Results 1 to 11 of 11

Math Help - set of subsequential limits

  1. #1
    Newbie
    Joined
    Nov 2012
    From
    Barcelona
    Posts
    5

    set of subsequential limits

    Let A be a closed subset of R. Prove that there exists a sequence of real numbers s.th. if B is the set of all subsequential limits, then B=A.

    I know how to do it for a closed interval, but not for an arbitrary closed set. Any ideas?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1

    Re: set of subsequential limits

    Quote Originally Posted by mobydick View Post
    Let A be a closed subset of R. Prove that there exists a sequence of real numbers s.th. if B is the set of all subsequential limits, then B=A.
    That statement is nonsense.
    Please restate the question in a clearly readable format.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2012
    From
    Barcelona
    Posts
    5

    Re: set of subsequential limits

    Quote Originally Posted by Plato View Post
    That statement is nonsense.
    Please restate the question in a clearly readable format.
    What exactly don't you understand. A is an arbitrary closed subset of R. I need to prove that there exists some sequence {Xn} of real numbers, such that the set of subsequential limits of {Xn} is equal to A.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: set of subsequential limits

    A subsequential limit is the limit of a subsequence. So he wants to know: if A is a closed subset of \mathbb{R}, is there a sequence (x_n)_{n=1}^\infty such that for every a\in{A}, there is a subsequence (x_{n_k})_{k=1}^\infty that converges to a.

    Let S be the set of all intervals of the form (r_1,r_2) where r_1 and r_2 are rational. This is a countable set. If you choose x_n to be a point from A\cap{(r_1,r_2)} if one exists, then you can prove that any neighborhood of a contains an element from the sequence. You would also have to prove that a limit point of the sequence is in A.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2012
    From
    Barcelona
    Posts
    5

    Re: set of subsequential limits

    Quote Originally Posted by hollywood View Post
    A subsequential limit is the limit of a subsequence. So he wants to know: if A is a closed subset of \mathbb{R}, is there a sequence (x_n)_{n=1}^\infty such that for every a\in{A}, there is a subsequence (x_{n_k})_{k=1}^\infty that converges to a.

    Let S be the set of all intervals of the form (r_1,r_2) where r_1 and r_2 are rational. This is a countable set. If you choose x_n to be a point from A\cap{(r_1,r_2)} if one exists, then you can prove that any neighborhood of a contains an element from the sequence. You would also have to prove that a limit point of the sequence is in A.

    - Hollywood
    Thanks, man but I don't think that works. What are the elements of the sequence?
    If you mean that (r_1,r_2) are intervals of rational numbers then if A is a singleton set of one irrational number your algorithm will fail. If you mean that (r_1,r_2) are the intervals of real numbers between these rational numbers, then if A=[0,1] your algorithm will not construct the required sequence, which is a sequence of all the rationals in the interval [0,1].
    Maybe I am missing something.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1

    Re: set of subsequential limits

    Quote Originally Posted by mobydick View Post
    A is an arbitrary closed subset of R. I need to prove that there exists some sequence {Xn} of real numbers, such that the set of subsequential limits of {Xn} is equal to A.
    Well that is certainly a clear discrimination.
    You want the set B to be countable, a sequence, the closure of which is A.

    You said you can do it for a closed interval. If you show us what you have in mind we may be able you to generalize that method.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2012
    From
    Barcelona
    Posts
    5

    Re: set of subsequential limits

    Quote Originally Posted by Plato View Post
    Well that is certainly a clear discrimination.
    You want the set B to be countable, a sequence, the closure of which is A.

    You said you can do it for a closed interval. If you show us what you have in mind we may be able you to generalize that method.
    A subsequential limit of a sequence {x_n} is a limit of some subsequence of {x_n}. B is the set of all such subsequential limits and it is not necessarily countable.

    For the interval [0,1]:
    The set of all rational numbers in the interval [0,1] is countable, so you can construct a sequence {x_n} with range equal to this set. Since the rationals are dense in R, every point in [0,1] is a subsequential limit of {x_n}.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1

    Re: set of subsequential limits

    Quote Originally Posted by mobydick View Post
    A subsequential limit of a sequence {x_n} is a limit of some subsequence of {x_n}. B is the set of all such subsequential limits and it is not necessarily countable.
    In the OP you asked for a sequence. A sequence is a countable set by definition. That is the confusing part of this question. Every point of A is the limit of a sequence of rationals. And you are right that the collection of those sequences may not the countable. But then it is not a sequence.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: set of subsequential limits

    Quote Originally Posted by mobydick View Post
    Thanks, man but I don't think that works. What are the elements of the sequence?
    If you mean that (r_1,r_2) are intervals of rational numbers then if A is a singleton set of one irrational number your algorithm will fail. If you mean that (r_1,r_2) are the intervals of real numbers between these rational numbers, then if A=[0,1] your algorithm will not construct the required sequence, which is a sequence of all the rationals in the interval [0,1].
    Maybe I am missing something.
    I meant to choose an element of A in (r_1,r_2). If no such element exists, skip that interval and move to the next one.

    In the case of A={a}, where a is irrational, you would choose a (the only possible choice) if r_1<a<r_2, and otherwise you would skip. There are infinitely many intervals containing a, so the sequence would be a, a, a, ....

    In the case of A=[0,1], yes, the result is not an enumeration of the rationals in A, but that's not the only possible sequence. Consider a point a\in{A}. Choose any point x_{n_1}<{a} (the same construction works for x_{n_1}>{a}). There are infinitely many intervals (r_1,r_2) with x_{n_1}<r_1<r_2<a, so choose one, and let x_{n_2} be the choice from that interval. In this way, a subsequence approaching a can be found.

    In fact, I think those two examples generalize to isolated points and limit points, so that's all the points of A.

    You would still need to prove that there is no b\not\in{A} that is the limit of a subsequence of my sequence.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Nov 2012
    From
    Barcelona
    Posts
    5

    Re: set of subsequential limits

    Quote Originally Posted by hollywood View Post
    I meant to choose an element of A in (r_1,r_2). If no such element exists, skip that interval and move to the next one.

    In the case of A={a}, where a is irrational, you would choose a (the only possible choice) if r_1<a<r_2, and otherwise you would skip. There are infinitely many intervals containing a, so the sequence would be a, a, a, ....

    In the case of A=[0,1], yes, the result is not an enumeration of the rationals in A, but that's not the only possible sequence. Consider a point a\in{A}. Choose any point x_{n_1}<{a} (the same construction works for x_{n_1}>{a}). There are infinitely many intervals (r_1,r_2) with x_{n_1}<r_1<r_2<a, so choose one, and let x_{n_2} be the choice from that interval. In this way, a subsequence approaching a can be found.

    In fact, I think those two examples generalize to isolated points and limit points, so that's all the points of A.

    You would still need to prove that there is no b\not\in{A} that is the limit of a subsequence of my sequence.

    - Hollywood
    This might work but we need to make it more precise. It is not at all clear how you choose the elements from the intervals.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: set of subsequential limits

    Since we don't know what A\cap{(r_1,r_2)} looks like, it's hard to be more specific. But I think the construction works regardless of which choice we make.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Subsequential Limits
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: October 9th 2012, 02:31 PM
  2. Subsequential limit problem
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: October 1st 2010, 02:37 PM
  3. Analysis subsequential limits
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: September 21st 2010, 03:27 AM
  4. Subsequential limits
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 19th 2009, 08:58 AM
  5. Subsequential limit Question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 2nd 2008, 11:57 PM

Search Tags


/mathhelpforum @mathhelpforum