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Math Help - Closed Subsets

  1. #1
    Senior Member bugatti79's Avatar
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    Closed Subsets

    Folks,

    Are the following subsets closed?

    1) S=(0,1]; in X=R

    2) S=\{x=(x_1,x_2) \in R^2 : x_1^2+x_2^2 \le 1 \} in X=R

    For 1) S=(0,1]; in X=R is not closed

    for if S_n is a convergent sequence with 0 < S_n \le 1 \forall n \in N

    S-N \le 1 \forall n \implies Lim S_n \le Lim 1 =1  as n \rightarrow \infty

    I dont know if this is right or how to continue.....?
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  2. #2
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    Re: Closed Subsets

    Quote Originally Posted by bugatti79 View Post
    Are the following subsets closed?
    1) S=(0,1]; in X=R
    For each n\in\mathbb{Z}^+ define x_n=\frac{1}{n}

    If it true that x_n\in S~? that is the limit of (x_n)~?
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  3. #3
    Senior Member bugatti79's Avatar
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    Re: Closed Subsets

    Quote Originally Posted by Plato View Post
    For each n\in\mathbb{Z}^+ define x_n=\frac{1}{n}

    If it true that x_n\in S~? that is the limit of (x_n)~?
    ok

    Let  \displaystyle x_n=1/n be a sequence \in S  \forall n  \in  Z^+

    ie \displaystyle 0 < 1/n \le 1 \forall n \in Z^+ but the lim_{ n \rightarrow \infty } 1/n= 0 \implies x_n \notin S \implies S=(0,1] is not closed...

    Correct?
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  4. #4
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    Re: Closed Subsets

    Quote Originally Posted by bugatti79 View Post
    Let  \displaystyle x_n=1/n be a sequence \in S  \forall n  \in  Z^+
    ie \displaystyle 0 < 1/n \le 1 \forall n \in Z^+ but the lim_{ n \rightarrow \infty } 1/n= 0 \implies x_n \notin S \implies S=(0,1] is not closed...Correct?
    NO indeed.
    The whole point is that \forall n,~x_n\in S but \displaystyle\lim_{n\to\infty}x_n\notin S.
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  5. #5
    Senior Member bugatti79's Avatar
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    Re: Closed Subsets

    Quote Originally Posted by Plato View Post
    NO indeed.
    The whole point is that \forall n,~x_n\in S but \displaystyle\lim_{n\to\infty}x_n\notin S.
    Does this still mean the set is not closed? Thanks
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  6. #6
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    Re: Closed Subsets

    Quote Originally Posted by bugatti79 View Post
    Does this still mean the set is not closed? Thanks
    You don't understand because you don't know what it means for a set to be closed. Do you? How do you expect to work any problem if you don't understand the underlying definitions?
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  7. #7
    Senior Member bugatti79's Avatar
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    Re: Closed Subsets

    Quote Originally Posted by Plato View Post
    You don't understand because you don't know what it means for a set to be closed. Do you? How do you expect to work any problem if you don't understand the underlying definitions?
    Definition: If S is a subset of an n.l.s, we say S is a closed set if for every

    convergent series (S_n) of points S_n \in S, the

    \displaystyle \lim_{n\to\infty}S_n \in S

    in our case \implies \lim_{n\to \infty} S_n \notin S

    Therefore the set is not closed...?
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  8. #8
    Senior Member bugatti79's Avatar
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    Re: Closed Subsets

    Quote Originally Posted by bugatti79 View Post
    Definition: If S is a subset of an n.l.s, we say S is a closed set if for every

    convergent series (S_n) of points S_n \in S, the

    \displaystyle \lim_{n\to\infty}S_n \in S

    in our case \implies \lim_{n\to \infty} S_n \notin S

    Therefore the set is not closed...?
    Yes, the set is not closed.

    Quote Originally Posted by bugatti79 View Post
    Folks,

    Are the following subsets closed?


    2) S=\{x=(x_1,x_2) \in R^2 : x_1^2+x_2^2 \le 1 \} in X=R
    We need to show ||x|| \le 1

    By Triangle inequality we have ||x|| \le ||x-x_n||+||x_n|| \forall n \in N
    but we know \lim_{n\to\infty}x_n=x

    so  \forall \epsilon > 0, \exists n_0 \in N s.t ||x-x_n|| < \epsilon

    In particular taking n=n_0

    ||x|| \le ||x-x_n_0||+||x_n_0|| < \epsilon +1

    I dont understand the above line, why is there a '+1' on the RHS?

    Thanks
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