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  1. #1
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    New post

    I am reading the same book and have similar confusions. Plato if you could please elaborate.

    To restate what we are trying to show. We are trying to show that if for any convergent sequence in S the limit p is also in S then S will be closed. So we are showing this by contradiction. Supposing that S is not closed...then the complement will not be open and so for a given point p in the complement of S the open ball around p will contain points from S. If I understand correctly at this point we are trying to use the points from S in this open ball around p to construct a sequence that converges to p (which is not in S) which contradicts our hypothesis that the limit of a convergent sequence in S is in S.

    The confusion is how do we know that the sequence will converge to p. If n is 100 then the given distance from p is 1/100 and we are asked to choose the p_n's from S from our original ball around p that fall within this distance. So the question is...since we really don't know anything about these p_n's how do we know that as n goes to infinity that there will be p_n's that are within 1/n of p? I.e. that the sequence will converge
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  2. #2
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    Re: Closed Sets and Convergent Sequences

    In other words what allows us to assume there will in fact exist p_n's less than 1/n for a given n?
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  3. #3
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    Re: New post

    Quote Originally Posted by qnohra View Post
    In other words what allows us to assume there will in fact exist p_n's less than 1/n for a given n?
    Always start a new post. Do not add to a thread that is several years old.

    We want to show that S is a closed set. So suppose that p is a limit point of S and p\notin S.

    That means that in any ball \exists p_{\delta}\in[\mathcal{B}(p;\delta)\cap (S\setminus\{p\})].

    So \exists p_{1}\in[\mathcal{B}(p;1)\cap (S\setminus\{p\})].
    Because p\ne p_1 we know that d(p,p_1)>0. So let \delta_2=\min\{d(p,p_1),1/2\}

    Again \exists p_{2}\in[\mathcal{B}(p;\delta_2)\cap (S\setminus\{p\})].

    If n\ge 3 let \delta_n=\min\{d(p,p_{n-1}),1/n\}.
    Again \exists p_{n}\in[\mathcal{B}(p;\delta_n)\cap (S\setminus\{p\})].

    We get a sequence of distinct points in S and by construction we have (p_n)\to p.
    Last edited by Plato; December 24th 2012 at 08:56 AM.
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    Re: New post

    Sorry about that I am new to this site I will remember next time.

    The third line of your reply "That means that in any ball..." is precisely where I am confused. In the book they do not explicitly say that p is a limit point in S but supposing it is, is that what allows us to assume that the following set you provided of the intersection of the ball of radius delta around p and S\{p} is non-empty? Also is there a reason for the \{p} since we assumed that p was not an element of S?

    Thank you for the help!
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  5. #5
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    Re: New post

    Quote Originally Posted by qnohra View Post
    Sorry about that I am new to this site I will remember next time.
    The third line of your reply "That means that in any ball..." is precisely where I am confused. In the book they do not explicitly say that p is a limit point in S but supposing it is, is that what allows us to assume that the following set you provided of the intersection of the ball of radius delta around p and S\{p} is non-empty? Also is there a reason for the \{p} since we assumed that p was not an element of S?
    Frankly the original post was so old that I had forgotten what was going on. It it the OP had said p\notin S.

    That is not necessary for this proof. All that is necessary is to show that S contains all of its limit points. What he was attempting is show that the complement is open. To find an open set containing p and no point of S would do that. So suppose no such open set exists. Then the construction I gave produces a sequence of points converging to p contrary to the given.

    That is why you don't want to add on to someone else's thread.
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