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Thread: Closure and accumlation point problem

  1. #1
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    Closure and accumlation point problem

    Suppose that the sequence $\displaystyle \{ x_n \} $ converges to a point x, prove that $\displaystyle x \in \ cl \{ x_1, x_2, x_3, . . . \} $

    In addition, when would x be an accumlation point.

    Proof so far.

    So since $\displaystyle x_n \rightarrow x $, $\displaystyle \forall \epsilon > 0 $, there exist $\displaystyle N \in \mathbb {N} $ such that whenever $\displaystyle n \geq N $, we would have $\displaystyle d(x,x_n) < \epsilon $.

    Now, I need to show that this x is in the closure, which is the intersection of all closed set containing $\displaystyle \{ x_1, x_2 , . . . , \} $.

    So my idea is, for any closed set that contains the sequence, suppose that x is not in them, then x is not in the set that the sequence is in, then but the completeness property, that is impossible.

    And x would be an accumlation point if the set is closed.

    Is this right?
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Suppose that the sequence $\displaystyle \{ x_n \} $ converges to a point x, prove that $\displaystyle x \in \ cl \{ x_1, x_2, x_3, . . . \} $
    In addition, when would x be an accumulation point?
    Here are some facts that may help you understand accumulation points.
    • No finite set can have an accumulation point.
    • If x is in the closure of S but not in S then x is an accumulation point of S.
    • If x is an accumulation point of S the there is a sequence of distinct points of S converging to x.
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  3. #3
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    If x is an accumulation point of S the there is a sequence of distinct points of S converging to x.

    Why is this statement true?

    Thanks.
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    If x is an accumulation point of S the there is a sequence of distinct points of S converging to x. Why is this statement true?
    It is easy to construct such. If a is an accumulation point of S any open set containing a contains a point of S different from a. If we call it b then there is an open set containing a but not b. Thus we can get a new point c of S different from a in the second set. So by a inductive process we can construct the sequence.

    Warning: That is a very loose outline on how we prove that statement.
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  5. #5
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    Here is what I have:

    Proof.

    Suppose that the sequence $\displaystyle \{ x_n \} $ is a finite set, then $\displaystyle x \in \{ x_n \} $, implies that $\displaystyle x \in cl \{ x_1 , x_2,... \} $.

    So suppose that $\displaystyle \{ x_n \} $ is not finite and that x is not included in the sequence.
    Claim: x is an accumulation point of the sequence.

    Let U be any open set that contains the point x, then there exist $\displaystyle \gamma > 0$ such that $\displaystyle D(x, \gamma ) \subset U $.

    Now, since the sequence $\displaystyle \{ x_n \} \rightarrow x$, pick $\displaystyle \epsilon = \frac { \gamma }{2}$, then there exist an $\displaystyle N \in \mathbb {N} $ such that whenever $\displaystyle n \geq N$, we have $\displaystyle d( x_n , x ) < \epsilon < \gamma \ \ \ \ \ \forall n $.

    So each points in the sequence is contained in the set U, which means x is an accumulation point of the sequence, implies that x is in its closure.

    Is this right? Thanks.
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