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Math Help - prove it is closed without refering to limit points

  1. #1
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    prove it is closed without refering to limit points

    I'm working on topology again and I wanted to know if this proof for a question that I came on is correct.

    The question was if the set { \left{\frac{1}{i}\right}} ^n_{i=1}, let's call it S, was closed in the Euclidean topology. From prior reading I knew it wasnt since it was missing it's limit point, but I couldn't use that here so I went with this.

    Take S' and suppose that it were open. There are smaller and smaller open intervals beside and not containing smaller and smaller rational numbers approaching zero.

    Let (a,b) be the open interval that surrounds zero. To the right of it there will be an open interval but to the left of that interval will be a point of S no matter how much you shrink (a,b).

    Therefore the set is not closed.
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  2. #2
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    Quote Originally Posted by magus View Post
    The question was if the set { \left{\frac{1}{i}\right}} ^n_{i=1}, let's call it S, was closed in the Euclidean topology. From prior reading I knew it wasnt since it was missing it's limit point, but I couldn't use that here so I went with this.
    You have given us this set S = \left\{ {\dfrac{1}{i}} \right\}_{i = 1}^n .
    That is a finite set.
    Therefore, it is closed in the Euclidean topology.
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  3. #3
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    My mistake, it should have been S = \left\{ {\dfrac{1}{i}} \right\}_{i = 1}^\infty
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  4. #4
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    Then the crucial point is: \displaytype \lim_{n\to\infty}\frac{1}{n}= 0 which is not of the form \frac{1}{n} for any integer n.
    That is essentially what you showed.
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  5. #5
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    Quote Originally Posted by magus View Post
    My mistake, it should have been S = \left\{ {\dfrac{1}{i}} \right\}_{i = 1}^\infty
    Then your title of the OP is totally misleading.
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  6. #6
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    I'm sorry. What I was trying to say was I needed an argument that did not include the fact that closed sets contain their limit points. Sorry.
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  7. #7
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    Quote Originally Posted by magus View Post
    I'm sorry. What I was trying to say was I needed an argument that did not include the fact that closed sets contain their limit points. Sorry.
    I am also sorry to tell you that I have no idea what that response means.
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  8. #8
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    Quote Originally Posted by Plato View Post
    I am also sorry to tell you that I have no idea what that response means.


    I understood what he meant, apparently, from the beginning: he needs to show that that set isn't closed

    without using at all limits points.

    For example, if \left\{\frac{1}{n}\right\}_{n=1}^\infty is closed, then its complement is open, but

    this isn't so since 0\in Comp.\left(\left\{\frac{1}{n}\right\}_{n=1}^\infty  \right) , but there is no

    open ball around zero which is completely contained in that complement. Q.E.D.

    Tonio
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