural N and all n>N, |x_n - x|<epsilon and hence x_n is in B(x,epsilon) and also in O for all n>N.

    Hence. . .

    (obviously if we assume that x is not a limit point of O, that is it is in the interior of O, the assumption that O is open is unnecessary)
    So, since (x_{n}) \in(x-\epsilon,x+\epsilon) for every n\geq\\N, only a finite number of the terms of (x_{n}) reside in the neighborhood centered around x.

    Thanks.
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  4. #4
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    Quote Originally Posted by Danneedshelp View Post
    So, since (x_{n}) \in(x-\epsilon,x+\epsilon) for every n\geq\\N, only a finite number of the terms of (x_{n}) reside in the neighborhood centered around x.
    Exactly
    Only the terms x_1,~x_2,\cdots,x_{N-1} may be outside that open set.
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