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Math Help - An exercise about limsup/liminf of a subset sequence

  1. #1
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    An exercise about limsup/liminf of a subset sequence

    Let E_{n} = $\{$x $\in$ $\[0,2$\pi$]$ : (sin(nx))/n > 0$\}$ with n natural number.
    Calculate: E', E'' where E' = liminf En, E'' = limsup En for n that goes to $\infty$

    Looking at the goniomethrical discus i should say that liminf is the empty set and that limsup is [0,2$\pi$] \ \{0,$\pi$, 2$\pi$\}, but how could i formalize it (if is correct) ?
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    Re: An exercise about limsup/liminf of a subset sequence

    What is the infimum of E_1? Of E_2? Of E_n? Given any n \in \mathbb{N}, choose any x \in \left(0,\dfrac{\pi}{2n}\right). Then 0<\dfrac{\sin(nx)}{n} < \dfrac{\sin\left(n\dfrac{\pi}{2n}\right)}{n} = \dfrac{\sin\left(\dfrac{\pi}{2}\right)}{n} = \dfrac{1}{n}. So, the infimum of E_n is 0 for all n. Hence, E' = 0.

    The suprema are a little different. How do you find the supremum of E_n? We need only look at the interval \left[ \dfrac{n-1}{n}2\pi,2\pi \right]. We know \sin(nx)> 0 on the interval \left(\dfrac{n-1}{n}2\pi, \dfrac{2n-1}{n}\pi\right) and \sin(nx)<0 on \left(\dfrac{2n-1}{n}\pi,2\pi\right). So, the supremum of E_n would be \dfrac{2n-1}{n}\pi. So, what is \lim_{n\to \infty} \dfrac{2n-1}{n}\pi?
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    Re: An exercise about limsup/liminf of a subset sequence

    the first answer should be (0,\dfrac{\pi}{n}\right $)$ and the other subsets depending on the periodicity so (\dfrac{2k\pi}{n}\right,\dfrac{(2k+1)\pi}{n}\right $)$ with the limitation that 2k+1/n < 2, 2k/n > 0 with k positive integer.

    the second answer is looking at the sign of the prime derivative of sin(nx)/n.

    the third answer should be 2$\pi$. So the limsup should be { 2$\pi$}

    P.S. When you say 0 for E' you mean the subset {0} i suppose, anyhow thanks a lot for your help. I was struggling about this one.
    Last edited by ayrcast; October 18th 2013 at 11:01 AM.
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    Re: An exercise about limsup/liminf of a subset sequence

    You seem a little confused about what an infimum or supremum is. It is a greatest lower bound or least upper bound respectively. You had an example before where you were looking at a set of sets. A set of sets can be viewed as a poset where the partial order is containment. A \le B if A\subseteq B. In that context, an infimum or supremum will be a set.

    In the context of the current problem, you do not have a set of sets. You have a set of points of the real line. In this context, a greatest lower bound or a least upper bound will be a point, not a set. So, the first answer is not the set \left(0,\dfrac{\pi}{n}\right). How do we know that 0 is the infimum of the set? We know it is a lower bound since 0 \le x for all x \in \left(0,\dfrac{\pi}{n}\right)\subseteq E_n \subseteq [0,2\pi]. Given any y > 0, 0<\dfrac{y}{2}<y which shows that y is not a lower bound. Hence, 0 is the greatest lower bound, and therefore the infimum for each E_n. So, \inf E_1, \inf E_2, \ldots = 0,0,\ldots is an infinite sequence of zeros. The limit of an infinite sequence of zeros is zero. That is what is meant by the limit inferior ( \liminf). So, when I say 0 for E', I mean 0, not {0}.

    Similarly for the supremum.
    Last edited by SlipEternal; October 18th 2013 at 06:38 PM.
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    Re: An exercise about limsup/liminf of a subset sequence

    Looking at my book of Mathematical Analysis III i found this definition: E' = \liminf_{k\rightarrow +\infty}\; E_k = \bigcup_{k=1}^{+\infty} \bigcap_{n=k}^{+\infty} E_n so is a set, not a value.
    I thought you was talking about \bigcap_{n=k}^{+\infty} E_n when you said infimum in this case.
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    Re: An exercise about limsup/liminf of a subset sequence

    Anyhow if i expressed myself bad what the exercise asks is to calculate E', E'' defined as E' = \liminf_{k\rightarrow +\infty}\; E_k = \bigcup_{k=1}^{+\infty} \bigcap_{n=k}^{+\infty} E_n, E'' = \limsup_{k\rightarrow +\infty}\; E_k = \bigcap_{k=1}^{+\infty} \bigcup_{n=k}^{+\infty} E_n for $\{E_{n}\}_{n\in\Bbb N}$

    P.S. i'm sorry for the double post but doesn't let me edit the previous one.
    Last edited by ayrcast; October 19th 2013 at 09:40 AM.
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    Re: An exercise about limsup/liminf of a subset sequence

    Using that definition, you are probably correct about E' = \emptyset, E'' = [0,2\pi] \setminus \{0,\pi,2\pi\}.

    Here is one possible approach to a formal proof:

    Obviously, \forall n\in\mathbb{N},\, \{0,\pi, 2\pi\} \cap E_n = \emptyset, so those points cannot be in E' or E''.

    If x \in \bigcap_{n\ge k}E_n, then x \in E_n for all n\ge k. Given x \in (0,2\pi), suppose that for every k\ge 1, there exists n\ge k with x \notin E_n. Then, obviously x \notin \bigcap_{n\ge k}E_n for any k \ge 1. That obviously implies x \notin \bigcup_{k\ge 1}\bigcap_{n\ge k} E_n. So, if you can show that given any x \in (0,2\pi) and any k\ge 1, there exists n\ge k with x \notin E_n, then you have proven E' \cap (0,2\pi) = \emptyset.

    For E'', if for all k\ge 1, there exists n\ge k with x \in E_n, then x \in E''. (This condition is dual to x \notin E' in a similar way to how intersections are dual to unions).

    If you need any more help setting up the argument, I have some ideas.
    Last edited by SlipEternal; October 19th 2013 at 01:44 PM.
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    Re: An exercise about limsup/liminf of a subset sequence

    thanks a lot SlipEternal
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