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Math Help - Closure and accumlation point problem

  1. #1
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    Closure and accumlation point problem

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

    In addition, when would x be an accumlation point.

    Proof so far.

    So since  x_n \rightarrow x ,  \forall \epsilon > 0 , there exist  N \in \mathbb {N} such that whenever  n \geq N , we would have 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  \{ 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  \{ x_n \} converges to a point x, prove that  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  \{ x_n \} is a finite set, then  x \in \{ x_n \} , implies that  x \in cl \{ x_1 , x_2,... \} .

    So suppose that  \{ 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  \gamma > 0 such that D(x, \gamma ) \subset U .

    Now, since the sequence  \{ x_n \} \rightarrow x, pick  \epsilon = \frac { \gamma }{2}, then there exist an  N \in \mathbb {N} such that whenever  n \geq N, we have  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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